CBSE Solutions For Class 10 Mathematics Chapter 12 Area Related To Circles

CBSE Solutions For Class 10 Mathematics Chapter 12 Area Related To Circles

Question 1. Find the area of a circle whose Circumference is 440m. 
Solution: 

Circumference of a circle = 440m

2πr = 440

⇒ \(r=\frac{440 \times 7}{2 \times 22}\)

r = 70m

Area of a circle = πr2 = \(\frac{22}{77} \times 70 \times 70\)

= 22 ×10×70

= 15400 m2

Question 2. Find the radius of a Circular sheet whose area is 55442″. 
solution: 

Area of Circular sheet

= 5544 m2

πr2 = 5544

πr2= \(\frac{5544 \times 7}{22}\)

r2 = \(\frac{38808}{22}\)

r2 = 1764

r = 42m

Question 3. The area of a Circular plot is 346.5m. Calculate the Cost of fencing the plot at the rate of ± 6 Per metre. 
Solution:

Area of plot = 346.5m2

⇒ πr2 = 346.5m2

⇒ r2 = \(\frac{3465 \times 7}{22}\)

⇒ r2= 110.25

⇒ r = 10.5m

Circumference of plot = 2πY = 2x \(\frac{22}{7}\) x 10.5 =  66m

Cost of fencing = Circumference X Cost of fencing per metre.

= 66 × 6 =7396

Question 4. The radii of the two Circles are 19 cm and 9cm respectively. Find the radius of the Circle which has a Circumference equal to the Sum of the circumferences of two circles. 
solution: 

Here, r1 =19 cm and r2 = 9 cm

Let the radius of the new circle =R Cm

Given that,

Circumference of given two new circles = Sum of the Circumferences of given two Circles

⇒ 2πR = 2πr1 + 2πr2

⇒ R = r1+r2 = 19+9 = 28cm

Question 5. The distance of a Cycle wheel is 28cm How many revolutions will it make in moving 13.2km? 
Solution: 

Distance travelled by the wheel in one revolution = 271 = 22 = x 28 = 88m Total distance travelled by the wheel = 13.2 x 1000 x 100 cm

Number of revolution made by the wheel = \(\frac{\text { total distance }}{\text { Circumference }}\)

⇒ \(\frac{13.2 \times 1000 \times 100}{88}\)

= 15000 revolutions.

CBSE Solutions For Class 10 Mathematics Chapter 12 Area Related To Circles

Question 6. The Circumference of a Circle exceeds the diameter by 16-8 cm. Find the radius of the Circle. 
Solution: 

let the radius of the Circle be r.

Diameter = 2r

Circumference of Circle =2πr

using the given Information, we have

2πr = 2r+ 16.8

⇒ \(2 \times \frac{22}{7} \times r\) = 2r+16·8

⇒ 44r = l4r+ 16.8×7

⇒ 30r = 117.6

⇒ \(r=\frac{117.6}{30}=3.92 \mathrm{~cm}\)

Question 7. Find the area of 15cm. ring whose outer and Inner radii are respectively 20cm and 
Solution: 

Outer radius R = 200m

Immer radius r = 15 Cm

Area of ring = πT (Rr2 – r2)

Area of ring = \(\frac{22}{7}\) [(20)2 = (15)2]

⇒ \(\frac{22}{7}\) (400-225)

⇒ \(\frac{22}{7}\)  x 175

⇒ 22×25

⇒ 550cm2

Question 8. A race track is in the form of a · ring with an inner circumference of 352m and an outer Circumference of 396m. Find the width of the track. 
Solution: 

Let R and r be the outer and inner radii of the Circle.

The width of the track = (R-r) (m

Now,  2πr = 352

⇒ 2 x  \(\frac{22}{7}\) x r  = 352

⇒ r = \(\frac{352 \times 7}{2 \times 22}\)

r = 7 x 8= 56m

Again, 2πR=396

⇒ \(2 x \frac{352 \times 7}{2 \times 22}\) x R = 396

⇒ R= \(\frac{396 \times 7}{2 \times 22}\) = 7×9= 63m

R= 63m, r=56m

width of the track = (R-r)m = (63-56)m = 7m

Question 9. Two Circles touch internally. The Sum of their areas is 116π (m2 and the distance between their Centres is 6cm. Find the radii of the circles. 
solution: 

let two circles with Centres ‘o’ and o  having radial R and r respectively touch each other at P.

It is given that θ = 6

⇒ R-r=6

⇒ R= 6+r2

Also, πR2+ πr2 = 116π

⇒ π(R2 + r2) = 116π

⇒ R2 + r2 = 116 → 2

From equations (1) and (2), we get (6+r)2 + r2 = 116

⇒ 36+r2+12r+r2=116

⇒ 27 2 +(20-80=0

⇒ r2 768-40=0

⇒ (1+(0)(x-4)=0

So r=-10 and r=4

But the radius Cannot be negative. So, we reject r=-10

r = 4cm

R=6+4=10cm

Hence, the radii of the two circles are 4cm and 10 Cm.

Question 10. The radius of a wheel of a bus is 5 cm. Determine its Speed in kilometres per hour, when its wheel makes 315 revolutions. Der minute. 
Solution: 

The radius of the wheel of the bus = 45 Cm

Circumference of the wheel = 2πr

⇒ \(=2 \times \frac{22}{7} \times 45=\frac{1980}{7} \mathrm{~cm}\)

Distance Covered by the wheel in One revolution = \(=\frac{1980}{7} \mathrm{~cm}\)

Distance Covered by the wheel in 315 revolution = \(\frac{1980}{7}\)

= 45 x 1980 = 89100 cm

⇒ \(\frac{89100}{1000 \times 100} \mathrm{~km}=\frac{891}{1000} \mathrm{~km}\)

Distance Covered in 60 minutes to Ihr  = \(\frac{891}{1000} \times 60=\frac{5346}{100}=\) 53.46km

Hence, Speed of bus  = 53.46 km/hr.

Question 11. A Square of the largest circle is lost as trimmings? area is cut out of a circle. What /% of the area of 
Solution: 

Let the radius of the Circle be v units,

Area of Circle = πr2 Sq. units

Let ABCD be the largest Square.

Length of diagonal = 2r= Side√2

Side = \(\frac{2 r}{\sqrt{2}}\) =√2r

Area (Square ABCD) = (Side)2 = (2x)2 = 2r2

Area of Circle lost by cutting out of a Square of largest area = 2r2

Required percentage of the area of c\(=\frac{2 r^2}{\pi r^2} \times 100=\frac{200}{\pi} \%\)

Question 12. The perimeter of a Semi Circular protractor is 32.4cm Calculate: 

  1. The radius of the protractor in Cm, 
  2.  The are of protractors in m2

Solution:

1 . let the radius of the protractor be 5cm.

Perimeter of semicircle protractor = (πr+21) Cm

r(π+2)=324

⇒ r \(\left(\frac{22}{7}+2\right)\) = 32.4

⇒ \(r \times \frac{36}{7}=32.4\)

⇒ \(\frac{324 \times 7}{36}\)

⇒ r=6.3 cm

Hence, the radius of the protractor = 6.3 cm

2) Area of Semi Circular protractor =\(\frac{1}{2} \pi r^2\)

⇒ \(\frac{1}{2} \times \frac{22}{7} \times 6.3 \times 6.3\)

⇒ 62.37 Cm2

Hence, the area of the protractor = 62.37 cm2

Question 13. The minute hand of a clock Is √21 cm long. Find the area described as the minute hand on the face of the clock between 6 am. and 605 am. 
Solution: 

In 60 minutes, the minute hand of a clock move through an angle of 360°:

In 5 minutes hand will move through an angle = 360° x 5 = 30°,

Now, r=√21 Cm and 0=30°

Area of Sector described by the minute hand between 6 am and 6.05 am.

⇒ \(\frac{\pi r^2 \theta}{360^{\circ}} \)

⇒ \(\frac{22}{7} \times(\sqrt{21})^2 \times 30^{\circ} \times \frac{1}{360^{\circ}}\)

⇒ \(\frac{22}{7} \times 21 \times \frac{1}{12}\)

Question 14.  In the adjoining figure, Calculate: 

  1. The length of minor arc ACB 
  2. Area of shaded Sector.

Solution: 

CBSE Solutions For Class 10 Maths chapter 12 The Area Of Shaded Sector

Here, θ =150°, r=14cm

1) Length of minor are = \(\frac{\pi r \theta}{180^{\circ}}\)

\(\frac{22}{7} \times 14 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

= 36.67 Cm

2) Area of shaded sector= \(\frac{\pi r^2 \theta}{360^{\circ}}\)

⇒ \(\frac{22}{7} \times(14)^2 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

⇒ \(\frac{22}{7} \times 4^2 \times 14 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

⇒ \(44 \times 14 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

= 256.67 cm2

Question 15. Find the area of a sector of a circle with a radius of 6 cm if the angle of the Sector is 60°. 
Solution: 

Here, the radius of the circle, r=6cm

The angle of Sector, θ = 60°

Area of Sector =\(\frac{\theta}{360^{\circ}} \times \pi r^2\)

⇒ \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6 \mathrm{~cm}^2\)

⇒ \(\frac{1}{6} \times \frac{22}{7} \times 6 \times 6 \mathrm{~cm}^2\)

⇒ \(=\frac{132}{7} \mathrm{~cm}^2\)

= 18.86cm2

Question 16. Find the area of a quadrant of a circle whose Circumference is 22cm. solution: 

Circumference of Circle, 2πr=22

⇒ \(2 \times \frac{22}{7} \times r=22\)

⇒ \(r=\frac{7}{2} \mathrm{~cm}\)

Now, the area of the quadrant of the Circle,

⇒ \(\frac{1}{4} \pi r^2\)

⇒ \(\frac{1}{4} \times \frac{2 \pi}{7} \times \frac{7}{2} \times \frac{7}{2}\)

⇒ \(\frac{1}{2} \times 11 \times \frac{1}{2} \times \frac{7}{2}\)

⇒ \(\frac{11}{4} \times \frac{7}{2}\)

⇒ \(\frac{77}{8} \mathrm{~cm}^2\)

Question 17. The length of the minute hand of a clock is 14cm. Find the area an Sqft take Swept by the Minute hand in 5 minutes. 
Solution: 

Length of a minute hand of clock = 14cm

Radius of Circle = 14cm

Angle Subtended by minute hand in 60 min  = 360°

Angle subtended by minute hand in Iminute =\(\frac{360^{\circ}}{60^{\circ}}=6^{\circ}\)

The angle subtended by minute hand in 5 minutes = 30°

From the formula,

Area of Sector of Circle = \(\frac{\theta \pi r^2}{360^{\circ}}\)

⇒ \(=30^{\circ} \times \frac{22 \times(14)^2}{7 \times 360^{\circ}}\)

⇒ \(\frac{22 \times 14 \times 2}{12}\)

⇒ \(\frac{616}{12}\)

⇒ \(\frac{154}{3} \mathrm{~cm}^2\)

Question 18. A chord of a circle of radius 10 cm Subtends a right angle at the  Centre. Find the area of the Corresponding: 

  1. minor Segment 
  2. major segment 

Solution: 

CBSE Solutions For Class 10 Maths chapter 12 Subtends A Right Angle At The  Centre

Given the radius of the Circle, A0-10cm.

The perpendicular is drawn from the Centre of the circle to the chord of the Circle to the chord of the circle which bisects this chord.

AD=DC

and LAOD = <COD  = 45°

LAOC = LAOD + LCOD

= 45° +45° =90°

In the right AAOD,

⇒ Sin45\(=\frac{A D}{A O} \Rightarrow \frac{1}{\sqrt{2}}=\frac{A D}{10}\)

⇒ AD=5√2 Cm

and cos 45° = \(\frac{O D}{A O} \Rightarrow \frac{1}{\sqrt{2}}=\frac{O D}{10}\)

⇒ OD = 5√2 Cm

Now, AC=2AD

= 2X5√2 = 10√2 cm

Now, the area of AAOC

⇒ \(\frac{1}{2}\) X AC X OD

⇒ \(\frac{1}{2}\) x 10√2 × 5√52

= 50cm2

Now, area of Sector

⇒ \(\frac{\theta \pi r^2}{360^{\circ}}=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2\)

⇒ \(\frac{314}{4}\)

=78.5cm2

(1)Area of minor Segment AEC

= area of sector OAEC – area of Aoc

= 78.5-50 =  28,50m2

(2)Area of major SegSector OAFGCO

= area of a circle – an area of sector OAEC = πr2-78.5

= 3-14x(10) 278.5

= 314-785

= 235.5cm2

Question 19. A chord of a circle of radius 12CM Subtends an angle of 120° at the Centre Find the area of the Corresponding Segment of the Circle. (use πT = 3.14 and √3 = 1-73) 
Solution: 

Here, the radius of the circle, r = 12 Cm

Angle Subfended by the chord at the Centre, 0=120°

Area of Corresponding minor segment

⇒ \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

⇒ \(r^2\left(\frac{\pi \theta}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

⇒ \(12 \times 12 \times\left(\frac{3.14 \times 120^{\circ}}{360^{\circ}}-\frac{1}{2} \times \sin 120^{\circ}\right)\)

⇒ \(144\left(\frac{3.14}{3}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)\)

⇒ \(144\left(\frac{3.14}{3}-\frac{1.73}{4}\right)\)

= 88,44 cm2

Question 20.  A Car has two wipers which do not overlap. Each wiper has a blade of length 25 cm Sweeping through each Sweep of the blades. an angle of 115 find the total area cleaned at 
Solution:

Given, length of wiper blade  = 25 cm = r (Say)

The angle formed by this blade, 0=115°

Area cleaned by a blade = area of Sector formed  by blade

⇒ \(\frac{\theta \pi r^2}{360^{\circ}}\)

⇒ \(115^{\circ} \times \frac{22}{7 \times 360^{\circ}} \times(25)^2\)

⇒ \(\frac{23 \times 22}{7 \times 72} \times 625\)

⇒ \(\frac{23 \times 11 \times 625}{7 \times 36}\)

Total area cleaned by two blades =  2x area cleaned by a blade

⇒ \(\frac{2 \times 158125}{252}\)

⇒ \(\frac{158125}{126} \mathrm{~cm}^2\)

CBSE Solutions For Class 10 Mathematics Chapter 10 Circles

CBSE Solutions For Class 10 Mathematics Chapter 10 Circles

Question 1. The radius of a Circle and 8 Cm. Calculate the length of a tangent down to this Circle from a point at a distance of 10 Cm from its Centre.
Solution:

Since the tangent is perpendicular to the radius through the paint of Contact

CBSE Solutions For Class 10 Maths chapter 10 The radius of a Circle

∠OTP = 90

In the right triangle OTP, we have

⇒  Op2 = OT2+ PT2

⇒  (10)2 = (8)2 + PT2

⇒ 100-64=PT2

⇒  PT2 = 36

⇒ PT = 6 Cm

Hence, the length of the tangent is 6cm

Question 2. Prove that the tangents drawn at the ends of the diameter of a Circle are Parallel.
Solution:

Let AB be the diameter of a circle with Centre O. PA and PB are the tangents to the Circle at pants A and B respectively.

CBSE Solutions For Class 10 Maths chapter 10 The Diameter Of The Circle

Now ∠PAB=90°

and∠QBA=90°

⇒  ∠PAB + ∠QBA = 90° +90° = 180°

PA ll QB

CBSE Solutions For Class 10 Mathematics Chapter 10 Circles

Question 3. prove that the perpendicular at the point of contact to the tangent to a Circler passes through the Centre.
Solution:

Given: A Circle with Centre 0 and a PQ tangent AQB and a Perpordicits is a dragon from point of contact Q to AB.

To prove: The perpendicular pa Passes through the Centre of the Circle.

Proof: AQ is the tangent of the Circle at point Q.

AQ will be the perpendicular to the radius of the circle.

⇒  PQ⊥AQ

⇒  The Centre of the Circle will lie on the line PQ.

Perpendicular PQ passes through the Centre of the Circle.

Question 4. A quadrilateral ABCD is drawn to Circumscribe a circle, and prove that AB+CD = AD+BC.
Solution:

As shown, the sides of a quadrilateral ABCD touch P a the Circle at P, Q, R and s. We know the tangents drawn from an external point to the Clucle are equal.

CBSE Solutions For Class 10 Maths chapter 10 A Quadrilateral

AP=AS, BP = BQ, CR = CQ, DR = DS

On adding, AP+BP + CR+DR

⇒  AB + BQ + CQ + DS

⇒ AB+CD= (AS + DS) + (BQ+CQ)

⇒  AB + CD = AP+BC

Hence proved.

Question 5. Ap is tangent to Circle 0 at point P. What is the length of OP?
solution:

Let the radius of the given Circle is r.

OP = OB = r

OA=2+r, OP=r, AP=4

∠OPA = 90°

CBSE Solutions For Class 10 Maths chapter 10 The Radius Of The Circle Point

In the right ∠OPA,

⇒  OA2= op2 +Ap2

⇒  (2+r)2 = r2+(4)2

⇒  4+r2+4r= r2+16

⇒  4r = 12 =) r=3

Op=3cm.

Question 6. If the angle between two tangents drawn from an external point p to a Clicle of radius ‘a’ and Centre 0, is 60°, then find the length of op .
Solution:

PA and PB are two tangents from an external point p such that

∠APB = 60°

∠OPA = ∠OPB = 30°

(tangents are equally inclined at the centre)

Also, ∠OAP=90°

Now, in right ∠OAP,

Sin 30° =\(=\frac{O A}{O P}\)

⇒ \(\frac{1}{2}=\frac{a}{o p}\)OP=2a units.

Question 7. In the given figure, if AB = AC, prove that BE = EC.
Solution:

We know that lengths of tangents from an external Point are equal.

CBSE Solutions For Class 10 Maths chapter 10 The Tangent

AD=AF

DB = BE

EC = FC

Now, it is given that

AB = AC

⇒  AD+DB = AF + EC

⇒  AD+DB = A8+EC

⇒  DB = EC

BE = EC

Question 8. In the given figure, AT is tangent to the Chicle with Centre 0 Such that Oto 4cm and LOTA = 30° Find the length of Segment AT.
Solution:

In the right ∠OAT,

CBSE Solutions For Class 10 Maths chapter 10 The Length Of Segment AT

Cos 30°\(=\frac{A T}{O T}\)

⇒ \( \frac{\sqrt{3}}{2}=\frac{A T}{4}\)

⇒  AT = 2√3 Cm

Question 9. The length of a tangent from point A at a distance of 5 cm from the Centre P 5cm of the Circle is ucm. Find the radius of the Circle.
Solution:

Let o be the Centre of the Circle and PQ is a tangent to the Circle from point P.

CBSE Solutions For Class 10 Maths chapter 10 The Length Of A Tangent

Given that, PQ=4cm and op=5cm

Now,∠OOP = 90°

In ∠OQP,

⇒  OQ2 = Op2= PQ2

= 52-42

=25-16=9

OQ = 3cm

Radius of Circle = 3cm

Question 10. Prove that the angle between the two tangents drawn from an external point to a circle is Supplementary to the angle Subtended by the line segment joining the points of contact at the Centre.
Solution:

PA and PB are the tangents of the Circle.

∠OAP = ∠OBP = 90°

In □ OAPB,

In □ OAPB,

∠OAP + ∠APB +∠OBP + ∠AOB = 360°

⇒ 90°+ ∠APB +90° +∠AOB = 360°

⇒  ∠APB + ∠AOB = 180°

⇒  ∠APB and ∠ADB are Supplementary

CBSE Solutions For Class 10 Mathematics Chapter 7 Co-Ordinate Geometry

CBSE Solutions For Class 10 Mathematics Chapter 7 Co-ordinate Geometry

Question 1. Find the distance between the following points.

1.  A(-6,4) and B(2,-2)

Solution:

Distance between the points (-6,4) and (2,-2)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \)

⇒  \(\sqrt{(2+6)^2+(-2-4)^2}\)

⇒  \(\sqrt{(8)^2+(-6)^2}\)

⇒  \(\sqrt{64+36}\)

⇒  \(\sqrt{100}\)

10 Units

2.A(-5,-1) and B (0,4)

Solution:

Distance between the points (-5,-1) and (0,4)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(0+5)^2+(4+1)^2}\)

⇒  \(\sqrt{(5)^2+(5)^2}\)

⇒  \(\sqrt{25+25}\)

⇒  \(\sqrt{50}\)

⇒  \(\sqrt{25 \times 2} \Rightarrow 5 \sqrt{2} \text { units }\)

CBSE Solutions For Class 10 Mathematics Chapter 7 Co-Ordinate Geometry

3. A(-4,-1) and B(7,3)

Solution:

Distance between the points (-4,-1) and (7,3)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(7-4)^2+(3+1)^2}\)

⇒  \(\sqrt{(3)^2+(4)^2}\)

⇒  \(\sqrt{9+16}\)

⇒  \(\sqrt{25}\)

5 Units

4. A(3,4) And B(5,2)

Solution:

Distance between the points (3,4) And (5,2)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(5-3)^2+(2-4)^2}\)

⇒  \(\sqrt{(2)^2+(-2)^2}\)

⇒  \(\sqrt{4+4}\)

⇒  \(\sqrt{8}\)

⇒  \(\sqrt{4 \times 2} \Rightarrow 2 \sqrt{2} \text { units }\)

Question 2. Find the Distance of the following points from the origin:

1. (3,-4)

Solution:

Distance between the points (3,-4) And (0,0)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(0-3)^2+(0+4)^2}\)

⇒  \(\sqrt{(-3)^2+(4)^2}\)

⇒  \(\sqrt{9+16}\)

⇒  \(\sqrt{25}\)

5 Units

2. (-8,6)

Solution:

Distance between the points (-8,6) And (0,0)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(0+8)^2+(0+6)^2}\)

⇒  \(\sqrt{(8)^2+(6)^2}\)

⇒  \(\sqrt{64+36}\)

⇒  \(\sqrt{100}\)

10 Units

Question 3. Find the Distance Between the points (a,b) and (-b, a)
Solution:

Distance between the points (-b, a) and (a,b)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(a+b)^2+(b-a)^2}\)

⇒  \(\sqrt{a^2+b^2+2 a b+b^2+a^2-2 a b}\)

⇒  \(\sqrt{2 a^2+2 b^2}\)

⇒  \(\sqrt{2\left(a^2+b^2\right)} \text { units }\)

Question 4. Find the Distance Between the points (2a,3a) and (6a,6a)
Solution:

Distance Between the points (2a,3a) and (6a,6a)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(6 a-2 a)^2+(6 a-3 a)^2}\)

⇒  \(\sqrt{(4 a)^2+(3 a)^2}\)

⇒  \(\sqrt{16 a^2+9 a^2}\)

⇒  \(\sqrt{25 a^2}\)

5a Units

Question 5. Find the Distance Between the points (a,-b)
Solution:

Distance Between the points (a,-b) and (0,0)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(0-a)^2+(0+b)^2}\)

⇒  \(\sqrt{a^2+b^2} \text { units }\)

Question 6. Find the Distance Between the points (6,0) and (0,y) is 10 units, and find the value of y.
Solution:

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(0-6)^2+(y-0)^2}\)

⇒  \(\sqrt{(6)^2+(y)^2}\)

⇒  \(\sqrt{36+y^2}\)

Given that,

⇒  \(\sqrt{36+y^2}\)

y2+36 = 100

y2= 100-36

y2 = 64

y = √64

⇒  \(y= \pm 8\)

Question 7. Find the Distance Between the points (3,x) and (-2,-6) is 13 units, and find the value of x.
Solution:

Distance Between the points (-2,-6) and (3,x)

⇒  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒  \(\sqrt{(3+2)^2+(x+6)^2} \Rightarrow \sqrt{(5)^2+x^2+36+12 x}\)

⇒  \(\sqrt{25+x^2+36+12 x}\)

⇒  \(\sqrt{x^2+|2 x+61}\)

Given that,

⇒  \(\sqrt{x^2+|2 x+61}\) = 13

⇒  \({x^2+|2 x+61}\)= 169

x2+12x+61-169=0

x2+12x+108=0

x2-6x+18x-108=0

x(x-6)(x+18)

(x-6)(x+18)=0

x-6=0 or x+18=0

x=6 or x= -18

Question 8. Prove that the following points are the vertices of a right-angled triangle:

1. A(-2,2) B(13,11) and C(10,14)

Solution:

Let the points are A(-2,2) B(13,11) and C(10,14)

AB2 = (13+2)2+(11-2)2

= (15)2+(9)2

= 225+81=306

BC2 = (10-13)2 + (14-11)2

= (3)2 +(3)2

= 9+9=18

CA2 = (10+2)2+(14-2)2

=(12)2+(12)2

=144+144=283

Therefore, AB = BC2+CA2

306=18+288

306=306

and AB2 = BC2+ CA2

ΔABC is a right-angled triangle.

2. A(-1,-6), B(-9,10), ((-7, ()
solution:

let the points are A(-1,-6), B(-9-10) and C(-7,6)

AB2 = (9+1)2 + (-10+6)2

= (-8)2 + (-4) 2

= 64+16= 80

Bc2= (-7+9)2 + (6+10)2

= (2) 4 (16) 2

= 4+256=260

AC2= (-7+1)2 + (6+6)2

= (-6)2+(12)2

=36+144 = 180

Therefore 260=80+180

and, BC2= AB2 +AC2

ΔABC is a right-angled triangle.

Question 9. Prove that the following points are the Vertices of an Isosceles right-angled triangle:

1. A(-8-9), B(0-3) and C(-6,5)

Solution: Let the points are A(-8-9), B(0,-3) and c(-6,5)

AB2 = (0+8) + (-3+9) 2

= (8)2+(6)2

= 64+36=100

BC2 = (6=0)2 + (5+3)2

= (-6)2 +(8)2

= 36 +64 = 100

(A2= (6+8)2 + (5+9)2

= (2)2 + (14)2

= 4+196=200

Therefore, AB = BC = √100

and AC2 = AB2+BC2

ΔABC is an isosceles right-angled triangle.

2. A (9-3), B(2,-1) and c(-2,-1)
Solution:

Let the points are A(9-3), B(2,-1) and c(-2,-1)

AB2 = (2-0)2 + (-1+3)2

=(2)2+(2)2

=4+4=8

BC2 = (-2-2)2 +(-1+1)2

(-4)2=16

Ac2 = (2-0)2 + (-1+3)2

= (-2)2 + (2)2

=4+4=8

Therefore, AB=AC=√5

and BC2 = AB2 + AC2

AABC is an isosceles right-angled triangle.

Question 10. Prove that the points A(1,1), B(-1,-1) and C(√3,-√3) are the vertices of an equilateral triangle.
solution:

Let the points are A(1, 1), B(-1,-1) and C(√3-√3)

⇒  \(AB=\sqrt{(-1-1)^2+(-1-1)^2}\)

⇒  \(AB=\sqrt{(-2)^2+(-2)^2}\)

\(AB=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\)

⇒  \(BC=\sqrt{(\sqrt{3}+1)^2+(-\sqrt{3}+1)^2}\)

⇒  \(BC=\sqrt{3+1+2 \sqrt{3}+3+3-2 \sqrt{3}}\)

⇒  \(CA=\sqrt{(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2}\)

⇒  \(CA=\sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}}\)

⇒  \(CA=\sqrt{8}=2 \sqrt{2}\)

AB=BC=CA

Question 11. Prove that the points (-1,-2), (-2,-5), (-4-6), and (-3,-3) are the vertices of a Parallelogram.
solution:

Let the points are A(-1,-2), B(-2,-5), c(-4,-6) and D(-3,-3)

⇒  \(AB=\sqrt{(-2+1)^2+(-5+2)^2}\)

⇒  \(AB=\sqrt{(-1)^2+(-3)^2}\)

⇒  \(AB=\sqrt{1+9}=\sqrt{10}\)

⇒  \(BC=\sqrt{(-4+2)^2+(-6+5)^2}\)

⇒  \(BC=\sqrt{(-2)^2+(-1)^2}\)

⇒  \(BC=\sqrt{4+1}=\sqrt{5} \)

⇒  \(CD=\sqrt{(-3+4)^2+(-3+6)^2}\)

⇒  \(CD=\sqrt{(1)^2+(3)^2}\)

⇒  \(CD=\sqrt{1+9}\)

⇒  \(CD=\sqrt{10}\)

⇒  \(AD=\sqrt{(-3+1)^2+(-3+2)^2}\)

⇒  \(AD=\sqrt{(-2)^2+(-1)^2}\)

⇒  \(AD=\sqrt{4+1}=\sqrt{5}\)

Therefore, AB = CD = √TO

BC = AD = √5

☐ ABCD is a parallelogram

Question 12. prove that the points (-4,-3), (3,2), (2,3) and (1,-2) are the vertices of a rhombus.
solution:

let the points are ·A(-4,-3), B(-3,2), C(2,3) and (1,-2)

⇒  \(A B=\sqrt{(-3+4)^2+(2+3)^2}\)

⇒  \(A B=\sqrt{(1)^2+(5)^2}=\sqrt{1+25}=\sqrt{26}\)

⇒  \(B C=\sqrt{(2+3)^2+(3-2)^2}\)

⇒  \(B C=\sqrt{(5)^2+(1)^2}=\sqrt{25+1}=\sqrt{26}\)

⇒  \(C D=\sqrt{(1-2)^2+(-2-3)^2}\)

⇒  \(C D=\sqrt{(-1)^2+(-5)^2}=\sqrt{1+25}=\sqrt{26}\)

⇒  \(D A=\sqrt{(1+4)^2+(-2+3)^2}\)

⇒  \(D A=\sqrt{(5)^2+(1)^2}=\sqrt{25+1}=\sqrt{26}\)

⇒  \(A C=\sqrt{(2+4)^2+(3+3)^2}\)

⇒  \(A C=\sqrt{(6)^2+(9)^2}=\sqrt{36+81}=\sqrt{117}\)

⇒  \(B D=\sqrt{(1+3)^2+(-2-2)^2}\)

⇒  \(B D=\sqrt{(4)^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}\)

Therefore, AB = BC= CD = DA = √26

AC and BD

〈〉 ABCD is a rhombus.

Question 13. Show that the following points are the vertices of a rectangle:

1. A(4,2), B(0,-4), c(-3,-2), D(14)

Solution.

Let the points are A (4,2), B(0,-4), C(-3,-2) and D(1,4)

⇒  \(A B=\sqrt{(0-4)^2+(-4-2)^2} \)

⇒  \(A B=\sqrt{(-4)^2+(-6)^2}=\sqrt{16+36}=\sqrt{52}\)

⇒  \(B C=\sqrt{(-3-0)^2+(-2+4)^2}\)

⇒  \(B C=\sqrt{(-3)^2+(-2)^2}=\sqrt{9+36} \sqrt{9+4}=\sqrt{13}\)

⇒  \(C D=\sqrt{(-3+1)^2+(4+2)^2}\)

⇒  \(C D=\sqrt{(-4)^2+(6)^2}=\sqrt{18+36}=\sqrt{52}\)

⇒  \(A D=\sqrt{(1-4)^2+(4-2)^2}\)

⇒  \(A D=\sqrt{(-3)^2+(2)^2}=\sqrt{9+4}=\sqrt{13}\)

Therefore, AB = CD=√52

BC=AD = √13

☐ ABCD is a rectangle

2. A(1,-1), B(2, 2), C(4,8), D(7,5)

solution:

Let the points are A(1,-1), B(-2,2), c(4,8) and D(7,5)

⇒  \(A B=\sqrt{(-2-1)^2+(2+1)^2}\)

⇒  \(A B=\sqrt{(-3)^2+(3)^2}=\sqrt{9+9}=\sqrt{18}\)

⇒  \(B C=\sqrt{(4+2)^2+(8-2)^2}\)

⇒  \(B C=\sqrt{(6)^2+(6)^2}=\sqrt{36+36}=\sqrt{72}\)

⇒  \(C D=\sqrt{(7-4)^2+(5-8)^2}\)

⇒  \(C D=\sqrt{(3)^2+(3)^2}=\sqrt{9+9}=\sqrt{18}\)

⇒  \(A D=\sqrt{(7-1)^2+(5+1)^2}\)

⇒  \(A D=\sqrt{(6)^2+(6)^2}=\sqrt{36+36}=\sqrt{72}\)

Therefore, AB = CD= √18

BC=AD = √72

☐ ABCD is a rectangle

Question 14. Show that the points A(2,1),B(0,3), C(-2,1) and D(0,-1) are the Vertices of a Square.
Solution:

Let the points are A(2,1), B(0,3), C(-2, 1) and D(0,-1)

⇒  \(A B=\sqrt{(3-2)^2+(3-1)^2}\)

⇒  \(A B=\sqrt{(-2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}\)

⇒  \(B C=\sqrt{(-2-0)^2+(1-3)^2}\)

⇒  \(B C=\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}\)

⇒  \(C D=\sqrt{(2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}\)

⇒  \(A D=\sqrt{(0-2)^2+(-1-1)^2}\)

⇒  \(A D=\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}\)

⇒  \(A C=\sqrt{(-2-2)^2+(1-1)^2}\)

⇒  \(A C=\sqrt{(-4)^2}=\sqrt{16}=4\)

⇒  \(B D=\sqrt{(0-0)^2+(-1-3)^2}\)

⇒  \(B D=\sqrt{(-4)^2}=\sqrt{16}=4\)

Therefore, AB = BC= CD=AD=√8

AC=BD=4

☐ ABCD is a Square

Question 15. Show that the points (1,1), (2,3) and (5,9) are collinear.
Solution:

⇒  \(A B=\sqrt{(2-1)^2+(3-1)^2}\)

⇒  \(A B=\sqrt{(1)^2+(2)^2}=\sqrt{1+4}=\sqrt{5}\)

⇒  \(B C=\sqrt{(9-3)^2+(5-2)^2}\)

⇒  \(B C=\sqrt{(6)^2+(3)^2}=\sqrt{36+9}=\sqrt{45}=\sqrt{9 \times 5}=3 \sqrt{5}\)

⇒  \(A C=\sqrt{(5-1)^2+(9-1)^2}\)

⇒  \(A C=\sqrt{(4)^2+(8)^2}=\sqrt{16+64}=\sqrt{80}\)

⇒  \(\sqrt{16 \times 5}=4 \sqrt{5}\)

Now, AB+BC = √5 +3√5 = 4√5 = AC

Points A, B, and care collinear.

Question 16. Show that the points (0,0), (5, 3), and (10,6) are Collinear.
Solution:

let the points are A(0,0), B(5, 3) and c(10,6)

⇒  \(A B=\sqrt{(5-0)^2+(3-0)^2}\)

⇒  \(A B=\sqrt{(5)^2+(3)^2}=\sqrt{25+9}=\sqrt{34} \)

⇒  \(B C=\sqrt{(10-5)^2+(6-3)^2}\)

⇒  \(B C=\sqrt{(5)^2+(3)^2}=\sqrt{25+9}=\sqrt{34}\)

⇒  \(A C=\sqrt{(10-0)^2+(6-0)^2} [/atex][/atex]\)

⇒  \(A C=\sqrt{(16)^2+(6)^2}=\sqrt{100+36}=\sqrt{136}=2 \sqrt{34}\)

Therefore, AB+BC= √34+ √34 = 2√34 = AC

Points A, B, and Care Collinear.

Question 17. Find the coordinates of a point that divides the line joining the points (5,3) and (10,8) in the ratio 2:3 internally.
Solution:

Let the Co-ordinates of the required point be (1,9).

Here, (X, Y1) = (5, 3) and (X2 Y2) = (10,8)

m : n = 2 : 3

⇒  \(x=\frac{m x_2+n x_1}{m+n}=\frac{2(10)+3(5)}{2+3}=\frac{20+15}{5}=\frac{35}{5}=7\)

⇒  \(y=\frac{m y_2+n y_1}{m+n}=\frac{2(8)+3(3)}{2+3}=\frac{16+9}{5}=\frac{25}{5}=45\)

Coordinates of required point = (7,5)

Question 18. Find the Coordinates of a point that divides the line joining the points (-1,2) and (3,5) in the ratio 3:5 internally.
solution:

Let the Co-ordinates of the required point be (x, y)

Here, (X1, y1) = (-1,2) and (x2,y2)=(3,5)

m:n = 3:5

⇒  \(x=\frac{m x_2+n x_1}{m+n}=\frac{3(3)+5(-1)}{3+5}=\frac{9-5}{8}=\frac{4}{8}=\frac{1}{2}\)

⇒  \(y=\frac{m y_2+n y_1}{m+n}=\frac{3(5)+5(2)}{3+5}=\frac{15+10}{8}=\frac{25}{8}\)

Co-ordinates of required point \(=\left(\frac{1}{2}, \frac{25}{8}\right)\)

Question 19. Find the Co-ordinates of a point that divides the line. Segment joining the points (1,3) and (4,6) in the ratio 2:1 internally.
Solution:

Let the Co-ordinates required point be (x,y)

Here, (X1,Y1) = (1,3) and (X2, y2) = (-4,6)

m:n = 2:1

⇒  \(x=\frac{m x_2+n x_1}{m+n}=\frac{2(-4)+1(1)}{2+1}=\frac{-8+1}{3}=\frac{-7}{3}\)

⇒  \(y=\frac{m y_2+n y_1}{m+n}=\frac{2(6)+1(3)}{2+1}=\frac{12+3}{3}=\frac{15}{3}=5\)

Coordinates of required point \( =\left(-\frac{7}{3}, 5\right)\)

Question 20. If point A lies on the line segment joining the points P(6,0) and $(0,0) Such that AP: AQ = 2:3, find the coordinates of point A.
Solution:

Let the Co-ordinates require point A (1,4)

Here, (X1,Y1,) = (6,0) and (x2,y2) = (0,8)

m:n=2:3

⇒  \(x=\frac{m x_2+n x_1}{m+n}=\frac{2(0)+3(6)}{2+3}=\frac{18}{5}\)

⇒  \(y=\frac{m y_2+n y_1}{m+n}=\frac{2(8)+3(0)}{2+3}=\frac{16}{5}\)

Co-ordinates of required point A\(\left(\frac{18}{5}, \frac{16}{5}\right)\)

Question 21. Find the ratio in which the Y-axis divides the line segment joining the points (3,4) and (-2,5).
Solution:

Let Y-axis divide the join of pants (3,4) and (-2,1) in the ratio k:1.

⇒  \(\frac{k \cdot x_2-1 \cdot x_1}{k+1}=0 \)

⇒  \(\frac{k(-2)+1 \cdot(3)}{k+1}=0\)

⇒  \(\frac{-2 k+3}{k+1}=0 \)

⇒  \(-2 k+3=0 \)

⇒  \(-2 k=-3 \)

⇒  \(k=\frac{3}{2}\)

required ratio = 3:2

Question 22. Find the Co-ordinates of the mid-point of the line joining the following points:

1. (2,4) and (6,2)

Solution: Co-ordinates of the mid-point of AB = \(\left(\frac{2+6}{2}, \frac{4+2}{2}\right)=\left(\frac{8}{2}, \frac{6}{2}\right)=(4,3)\)

2. (0,2) and (2,-4)

Solution: Co-ordinates of mid-point of AB = \(\left(\frac{0+2}{2}, \frac{-4+2}{2}\right)=\left(\frac{2}{2}, \frac{-2}{2}\right)=(1,-1)\)

3. (a+b, a-b) and (b-a, a+b)

Solution: Co-ordinates of mid-point of AB= \(\left(\frac{a+b+b-a}{2}, \frac{a-b+a+b b}{2}\right)=\left(\frac{2 b}{2}, \frac{2 a}{2}\right)\) = (b,a)

4. (3,-5) and (-1,3)

Solution: Co-ordinates of mid-point of AB= \(=\left(\frac{3-1}{2}, \frac{-5+3}{2}\right)=\left(\frac{2}{2}, \frac{-2}{2}\right)=(1,-1)\)

Question 23. The coordinates of the endpoints of the diameter of a Circle are (3,-2) and (-3,6). Find the Co-ordinates of the Centre and radius.
solution: Let the points be (3,-21) (-3,6)

⇒  \(\left(\frac{3-3}{2}, \frac{-2+6}{2}\right)\)

⇒  \(\left(0, \frac{4}{2}\right)=\left(0, \frac{4}{2}\right)\)

Center = (0,2)

Distance d = \(d=\sqrt{(-3-3)^2+(6+2)^2}\)

⇒  \(d=\sqrt{(-6)^2+(8)^2} \)

⇒  \(d=\sqrt{36+64}\)

⇒  \(d=\sqrt{100}=10\)

radius=\(\frac{\text { diameter }}{2}=\frac{10}{2}=5\)

Co-ordinates of the Centre (0,2) and radius = 5.

Question 24. The Co-ordinates of the vertices of a 4ABC are A(1, 0), B(3,6) and ((3,2). Find the length of its medians.
Solution:

let the points are A(1,0), B(3,6) and ((3,2)

let Ap be the median drawn from Vertex A.

The midpoint of BC is p.

Now, the Co-ordinates of P

⇒  \(\left(\frac{3+3}{2}, \frac{6+2}{2}\right)=\left(\frac{6}{2}, \frac{8}{2}\right)=(3,4)\)

⇒  \(AP =\sqrt{(3-1)^2+(4-0)^2}\)

⇒  \(\sqrt{(2)^2+(4)^2}\)

⇒  \(\sqrt{4+16} \)

⇒  \(\sqrt{20}\)

⇒  \(\sqrt{5 \times 4}=2 \sqrt{5}\)

let Bp be the median drawn from vertex Vertex B.

The mid-point Ac is p \(\left(\frac{1+3}{2}, \frac{0+2}{2}\right)=\left(\frac{4}{2}, \frac{2}{2}\right)=(2,1)\)

⇒  \(\text { and } B p=\sqrt{(2-3)^2+(1-6)^2}\)

⇒  \(\sqrt{(-1)^2+(-5)^2}\)

⇒  \(\sqrt{1+25}=\sqrt{26}\)

Let Cp be the median drawn from Vertex C.

The mid-point AB is P

⇒  \(\left(\frac{1+3}{2}, \frac{0+6}{2}\right)=\left(\frac{4}{2}, \frac{6}{2}\right)=(2,3)\)

and \(C p=\sqrt{(2-3)^2+(3-2)^2}\)

⇒  \(\sqrt{(-1)^2+(1)^2} \)

⇒  \(\sqrt{2}\)

Question 25. The coordinates of three consecutive vertices of a Parallelogram are (2,0), (4,1) and (6,4). Find the Coordinates of its 4th vertex.
Solution: Let A(2,0), B(4,1), c(6,4), and D(X, Y) be the vertex of a parallelogram ABCD.

We know that the diagonals of a parallelogram bisect each other.

Co-ordinates of the mid-point of AC = Co-ordinates of the mid-point of BD

⇒  \(\left.\left(\frac{2+6}{2}\right), \frac{0+4}{2}\right)=\left(\frac{4+x}{2}, \frac{1+4}{2}\right)\)

⇒  \(\left(\frac{8}{2}, \frac{4}{2}\right)=\left(\frac{4+x}{2}, \frac{1+y}{2}\right)\)

⇒  \((4,2)=\left(\frac{4+x}{2}, \frac{1+y}{2}\right)\)

⇒  \(4=\frac{4+x}{2} \text { and } 2=\frac{1+y}{2}\)

8=4+x and 4=1+y

x=4 and y=3

Co-ordinates of fourth vertex = (4,3)

Question 26. Find the Coordinates of the points of trisection of the line segment joining the points (2,5) and (6-2).
Solution:

Let Plaib) and Q(c,d) trisect the line joining the points A(2,5) and B(6,-2).

Now, Point Pla,b) divides the line AB in the ratio 1:2.

⇒  \(a=\frac{1(2)+2(6)}{1+2}=\frac{2+12}{3}=\frac{14}{3}\)

⇒  \(b=\frac{1(5)+2(-2)}{1+2}=\frac{5-4}{3}=\frac{1}{3}\)

Therefore, co-ordinate of point p= \(\left(\frac{14}{3}, \frac{1}{3}\right)\)

Q(c,d) divides the line AB in the ratio 2:1

⇒  \(c=\frac{2(2)+1(6)}{2+3}=\frac{4+6}{3}=\frac{10}{3}\)

⇒  \(d=\frac{2(5)+1(-2)}{2+1}=\frac{10-2}{3}=\frac{8}{3}\)

Therefore, Co-ordinates of Q =\(\left(\frac{10}{3}, \frac{8}{3}\right)\)

Co-ordinates of points of trisection of AB \(=\left(\frac{14}{3}, \frac{1}{3}\right) \text { and }\left(\frac{10}{3}, \frac{8}{3}\right) \text {. }\)

Question 27. Find the Coordinates of the points of trisection of the line segment joining the points (-2,0) and (4,0).
Solution:

Let P(a,b) and Q(Cd) trisect the line joining the points A(-2,0) and B(4,0).

Now, Point (a,d) divides the line AB in the ratio 1:2.

⇒  \(a=\frac{1(-2)+2(4)}{1+2}=\frac{-2+8}{3}=\frac{6}{3}=2\)

⇒  \(b=\frac{1(0)+2(0)}{1+3}=0\)

Therefore, the Co-ordinate of point P = (2,0)

Q(c,d) divides the line AB in the ratio 2:1

⇒  \(c=\frac{2(-2)+1(4)}{1+2}=\frac{-4+4}{3}=0\)

⇒  \( d=\frac{2(0)+1(0)}{1+2}=0\)

Therefore, Co-ordinates of Q = (0,0)

Co-ordinates of points of trisection of AB =(2,0) and (0,0)

Question 28. Find the ratio in which the join of points (3,-1) and (8,9) is divided by the line y-x+2=0.
solution:

let the line y-x+2=0 divide the line segment joining the points (3-1) and (8,9) in the ratio k:1.

Co-ordinates of p= \(\left(\frac{8 k+3}{k+1}, \frac{9 k-1}{k+1}\right)\)

but this point p lies on the line y-x+2=0

⇒  \(\frac{9 k-1}{k+1}-\frac{8 k+3}{k+1}+2=0\)

9 k-1-8 k+3+2 k+2=0

3k=2

⇒  \(k=\frac{2}{3}\)

Required ratio \(=\frac{2}{3}\):1 – 2:3

Question 29. Find the area of that triangle whose vertices are (2,3), (-3,4), and (7,5).
Solution:

Area of triangle =\(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \(\frac{1}{2}[2(4-5)-3(5-3)+7(3-4)]\)

⇒  \(\frac{1}{2}[2(-1)-3(2)+7(-1)]\)

⇒  \(\frac{1}{2}[-2-6-7]\)

⇒  \(\frac{-15}{2}\)

But the area of the triangle Cannot be negative

Area of triangle = \(\frac{-15}{2}\) Square units

Question 30. Find the area of that triangle whose vertices are (1,1), (-1,4) and (3,2).
Solution:

Area of triangle=\( \frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \(\frac{1}{2}[1(4-2)-1(2-1)+3(1-4)]\)

⇒  \(\frac{1}{2}(1(2)-1(1)+3(-3)]\)

⇒  \(\frac{1}{2}[2-1-9]\)

⇒  \(\frac{1}{2}[-8]\)

= -4

But the area of the triangle Cannot be negative.

Area of triangle = 4 Square units.

Question 31. Find the area of that triangle whose vertices are (5,2), (-4,3), and (-2,1)
Solution:

Area of triangle = \( \frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+y_3\left(y_1-y_2\right)\right] \)

⇒  \(\frac{1}{2}[5(3-1)-4(1-2)-2(2-3)] \)

⇒  \(\frac{1}{2}[5(2)-4(-1)-2(-1)]\)

⇒  \(\frac{1}{2}[10+4+2]\)

⇒  \(\frac{16}{2}=8\)

Question 32. Find the area of that triangle whose vertices are (b+c, a), (b-ca), and (9, -a),
Solution:

Area of triangle =\( \frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \(\frac{1}{2}[b+c(a+a)+b-c(-a-a)+a(a-a)]\)

⇒  \( \frac{1}{2}\left[b a+b b+c a+c a-b b-b a+c a+c a+a^2-\alpha^2\right]\)

⇒  \( \frac{1}{2}[4 a c]\)

= 2ac

Area of triangle = 2ac Square units

Question 33. Prove that the following points are Collinear:

1. (2,1), (4,3), and (3,2)

Solution: Area of triangle = \( \frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \( \frac{1}{2}[2(3-2)+4(2-1)+3(1-3)]\)

⇒  \( \frac{1}{2}[2(1)+4(1)+3(-2)]\)

⇒  \( \frac{1}{2}[2+4-6]\)

⇒  \( \frac{1}{2}[6-6]\)

⇒  \( \frac{1}{2}(0)=0\)

Therefore, the given points are collinear.

2. (9,6), (1,6) and (-7,-6)

solution: Area of triangle =\( \frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \(\frac{1}{2}[9(0+6)+1(-6-6)-7(6-0)]\)

⇒  \(\frac{1}{2}[9(6)+12-7(6)]\)

⇒  \(\frac{1}{2}[54-12-42]\)

⇒  \(\frac{1}{2}[54-54]\)

⇒  \(\frac{1}{2}(0)=0\)

Therefore, the given points are collinear

3. (b+c, 2), (Cta, b), and (a+b, c)

Solution: Area of triangle = \( \frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \( \frac{1}{2}[b+c(b-c)+c+a(c-a)+a+b(a-b)]\)

⇒  \( \frac{1}{2}\left[b^2-b c+c b-c^2+c^2-ca+ac-a^2+ a^2-ab+ba-b^2\right]\)

⇒  \( \frac{1}{2}\)

These fore, the given points are collinear.

4. (5,6), (-1,4) and (2,5)

Solution:

Area of triangle =\( \frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \(\frac{1}{2}[5(4-5)-1(5-6)+2(6-4)]\)

⇒  \(\frac{1}{2}[5(-1)-1(-1)+2(2)]\)

⇒  \(\frac{1}{2}[-5+1+4]\)

⇒  \(\frac{1}{2}[-5+5]\)

⇒  \(\frac{1}{2}(0)=0\)

Therefore given points are Collinear.

Question 34.1. If the points (2,3), (5,k), and (6,7) are Collinear, find the value of k.

Solution: Given points are Collinear

Area of triangle = 0

⇒  \(\frac{1}{2}[2(k-7)+5(7-3)+6(3-7)\)

⇒  \(\frac{2}{2}[2 k-14+20-24]=0\)

⇒  \(\frac{1}{2}[2 k-13]=0\)

⇒  2 k-18=0

⇒  2 k=17

⇒  \(k=\frac{18}{2}\)

⇒  K = 9

2. If the points A(k+1,2k), B(3k, 2k+3) and C(SK-1,5k) are collinear, find the Value of k.

Solution: Given points are Collinear

Area of Triangle = 0

⇒  \(\frac{1}{2}[k+1(2 k+3-5 k)+3 k(5 k-2 k)+5 k-1(2 k-2 k-3)]=0\)

⇒  \(\frac{1}{2}[k+1(-3 k+3)+3 k(3 k)+5 k-1(-3)]=0\)

⇒  \(\frac{1}{2}\left[-3 k^2-3 k+3 k+3+9 k^2-15 k+3\right]=0\)

⇒  \(\frac{1}{2}\left[6 k^2-15 k+6\right]=0 \)

⇒  \(\frac{3}{2}\left[2 k^2-5 k+2\right]=0\)

⇒  2x2=5k+2=0

⇒  2k2-4k-k+2=0

⇒  2k(k-2)-(K-2)=0

⇒ (2k-1) (k-2)=0

⇒  2k-1=0 and K-2=0

⇒  2k = 1

⇒  k= \(\frac{1}{2}\)

Question 35. If the points (x, y), (-1,3) and (5,-3) are Collinear, then Show that x+y=2.
Solution:

Given points are Collinear

Area of triangle = 0

⇒  \(\frac{1}{2}[x(3+3)-1(-3-y)+5(y-3)]=0 \)

⇒  \(\frac{1}{2}[x(6)+3+y+5 y-3(5)=0\)

⇒  \(\frac{1}{2}[6 x+6 y-12]=0\)

⇒  \(\frac{6(x+y-2)}{2}=0\)

⇒  x+y=2

Question 36. Find the Values of y for which the distance between the points A(3,-1) and B(11,y) is 10 units.
Solution:

Distance between the points A(3,-1) B(11,y)

⇒  \(AB =\sqrt{(11-3)^2+(y+1)^2}=10\)

⇒  \((8)^2+y^2+1+2 y=100\)

⇒  \( 64+y^2+1+2 y-100=0 \)

⇒  \( y^2+2 y-35=0\)

⇒  \(y^2-5 y+7 y-35=0\)

⇒  y(y-5)+7(9-5)=0

⇒   (y+7) (4-5)=0

⇒  4+7=0 or 4-5=0

⇒  y=-7 or y=5

Question 37. Find the relation between x and y Such that the point p(x,y) is equidistant from the points A(1,4) and B (-1,2).
solution:

Given that P(x,y) A(1,4) and B(-12)

⇒  PA= PB ⇒ PA2+ PB2

⇒  \( (x-1)^2+(y-4)^2=(y+1)^2+(y-2)^2 \)

⇒  \(x^2+\left(-2 x+y^2+16-8 y=x^2+1+2 x+y^2+4-4 y\right.\)

⇒  -2x-8y+17-2x+4y-5=0

⇒  -4x-4y+12=0 -4(x+y-3)=0

⇒  x+4=3

Question 38. Find the point on the y-axis which is equidistant from the points (-512) and (9,-2).
Solution:

let the required point on the y-axis be p(0,4) and the given points be A(-5,2) and B(9,-2).

Now, given that

PA=PB ⇒ PA2 = PB2

⇒  (0+5)2 + (9-2)2 = (6-9) + (Y+2)2

⇒  25+4 + 4-4y= 81+ y2+4+499

⇒  -44+29-44-85= 0

⇒  -8y-16=0

⇒  -8y=16

⇒  \(y=\frac{-16}{8}\)

y= -2

Question 39. Show that the pants (1,1), (1,5), (7,9) and (9,5) taken in that order, are the Vertices of a rectangle
Solution:

Given points are A(1, 1), B(-1,5), ((7,9) and D(9,5)

⇒  \(A B=\sqrt{(-1-1)^2+(5-1)^2}\)

⇒  \(\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\)

⇒  \(B C=\sqrt{(7+1)^2+(9-5)^2}\)

⇒  \(\sqrt{(8)^2+(4)^2}=\sqrt{64+16}=\sqrt{80}\)

⇒  \(C D=\sqrt{(9-7)^2+(5-9)^2}\)

⇒  \(\sqrt{(2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}\)

⇒  \(A D=\sqrt{(9-1)^2+(5-1)^2}\)

⇒  \(\sqrt{(8)^2+(4)^2}=\sqrt{64+(6}=\sqrt{80}\)

⇒  \(A B=C D=\sqrt{20}\)

⇒  \(B C=A D=\sqrt{80}\)

☐ ABCD is a rectangle.

Question 40. Show that the points ‘A (3,5), B( 6,01, C(,-3), and D(-2,2) are the Vertices of a Square ABCD.
Solution:

Given that

⇒  \(A B=\sqrt{(6-3)^2+(0-5)^2}=\sqrt{(3)^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}\)

⇒  \(B C=\sqrt{(1-6)^2+(-3-0)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{25+9}=\sqrt{34}\)

⇒  \(C D=\sqrt{(-2-1)^2+(2+3)^2}=\sqrt{(-3)^2+(5)^2}=\sqrt{9+25}=\sqrt{34}\)

⇒  \(A D=\sqrt{(-2-3)^2+(2-5)^2}=\sqrt{(-5)^2+(-3)^2}=\sqrt{25+9}=\sqrt{34}\)

⇒  \(A C=\sqrt{(1-3)^2+(-3-5)^2}=\sqrt{(-2)^2+(-8)^2}=\sqrt{4+64}=\sqrt{68}\)

⇒  \(B D=\sqrt{(-2-6)^2+(2-0)^2}=\sqrt{(-8)^2+(2)^2}=\sqrt{64+4}=\sqrt{68}\)

⇒  \(A B=B C=C D=A B=\sqrt{34}\)

⇒  \(A C=B D=\sqrt{63}\)

CBSE Solutions For Class 10 Mathematics Chapter 5 Arithmetic Progression

Arithmetic Progression

Question 1. The nth term of a Sequence In defined as follows. Find the first four terms:

1. an=3n+1
Solution:

3n=3nt

Put n= 1,2,3,4, we get

a1 = 3×1+1=4

a2 = 3×2+1=7

a3 = 3×3 +1 = 10

a4=3×4+1= 13

The first four terms of the Sequence are 4,7, 10,13.

2. an= n2+3
Solution:

an=n2+3

Put n=1,2,3,4 we get

a1= (1)2+3 = 4

a2=(2)2 +3= 7

a3 = (3)2+3 = 12

a4=(4)2 +3= 19

The first four terms of the Sequence are 4, 7, 12, and 19.

3. an = n(n+1)
Solution:

an=n (n+1)

Put n= 1,2,3,4 we get

a1 = 1(1+1)=2

a2 = 2(2+1)=6

a3=3(3+1)= (2

a4 = 4(4+1)=20

The first four terms of the Sequence are 2,6,12,20

4.\(a_n=n+\frac{1}{n}\)
Solution:

⇒ \(a_n=n+\frac{1}{n}\)

Put n=1,2,3,4 we get

⇒ \(a_1=1+\frac{1}{1}=2 \)

⇒ \(a_2=2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2} \)

⇒ \(a_3=3+\frac{1}{3}=\frac{9+1}{3}=\frac{10}{3} \)

⇒ \(a_4=4+\frac{1}{4}=\frac{16+1}{3}=\frac{17}{3}\)

First four terms of the Sequence are 2,\( \frac{5}{2}, \frac{10}{3}, \frac{17}{3}\)

5. an= 3n
Solution:

an=3n

Put n=1,2,3,4 we get

a1=31=3

a2=32 = 9

a3=33=27

a4=34=81

The first four terms of the Sequence are 3, 9, 27, 81

Question 2. The nth term of a Sequence is (3n-7). Find its 20th term.
solution:

3n-7

n=20

3(20)-7

60-7

57

The 20th term of a Sequence is 57

Question 3. Which of the following are A.p.’s? If they form an A.P., find the Common difference ‘d’ and write three more terms:

1. -10, -6, -2, 2,…….
Solution:

Here a=-10,

d=-6-10=4

-10, -6, -2, 2,4,6,10,14

Yes d= 4, next three items = 6, 10, 14

2. 3,3+ √2, 3+252, 3+3√2,…..
Solution:

Here a=3

d=3+√2-3 =) √2

3, 3+ √2, 3+2√2, 3+3√2, 3+4√2,3+5√2

Yes d=52, next three terms = 3+3√2, 3+4√2,3+5√2.

3. 0, -4, -8,-12,
Solution:

Here a=0

d=0-4=-4

0,-4,-8,-12-16,-20,-24

Yes do, next three terms = -16,-20,-24

Question 4. For the following A.-P. ‘S write the first terms and Common differences:

1. 2,5, 8, 11,..
Solution:

2, 5, 8, 11,

Her first term a=2

Common difference = 5-2 = 3

2.  -5,-1, 3, 7,
Solution: -5,-1, 3, 7

Her first term a=-5

Common difference = -541 = 4

CBSE Solutions For Class 10 Mathematics Chapter 5 Arithmetic Progression

Question 5. write the first four terms of the Ap., when the first term ‘a’ and the Common difference ‘d’ are given as follows:

1.  a=5, d=3
solution:

a=5, d= 3

a1 =5

a2=5+3=8

a3=8+3=11

a4=11+3= 14

The first four terms are 5, 8, 11, 14

2.  a=-2,d=4
Solution:

a=-2, d=4

a1 =-2

a2=-2+4=2

a3=2+4=6

a4=6+4= 10

The first four terms are -2,2,6,10

Question 6. Find the 10th term of the AP: 1,3,5,7,..
Solution:

Here, a=1

d=3-1=2,

n = 10

an = a+ (n-1)

90= 1+ (10-1)2

⇒96=1+18

⇒ 210=19

10th term of the given Ap=19

Question 7. Find the 7th term of the AP 80, 77,74,71,
Solution:

Here a=80

d=77-80=-3,

n=7

a7 = a + (n-1)d

a7=80+(7-1)-3

a7=80-18

⇒ a7 = 62

7th term of the given A.p. =62

Question 8. Find the nth term of the A⋅p: -5, -3, -1, 1, —-
solution:

Here a = -5

d=-3-5=2

n = n

an = a+ (n-1) d

an=-5+ (n-1)2

⇒ an=-5+2n-2

⇒ an= 2n-7

nth term of the given A.p. = (2n-7)

Question 9. Which term of the A-P. 4, 8, 12, is 76?
Solution:

Here, a = 4,

d=8-4=4

Let an=76

=) 4+ (n-1)4=76

= 4+4n-4=76

4n=76

⇒ \(n=\frac{76}{4} \Rightarrow n=19\)

19thterm of the given A.P. is 76

Question 10. which term of the Ap. 36, 33, 30, is Zero?
Solution:

Here a=36

d=33-36=-3

let an = 0

36+ (n-1)-3=0

36-3n+3=0

-3n=-39

\(n=\frac{39}{3} \Rightarrow n=13\)

The 13th term of the given A.P. is zero

Question 11. which term of the\(\frac{3}{4}, 1, \frac{5}{4}, \ldots \text {. is } 12 ?\)
Solution:

⇒ \(\text { Here } a=\frac{3}{4} \)

⇒ \(d=1-\frac{3}{4}=\frac{1}{4} \)

⇒ \(\text { let } a_n=12 \)

⇒ \(\frac{3}{4}+(n-1) \frac{1}{4}=12 \)

⇒ \(\frac{3+(n-1)}{4}=12\)

3+n-1=48

⇒ n+2=48

⇒n=46

46th term of the given AP is 12.

Question 12. Find the number of terms in the Ap. 8, 12, 16,
solution:

Here a=8

d=12-8=) 4

let an=124

=) 8+ (n-1)4=124

=) 8+40-4=124

=) 4n=124-4

4n = 120

⇒ n = \(n=\frac{120}{4} \Rightarrow 30\)

30th term of the given A.P. is 124.

Question 13. Find the number of terms in the A.p. 75, 70, 65, 15
Solution:

Here a=75

d=70-75=-5

let an=15

75+ (n-1)-5=15

75-5n+5=15

70-5n=15

-517=15-70

-5n=-55 ⇒ n=11

11th term of the given A. p. is 15.

Question 14. Find the 10th term from the end of the A.P. 82, 79, 76, —-,4.
Solution:

Here aa 1=4

d=79-82=-3

n=10

(-(10-1)defined

– 4-(10-1)-3

=4727

=31

10th term from the end = 31

Question 15. Find the 16th term from the end of the A.P. 3,6,9,99
solution:

Here, 1=99

d= 6-3 = 3,

n=16

⇒ 99-(16-1) 3

⇒ 99-45

⇒ 54

16th  term from the end = 54

Question 16. Find the Sum of the following A.p.: 3,8, 13, to 20 terms
Solution:

S1 =3+8+13,

a1=3

d=8-3=5

n=20

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d]] \)

⇒ \(S_{20}=\frac{20}{2}[2(3)+(20-1) 5]\)

⇒ \(S_{20}=10[6+95]\)

⇒ \(S_{20}=10[101]\)

⇒ \(S_{20}=1010 \)

Question 17. Find the sum of the following A.p. 5: 1,4,7,– to 50 terms
Solution:

S = 1+4+7+—–50

a=1

d=4-1= 3

n=50

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d] \)

⇒ \(S_{50}=\frac{50}{2}[2(1)+(50-1) 3] \)

⇒ \(S_{50}=25[2+147] \)

⇒ \(S_{50}=25[(49]\)

⇒ \(S_{50}=3,725\)

Question 18. Find the Sum given below: 3+6+9+ …..+96
Solution:

S=3+6+9+…… +96

a1 =3, d= 6-3 =) 3

an=96

a1+ (n-1) d=96

3+(n-1)3=96

3+3n-3=96

3n=96

⇒ \(n=\frac{96}{3} \Rightarrow n=32\)

S1 = Sum of 32 terms with first 3 terms and last term 96

⇒ \(S_1=\frac{32}{2}[3+96]\)

⇒ \(S_1=\frac{32}{2}[99]\)

⇒ \(S_1=16[99]\)

⇒ \(S_1=1584\)

Question 19. Find the Sum given below! 2 + 4+ 6+.
solution:

S=2+4+6+—–+50

a1 =2,

d=4-2 =) 2 Q1=2,

an=50

a1+ (n-1)d=50

2+(n-1)2=50

2+2n=2=50

⇒ \(n=\frac{50}{2} \Rightarrow n=25\)

S1 = Sum of 25 terms with first 2 terms and last terms so

⇒ \(S_1=\frac{50}{2}[2+50] \)

⇒ \(S_1=\frac{50}{2}[526] \)

⇒ \(S_1=50[26] \)

⇒ \(S_1=650\)

Question 20. In an A.p.: given a=2, d= 3, Qn = 50, find n and Sn.
Solution:

Given a=2, d=3, an =50

a+(n-1)d=an

2+(n-1)3=50

2+3n-3=50

3n=5041

⇒ \(n=\frac{51}{3} \Rightarrow n=17 \)

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d] \)

⇒ \(5_n=\frac{17}{2}[2(2)+(17-1) 3] \)

⇒ \(S_n=\frac{17}{2}[2(2)+(17-1) 3] \)

⇒ \(S_n=\frac{17}{2}[4+48] \)

⇒ \(S_n=\frac{17}{2}\left[S_2\right] \)

⇒ \(S_n=17[26] \)

⇒ \(S_n=442\)

Question 21. Find the Value of x for which (x+2), 2x, (2x+3) are three consecutive terms of A.P.
solution:

(x+2), 2x, (2x+3) are three consecutive terms of A.p.

\(2 x=\frac{(x+2)+(2 x+3)}{2}\)

4x= x+2+2x+3

4x=3x+5

4x-3x=5

x=5

CBSE Solutions For Class 10 Mathematics Chapter 4 Quadratic Equations

Quadratic Equations

Question 1. Which of the following are quadratic equations?

1.  X28x+12=0
Solution:

Given Solution is x2-8X+12=0

⇒ x2-6x-2x+12=0

⇒ x(x-6)-2(x-6) = 0

⇒(x-2)(x-6)=0

⇒x-2=0 or 2-6=0

⇒ x=2 Or x=6

Hence, x=2 and x=6 are the solutions.

2. 5x2-7x=3x2-7x+3

Solution:

Given Solution is 5x2-7x=3x2-7x+3

5x2-7x-3x2+7x-3=0

2x2-3=0

2x2=3

x2 = \(\frac{3}{2}\)

⇒ \(x=\sqrt{\frac{3}{2}}\)

Hence , \(x=\sqrt{\frac{3}{2}}\) are the solutions.

3. \(\frac{1}{4} x^2+\frac{7}{6} x-2=0\)
Solution:

Given equation is \(\frac{1}{4} x^2+\frac{7}{6} x-2=0\)

⇒ \(\frac{3 x^2+14 x-24}{12}=0\)

3x2+ 14x-24= 0 (1)

Equation in the form od ax2 + bx + c =0

a = 3, b = 14, c = -24

⇒ \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-14 \pm \sqrt{(14)^2-4(3)(-24)}}{2(3)}\)

⇒ \(\frac{-14 \pm \sqrt{196+288}}{6}\)

⇒ \(\frac{-14 \pm \sqrt{484}}{6}\)

⇒ \(\frac{-14 \pm 22}{6}\)

⇒ \(\frac{-14+22}{6} \text { or } \frac{-14-22}{6}\)

⇒ \(\frac{8}{6} \text { or } \frac{-36}{6}\)

⇒ \(\frac{4}{3} \text { or }-6\)

Hence \(x=\frac{4}{3} \text { and } x=-6\) are the solutions.

Question 2. Which of the following are roots of 4x2-9x-100=0?

  1. -4
  2. \(\frac{3}{4}\)
  3. \(\frac{25}{4}\)

Solution:

The given equation is 4x2-9x-100=0

On substituting x=-4 in the given equation

L.H.S= 4(-4)2-9(-4) – 100 = 0

64+34-100= 0 ⇒ = 0 = R.H.S

∴ x=-4 is a Solution of 4×2-9x-100.

On Substituting x=2/14 in the given equation

L.H.S= \(4\left(\frac{3}{4}\right)^2-9\left(\frac{3}{4}\right)-100=0\)

⇒ \(4\left(\frac{9}{16}\right)-\frac{27}{4}-100=0\)

⇒ \(\frac{36-108-1000}{16}=0\)

= 36-208=0

= 178 not equal to R.H.S

⇒ \(x=\frac{3}{4}\) is not a solutions of 4×2 9x – 100=0

On substituting \(x=\frac{25}{4}\)

⇒ \(\text { L.H.S }=4\left(\frac{25}{4}\right)^2-9\left(\frac{25}{4}\right)-100=0\)

⇒ \(4\left(\frac{625}{16}\right)-\frac{225}{4}-100=0\)

⇒ \(\frac{2500-900-1600}{16}=0\)

2500-2500=0 = R.H.S

⇒ \(x=\frac{25}{4}\) is a solution of 4×2 is a solution.

CBSE Solutions For Class 10 Mathematics Chapter 4 Quadratic Equations

Question 3. If one root of the quadratic equation 6x2-x-k=0 is 2, find the k value.
Solution:

Since, \(x=\frac{2}{3}\) is a solution of 6x2-x-k=0

⇒ \(6\left(\frac{2}{3}\right)^2-\frac{2}{3}-k=0\)

⇒ \(6\left(\frac{4}{9}\right)-\frac{2}{3}-k=0\)

⇒ \(\frac{24-6-9 k}{9}=0\)

18-9k = 0

18= 9k

⇒ \(k=\frac{18}{9}\)

k=2

Hence k=2 of the solution.

Question 4. 3x2-243=0
Solution:

Given equation is 3x2-243=0

3x2=243

⇒ \(x^2=\frac{243}{3}\)

x⇒ = 81

⇒ \(x=\sqrt{81}\)

⇒ \(x= \pm 9\)

x = 9 or x = -9

Hence x = 9 and x = -9 are the solutions.

Question 5. 5x2+4x=0
Solution:

Given equation is 5x2+4x=0

x(5x+4)= 0

x = 0 or 5x+4 – 0

5X=-4

x= \(\frac{-4}{5}\)

Hence x=0 and x = \(\frac{-4}{5}\) are the Solutions.

Question 6. x2 +12x+35=0
Solution:

The given equation is x2 +12x+35=0

x2 +5x+7x+35=0

x(x+5)+7(x+5)=

(1+7)(x+5)=0

x+7=0 or x+5=0

x=-7 or x=-5

Hence 2=-7 and x=-5 are the solutions.

Question 7. 2x2=5x+3=0
solution:

Given equation is 2x2-5x+3=0

2x2 =5x+3=0

2x2 =3x-2x+3=0

x(2x-3)-1(2x-3)=0

(x-1)(2x-3)=0

x-1=0 or 2x-3=0

x=1 or 2x-3=0

2x=3

⇒ x = \(=\frac{3}{2}\)

Hence x=1 and x = \(=\frac{3}{2}\) are the Solutions.

Question 8. 6x2-x-2=0
Solution:

Given equation is 6x2-x-2=0

⇒ 6x2-4x+3x-2=0

⇒ 2x(3x-2)+1(3x-2)=0

⇒ (x+1)(3x-2)=0

⇒ 2x+1=0 Or 3x-2=0

⇒ 2x=-1 or 3x=2

⇒ \(x=-\frac{1}{2}\) Or \(x=\frac{2}{3}\)

Hence \(x=-\frac{1}{2}\) and \(x=\frac{2}{3}\) are the Solutions.

Question 9. 8x2-2x-21=0
Solution:

Given equation are 8x2-2x-21=0

8x2+6x-28x-21=0

2x(4x+3)-7(4x+3)=0

(2x-7)(4x+3)= 0

2x-7=0 4x+3=0

2x=7 Οr 4x=-3

⇒ \(x=\frac{7}{2}\) or \(x=\frac{-3}{4}\)

Hence \(x=\frac{7}{2}\) and \(x=\frac{-3}{4}\)

Question 10.6x+40=31x
Solution:

Given equation are 6×2-31x+40=0

⇒ 62x= 18x-16x+40=0

⇒ 3x(2x-5)-8(2x-5)=6

⇒ (3x-8)(2x-5)=6

⇒ 3x-8=0 or 2x-5=0

⇒ 3x=8 Οr 2x = 5

⇒ \(x=\frac{8}{3}\) Or \(x=\frac{5}{2}\)

Hence \(x=\frac{8}{3}\) and \(x=\frac{5}{2}\) are the Solutions:

Question 11. \(\sqrt{3} x^2-11 x+8 \sqrt{3} x=0\)
Solution:

Given Equation is √3x2=11x+8√3=0

Equation in the form ax+ bx+c

a=√3, b=-11, C=8√3

⇒ \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{11 \pm \sqrt{(-11)^2-4(\sqrt{3})(8 \sqrt{3})}}{2 \sqrt{3}}\)

⇒ \(\frac{11 \pm \sqrt{121-96}}{2 \sqrt{3}}\)

⇒ \(\frac{11 \pm \sqrt{25}}{2 \sqrt{3}}\)

⇒ \(\frac{11 \pm 5}{2 \sqrt{3}}\)

⇒ \(\frac{11+5}{2 \sqrt{3}} \text { or } \frac{11-5}{2 \sqrt{3}}\)

⇒ \(\frac{16}{2 \sqrt{3}} \text { or } \frac{6}{2 \sqrt{3}}\)

⇒ \(\frac{8}{\sqrt{3}} \text { or } \frac{3}{\sqrt{3}}\)

⇒ \(\frac{8 \times 3}{\sqrt{3}} \text { or } \sqrt{3}\)

⇒ \(8 \sqrt{3}\)

x=8√3 and x =√3

Hence x=8√3 and x =√3 are the solutions.

Question 12.3x2 – 256x+2=0
Solution:

Given equation is 3x2-2√6x+2=0

Equation in the form of an 7 bx + c = 0

a = 3, 6=-256, C= 2

⇒ \(\Rightarrow \frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{2 \sqrt{6} \pm \sqrt{(-2 \sqrt{6})^2-4(3)(2)}}{2(3)}\)

⇒ \(\frac{2 \sqrt{6} \pm \sqrt{26-26}}{6}\)

⇒ \(\frac{2 \sqrt{6}}{6}\)

⇒ \(\frac{2}{\sqrt{6}}\)

⇒ \(\frac{2}{\sqrt{2 \times 3}}\)

⇒ \(\frac{\sqrt{2}}{\sqrt{3}} \Rightarrow \sqrt{\frac{2}{3}}\)

Hence \(x=\sqrt{\frac{2}{3}}\) are the solution.

Question 13. x2 + 5= \(\frac{9}{2} x\)
Solution:

Given equation is x2 + 5 – \(\frac{9}{2} x\)

2x2 +10-9x=0

⇒ 2x2=9x+10=0

⇒ 2x2=4x-5x+10=0

⇒ 2x(x-2)-5(x-2)=0

⇒ (2x-5)(x-2)=0

⇒ 2x-5=0 or x-2=0

⇒ 2x=5 or x=2

⇒ \(x=\frac{5}{2}\)

Hence \(x=\frac{5}{2}\) and x= 2 are the solution

Question 14. \(x=\frac{3 x+x}{}\)
Solution:

Given equation is x= \(x=\frac{3 x+x}{}\)

4x2-3x+1

4x2-37-1=0

4x2=4x-x-1=0

4x(x-1)+(x-1)=6

(4x+1)(x-1)=0

4×71=0 or x-1=0

4x=-1 or x=1

⇒  \(x=\frac{-1}{4}\)

Hence \(x=\frac{-1}{4}\) and x=1 are the solution.

Question 15. \(5 x-\frac{35}{x}=18, x \neq 0\)
Solution:

Given equation is \(5 x-\frac{35}{x}=18\)

5×2-18x-35=0

⇒ 5×2+7x-25x-35=0

⇒ x(5x+7)-5(5x+7)= 0

⇒ x-5=0 or 5x+7=0

⇒ x = 5 or 5x=-7

⇒ \(x=\frac{-7}{5}\)

Hence x = 5 and \(x=\frac{-7}{5}\) are the solution.

Question 16. \(\frac{2}{x^2}-\frac{5}{x}+2=0, x \neq 0\)
Solution:

Given equation is \(\frac{2}{x^2}-\frac{5}{x}+2=0\)

\(\frac{2-5 x+2 x^2}{x^2}=0\)

2x2-5x+2=0

2x2-4x-x+2=0

2x2-4x-x+2=0

2×(x-2)-(X-2) = 0

(2x-1)(x-2)=0

2x-1=0 or x=2=0

2x = 1 Or x = 2

⇒ \(x=\frac{1}{2}\)

Hence \(x=\frac{1}{2}\)or x=2 are the solution.

Question 17. a2x2+2ax+1= 0
Solution:

Given equation is a2x2+2ax+1= 0

a2x2+2ax+1= 0

a2x2+ax+ax+1= 0

ax(ax+1)+(ax+1)=0

(ax+1)(ax+1)=0

ax+1=0 or ax+1=0

ax=-1 Or ax = -1

⇒ \(x=\frac{-1}{a} \text { or } x=\frac{-1}{a}\)

Hence \(x=\frac{-1}{a} \text { or } x=\frac{-1}{a}\) are the solution.

Question 18. x2 – (p+q)x+pq=0
Solution:

Given equation is x2 – (p+q)x+pq=0

x2 – qx – Px +Pq=0

x(x-2)-P(x-2)=0

(X-P) (x-2)=0

X-P=O or x-2=0

X=P or x=2

Hence x=P and x=q are the solutions.

Question 19. 12abx2-(9a2-8b2)x-6ab=0
Solution:

Given Equation is 12 abx (992-867)x-6ab=0

12 abx=99x+86-x=6ab=0

3ax (4bx-3a)+2b (4bx-3a)=0

(3ax+26) (46x-3a)=

3ax+2b=0 or 4bx-3a=0

3ax=-2b or 4bx=3a

Hence x = -2b and n=39 are the solutions.

⇒ \(x=\frac{-2 b}{3 a}\) Or \(x=\frac{3 a}{4 b}\)

Hence \(x=\frac{-2 b}{3 a}\) and \(x=\frac{3 a}{4 b}\) are the solution.

Question 20.4×2-4ax+(a2-b2)=0
Solution:

Given Equation is 4x2 – yax + (a2 – b2) =0

4×2-4ax + (a2-b2)=0

4×2 = 49x+a2= b2=0

⇒ \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{4 a \pm \sqrt{(4 a)^2-4(4)\left(a^2-b^2\right)}}{2(4)}\)

⇒ \(\frac{4 a \pm \sqrt{16 a^2-16 a^2+16 b^2}}{8}\)

⇒ \(\frac{4 a \pm \sqrt{(4 b)^x}}{8}\)

⇒ \(\frac{4 a+4 b}{8}\)

⇒ \(\frac{4 a+4 b}{8} \text { or } \frac{4 a-4 b}{8}\)

⇒ \(\frac{4(a+b)}{8} \text { or } \frac{4(a-b)}{8}\)

⇒ \(\frac{a+b}{2} \text { or } \frac{a-b}{2}\)

Hence \(x=\frac{a+b}{2} \text { and } x=\frac{a-b}{2}\) are the solution.

Find the roots of the following quadratic equations by the method of  Completing the Square.

Question 21. Find the roots of the following quadratic equations by the method of  Completing the Square x2=10x-24=0
Solution:

Given equation is x2-10x-24=0

on Comparing with ax2+ bx+ c=0, we get

a=1, b=-10, c= -24

Discriminant, D= 2-4ac

⇒ D= (-10)2- 4(1)(-24)

⇒ D= 100+96

⇒ D = 196

Hence, the given equation has two real roots.

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{+10 \pm \sqrt{196}}{2}\)

⇒ \(x=\frac{10 \pm 14}{2}\)

⇒ \(x=\frac{10+14}{2} \text { or } \frac{10-14}{2} \text {, }\)

⇒ \(x=\frac{24}{2} \text { or }-\frac{4}{2}\)

x= 12 Or – 2

x= 12,-2 are the roots of the equation.

Question 22. 2x2-7x-39=0
Solution:

The given equation is 2x2-7x-39=0

on Comparing that ax + bx+C=0, we get

=2, 6=-7, C=-39

Discriminant D= b2-4ac

⇒ D= (-7)==4(2)(-39)

⇒ D= 49+312

⇒ D = 361

Hence, the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{7 \pm \sqrt{361}}{2(2)}\)

⇒ \(x=\frac{7+19}{4} \text { or } \frac{7-19}{4}\)

⇒ \(x=\frac{13}{2} \text { or }-3\)

⇒ \(x=\frac{13}{2} \text { or }-3\) are roots of the equation.

Question 23. 5×2+6x-8 = 0
Solution:

Given equation is 5×2+6x-8 = 0

on Comparing that ax2+ bx+c=0, we get

a=5, b=6, C=-8

Discriminant D= b2– 4ac

⇒ D= (6)2-4(5)(-8)

⇒ D = 36 + 160

⇒ D= 196

Hence, the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-6 \pm \sqrt{196}}{2(5)}\)

⇒ \(x=\frac{-6 \pm 14}{10}\)

⇒ \(x=\frac{-6+14}{10} \text { or } \frac{-6-14}{10}\)

⇒ \(x=\frac{4}{5} \text { or }-2\)

⇒ \(x=\frac{4}{5} \text { or }-2\) are the real roots.

Question 24. \(\sqrt{3} x^2+11 x+6 \sqrt{3}=0\)
Solution:

Given Equation is \(\sqrt{3} x^2+11 x+6 \sqrt{3}=0\)

a=√3, b=ll, C=6√3

Discriminant D= b2=yac

⇒ D= (11)2 – 4(√3)(6√3)

⇒ D= 121-72

⇒ D = 49

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-11 \pm \sqrt{49}}{2 \sqrt{3}}\)

⇒ \(x=\frac{-11 \pm 7}{2 \sqrt{3}}\)

⇒ \(x=\frac{-11+7}{2 \sqrt{3}} \text { or } \frac{-11-7}{2 \sqrt{3}}\)

⇒ \(x=\frac{-4}{2 \sqrt{3}} \text { or } \frac{-18}{2 \sqrt{3}}\)

⇒ \(x=\frac{-2 \sqrt{3}}{3} \text { or }-3 \sqrt{3}\)
are the roots.

Question 25. 2×2 -9x+7=0
Solution:

Given equation is 2×2=9x+7=0

2×2-9x+7=0

on Comparing that Qx2+6x+(=0, we get

a=2, 6=-9, c=7

Discriminant D=6=42C

⇒ D=(-9)=4(2)(7)

⇒ D=81-56

⇒ D=25

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{9 \pm \sqrt{25}}{2(2)}\)

⇒ \(x=\frac{9 \pm 5}{4}\)

⇒ \(x=\frac{9+5}{4} \text { or } \frac{9-5}{4}\)

⇒ \(x=\frac{14}{4} \text { or } \frac{4}{4}\)

⇒ \(x=\frac{7}{2} \text { or } 2\)

⇒ \(x=\frac{7}{2} \text { or } 2\)are the real roots.

Question 26.5x2-9x+17=0
Solution:

Given equation is 5x2-9x+17=0

on Comparing that ax2 + bx + c =0, we get

a=5, b=-19, C=17

Discriminant D=b2-4ac

⇒ 3D = (19)2 – 4(5)(17)

⇒ D= 361-340

⇒ D = 21

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{19 \pm \sqrt{21}}{2(5)}\)

⇒ \(x=\frac{19+\sqrt{21}}{10} \text { or } \frac{19-\sqrt{21}}{10}\)

⇒ \(x=\frac{19+\sqrt{21}}{10} \text { or } \frac{19-\sqrt{21}}{10} \)are real roots.

Question 27. x2-18x+77=0
Solution:

Given equation is x2-18x+77=0

On Comparing that an’+ bx+C=0, we get

a=1, b=-18, c=77

Discriminant D= b2 – 4ac

⇒ D=(-18)-4(1)(77)

⇒ D= 394 308

⇒ D= 3·16

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{18 \pm \sqrt{16}}{2(1)}\)

⇒ \(x=\frac{18 \pm 4}{2}\)

⇒ \(x=\frac{18+4}{2} \text { or } \frac{18-4}{2}\)

⇒ \(x=\frac{22}{2} \text { or } \frac{14}{2}\)

x=11or7

x= 11,7 are two real numbers.

Question 28. \(\frac{2}{3} x=-\frac{1}{6} x^2-\frac{1}{3}\)
Solution:

Given equation is \(\frac{2}{3} x=-\frac{1}{6} x^2-\frac{1}{3}\)=0

On comparing that at ax2+bx+C=0, we get

⇒ \(a=\frac{1}{6}, b=\frac{2}{3}, c=\frac{1}{3}\)

⇒ \(\text { Discriminant } D=b^2-4 a c\)

⇒ \(\Rightarrow D=\left(\frac{2}{3}\right)^2-4\left(\frac{1}{6}\right)\left(\frac{1}{3}\right)\)

⇒ \(D=\frac{4}{9}-\frac{4}{18}\)

⇒ \(D=\frac{8-4}{18}\)

⇒ \(D=\frac{4}{18} \Rightarrow D=\frac{2}{9}\)

Hence the given equation has two real roots.

⇒ \(x=\frac{-b \pm \sqrt{0}}{2 a}\)

⇒ \(x=\frac{-\frac{2}{3} \pm \sqrt{\frac{2}{9}}}{2\left(\frac{1}{6}\right)}\)

⇒ \(x=\frac{\frac{-2}{3} \pm \frac{\sqrt{2}}{3}}{\frac{2}{6}}\)

⇒ \(x=\frac{\frac{-2}{3} \pm \frac{\sqrt{2}}{3}}{1 / 3}\)

⇒ \(x=\frac{(-2 \pm \sqrt{2}) \times 3}{\not 2}\)

⇒ \(x=-2 \pm \sqrt{2}\)

⇒ \(x=-2+\sqrt{2} \text { or }-2-\sqrt{2}\)

⇒ \(x=-2+\sqrt{2},-2-\sqrt{2}\) are two real roots.

Question 29. \(\frac{1}{15} x^2+\frac{5}{3}=\frac{2}{3} x\)
Solution:

Given equation is \(\frac{1}{15} x^2-\frac{2}{3} x+\frac{5}{3}=0\)

⇒ \(a=\frac{1}{15}, b=\frac{-2}{3}, c=\frac{5}{3}\)

Discriminant \(D=b^2-4ac\)

⇒ \(D=\left(\frac{-2}{3}\right)^2-4\left(\frac{1}{15}\right)\left(\frac{5}{3}\right)\)

⇒ \(D=\frac{4}{9}-\frac{20}{45}\)

⇒ \(D=\frac{20-20}{45}\)

⇒ D = 0

Hence the given equation is two real roots.

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{\frac{2}{3} \pm 0}{2\left(\frac{1}{15}\right)}\)

⇒ \(x=\frac{x}{3} \times \frac{15^{5}}{x}\)

x = 5,5 are two real roots.

Question 30. \(\sqrt{6} x^2-4 x-2 \sqrt{6}=0\)
Solution:

Given equation is √6x=4x2-2√6=0

On Comparing that an2+ bx+c=0, we get

Discriminant D= 2-42c

⇒ D= (4)2=4(√c)(-2√2)

⇒ D= 16+48

⇒ D= 64

Hence Given equation has two real roots.

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{4 \pm \sqrt{64}}{2 \sqrt{6}}\)

⇒ \(x=\frac{4 \pm 8}{2 \sqrt{6}}\)

⇒ \(x=\frac{4+8}{2 \sqrt{6}} \text { or } \frac{4-8}{2 \sqrt{6}}\)

⇒ \(x=\frac{12}{2 \sqrt{6}} \text { or } \frac{-4}{2 \sqrt{6}}\)

⇒ \(x=\frac{\not 2 \times 6}{\not 2 \sqrt{6}} \text { or } \frac{-\not 2 \times 2}{\not 2 \sqrt{6}}\)

⇒ \(x=\sqrt{6} \text { or } \frac{-2}{\sqrt{6}}\)

⇒ \(\frac{-2}{\sqrt{2 \times 3}}\)

⇒ \(\frac{-6 \times 2}{2 \sqrt{6}}\)

⇒ \(\frac{-\sqrt{6}}{3}\)

⇒ \(x=\sqrt{6},-\frac{\sqrt{6}}{3}\) are two real roots.

Question 31. 256 x 2 – 32x + 1 = 0
Solution:

Given equation is 256 x2 = 32x+1=0

On Comparing that an’ + bx + c = 0, we get

a=256, b=-32, C=1

Discriminant D=6=4ac

⇒ D= (32) = 4(256)(1)

⇒ D= 1024-1024

⇒ D= 0

Hence given equation is two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{32 \pm \sqrt{6}}{2(256)}\)

⇒ \(x=\frac{32}{512}\)

⇒ \(x=\frac{1}{16}\)

⇒ \(x=\frac{1}{16}\) are two real roots.

Question 32. (2x+3)(3x-2)+2=0
Solution:

Given equation is (2x+3)(3x-2)+2=0

6x2-4x+9x-6+2=0

6x2+5x-4=0

On Comparing that ax2+bx+C=0, we get

a=6, b=5, C=-4

Discriminant D= b2 – 4ac

⇒ D=(C)2=4(6)(-4)

⇒ D= 25+96

⇒ D = 121

Hence given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-5 \pm \sqrt{127}}{2(6)}\)

⇒ \(x=\frac{-5 \pm \sqrt{121}}{12}\)

⇒ \(x=\frac{-5+11}{12} \text { or } \frac{-5-11}{12}\)

⇒ \(x=\frac{6}{12} \text { or }-\frac{16}{12}\)

⇒ \(x=\frac{1}{2} \text { or }-\frac{4}{3}\)

⇒ \(x=\frac{1}{2} \text { or }-\frac{4}{3}\) are two real roots.

Question 33. x2-16=0
Solution:

Given equation is x2-16=0

x2-42=0

(x-4)2=0

x2+4-4x=0

On comparing that ax2+6x+C=0, we get

a=1, 6=-4, c=4

Discriminant D=b2-4ac

⇒ D=(-4)2-4(1)(4)

⇒ D = 16-16

⇒ D = 0

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{4 \pm \sqrt{0}}{2}\)

⇒ \(x=\frac{4}{2}\)

x = 2

Question 34. 36x2 – 129x+(a2-b2)=0
Solution:

Given equation is 36x – 12ax + (a2 – b)=0

On comparing that ax2+bx+c=0, we get

a=36, b=-12a, C=(a2– b2)

Discrimanant =) D= b2-4ac

⇒ D=(12a)2-4(36)(a2-6-b2)

⇒ D= 144a2 – 144(a2-b2)

⇒ D= 144a2 – 144a2+144b2

⇒ D=144b2

Hence the given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{12 a \pm \sqrt{144 b^2}}{2(36)}\)

⇒ \(x=\frac{12 a \pm\not \sqrt{(12 b)\not^2}}{72}\)

⇒ \(x=\frac{12 a \pm 12 b}{72}\)

⇒ \(x=\frac{12(a \pm b)}{72}\)

⇒ \(x=\frac{12(a+b)}{72} \text { or } \frac{12(a-b)}{72}\)

⇒ \(x=\frac{a+b}{6} \text { or } \frac{a-b}{6}\)

⇒ \(x=\frac{a+b}{6} \text { or } \frac{a-b}{6}\) are two real roots.

Question 35. P2 x2 + (p2– q2)x-q2=0
Solution:

Given equation is p2x2 + (p2 -q2)x-q2=0

On Comparing that ax2+ bx + c = 0, we get

a=p2, b= (p2 q2), c = -q2

Discriminant D= b2-4ac

⇒ D = (p2 q2)2 – 4(p2) (−22)

⇒ D= p2 q4 + 4p2 q2 – 2p2q2

⇒ D = p2 + 2 p2 q 2 + q 2 =) (P2+q2) 2

Hence the given equation has two real roots.

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-\left(p^2-q^2\right) \pm \sqrt{p^2+q p^2 q^2+q^2}}{2 p^2}\)

⇒ \(x=\frac{-p^2+q^2 \pm \not\sqrt{\left(p^2+q^2\right)\not ^2}}{2 p^2}\)

⇒ \(x=\frac{-p^2+q^2 \pm\left(p^2+q^2\right)}{2 p^2}\)

⇒ \(x=\frac{\not -p^2+q^2+\not p^2+q^2}{2 p^2} \text { or }-\frac{-p^2+\not q^2-p^2-\not q^2}{2 p^2}\)

⇒ \(x=\frac{2 q^2}{2 p^2} \quad \text { or } \quad \frac{-2 p^2}{2 p^2}\)

⇒ \(x=\frac{q^2}{p^2} \quad \text { or }-1\)

⇒ \(x=\frac{q^2}{p^2} \quad \text { or }-1\) are two real roots.

Question 36. abx2+(b2-ac)x-bc=0
solution:

The given equation is abx2+ (6-ac)x-bc=0

on Comparing that ax2 + bx + c=0, we get

a= ab, b= (b2ac), c=-bc

Discriminant =) D= b2 – 4ac

⇒ D= (b2-ac)2 +4(ab) (-bc)

⇒ D= b4 + a2c2 = 2b2ac+4ab2c

⇒ D= b4+ a2c2+2ab2c

⇒ 0= (b2+ac)2

Hence given equation has two real roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-\left(b^2-a c\right) \pm \sqrt{\left(b^2+a c\right)^2}}{2 a b}\)

⇒ \(x=\frac{\not -b^2+a c+\not b^2+a c}{2 a b} \text { or } x=\frac{-b^2+\not a c-b^2-\not \ a c}{2 a b}\)

⇒ \(x=\frac{2 a c}{2 a b} \quad \text { or } x=\frac{-2 b^2}{2 a b}\)

⇒ \(x=\frac{c}{b} \quad \text { or } x=\frac{-b}{a}\)

⇒ \(x=\frac{c}{b} \quad \text { and} x=\frac{-b}{a}\) are two roots.

Question 37. 12abx2 – (9a2-8b2) x-6ab=0
solution:

Given equation is 12abx – (9a2-8b2)x-6ab=0

on comparing that ax2+ bx + c = 0, we get

a=12ab, b=-(9a2 =8b2), c=-6ab

Discriminant ⇒ D= b2-4ac

⇒ D= (-(9a2 = 8b2)2 – 4(12ab) (-6ab)

⇒D=81a464b4-144a2b2+288a2b2

⇒ D= 81a4+64b4+ 144a2b2

⇒ D= (9a2 +8b2)2

Hence the given equation has two roots

⇒ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{\left(9 a^2-8 b^2\right) \pm \sqrt{\left(9 a^2+8 b^2\right)^2}}{24 a b}\)

⇒ \(x=\frac{9 a^2-8 b^2 \pm\left(9 a^2+8 b^2\right)}{24 a b}\)

⇒ \(x=\frac{9 a^2-8 b^2+9 a^2+8 b^2}{24 a b}\) Or \(x=\frac{9 a^2-8 b^2-9 a^2-8 b^2}{24 a b}\)

⇒ \(x=\frac{18 a^x}{24 a b} \quad \text { or } x=\frac{-16 b^x}{24 a b}\)

⇒ \(x=\frac{3 a}{4 b}\) or \(x=\frac{-2 b}{3 a}\)

⇒ \(x=\frac{3 a}{4 b},-\frac{2 b}{3 a}\) are two real roots.

Question 38. Determine the nature of the roots of the following quadratic equations: 2x2+5x-4=0
Solution:

The given equation is 2x2+5x-4=0

9x2-6x+1=0

Comparing with ax + bx+c=0

a=2, 6:5, C=-4

Discriminant ⇒ D= b2– 4ac

⇒ D= (5)=4(2)(-4)

⇒ D=25+32

⇒ D= 47

⇒ D>0

Hence the equation has real and distinct roots.

Question 39. 9x2-6x+1=0
solution:

The given equation is 9x2-6x+1=0

Comparing with ax2 + bx +C=0

a=9, 6=-6, C=1

Discriminant ⇒ D= b2-4ac

⇒ D= (-6)2-(9)(1)

⇒ D= 36-36

⇒ D= 0

Hence the given equation has real and equal roots.

Question 40. Find the Value of k for which the equation 12×2+4kx+3=0 has real and equal roots.
Solution:

Given equation is 12x2 + 4kx+3=0

Comparing with ax2+ bx+c=0,

a=12, b=4k, c=3

For real and equal roots,

Discriminant (D)=0 ⇒ D= b2-4ac=0

⇒(4K)2=-4(12)(3)=0

⇒ 16K2-144=0

⇒ 16k2=144

⇒ k2=\(\frac{144}{16}\)

⇒ K2 = 9

⇒ k = √9

⇒ k= ±3

Question 41. Find the value of k for which the equation 2×2+5x-k=0 has real roots.
solution:

Given equation is 2x2+5x-k=0

Comparing with ax2+bx+C=0

a=2, b=5, C=-k

For real roots, Discriminant (D) =20

(5)2+4(2)(18)20

⇒ \(k\frac{-25}{8}b^2-4 a c \geq 0\)

Question 42. The sum of a number and its reciprocal is \(\frac{10}{3}\), find the number (5).
Solution:

let the number be

According to the given Statement \(x+\frac{1}{x}=\frac{10}{3}\)

⇒ 3x2+3=10x

⇒ 3x2 – 10x+3=0

⇒ 3x2-9x+x+3=0

⇒ 3x(x-3)-(x-3)=0

⇒ (3x-1)(x-3) = 0

when 3x-1=0, x = = = =

and when x-3=0, x = 3

Hence the number of (5) are 3 and 1.

⇒ \(\frac{x^2+1}{x}=\frac{10}{3}\)

⇒ \(3 x-1=0, x=\frac{1}{3}\)

and when x-3=0 , x=3

⇒ latex]3 x^2+3=10 x/latex]

Hence the numbers of are \(\frac{1}{3}\)

CBSE Solutions For Class 10 Mathematics Chapter 3 Linear Equation In Two Variables

Linear Equation In Two Variables

Question 1. Solve for x and y:

7+Y=17,

7-4=1

Solution: Given

x+y=17  → (1)

x-4=1    → (2)

From equation 1 we get y= 17-x → R

Substituting the value of from equation in 2, we get

x-(17-x)=1

2-17+2=1

2x=1+17

2x = 18

x = \(\frac{18}{2}\) ⇒ x=9

Substituting the value of x in an equation, we

y= 17-9

y= 8

Solution is x=9

y=8

2. x+2y=19

-x+2y=1

Solution: Given x+2y= 190 → (1)

-x+2y=1 → (2)

From equation 1 we get x=19-2y → 3

Substituting the value of and from equations (3)and (2) we get

-(19-2y) +2y= 1

– 19+2y+24=1

4y = 1+19

4y = 20

y = \(\frac{20}{4}\)

y = 5

Substitute value of y in equation 3 we get

x=19-25

X=19-2(5)

1=19-10

x=9

Solution is x =9

y=5

3. x – y= 0.9

\(\frac{11}{2(x+y)}=1\)

Solution: Given x – y= 0.9 → (1)

⇒ \(\frac{11}{2(x+y)}=1\)

⇒ \(\frac{11}{2 x+2 y}=1\)

2x + 2y = 11 → (2)

From equation we get x = 0-9 + y → (3)

Substituting the value of x from the equation, we get

2(0.9+4)+2y=11

1.8 +25+24=11

4y = 11-1-1

4y= 9, 2

y = \(\frac{9.2}{4} \Rightarrow y=2.3\)

Substituting the value of the y value in the (3) equation we get

2=0·9+2·3

X = 3.2

Solution is 2=3.2

y=2-3

4. 3x-2y=6

\(\frac{x}{3}-\frac{y}{6}=\frac{1}{2}\)

Solution: Given 3x-2y=6

⇒ \(\frac{x}{3}-\frac{y}{6}=\frac{1}{2}\)

⇒ \(\frac{6 x-y}{6}=\frac{1}{2}\)

2(6x-4)=6

12x-2y=6

From equation 1 we get 3x-2y=6

3x=6+24

⇒ \(x=\frac{2 y+6}{3} \rightarrow \text { (3) }\)

Substituting the value of x from equation 3 in 2, we get

⇒ \(12\left(\frac{2 y+6}{3}\right)-2 y=6\)

8y+824-2y=6

6y=6-24

6y=-18

⇒ \(y=\frac{-18}{6} \Rightarrow y=-3\)

Substituting the Value of y in Equation 3

⇒ \(x=\frac{2(-3)+6}{3}\)

⇒ \(x=\frac{-6+6}{3} \Rightarrow x=0\)

Solution is x=0

y=-3

5. 0.4x+0.34 = 1.7

07x-0-2y=0·8

Solution: Given 0.4x+0-3y=1-7 → (1)

0.72-0·24=0·8 → (2)

From equation (1) we get 0.4x=1.7-0.34

⇒ \(x=\frac{1.7-0.3 y}{0.4} \rightarrow \text { (3) }\)

Substituting the value of x from equation (3) in (2)we get

⇒ \(0.7\left(\frac{1.7-0.3 y}{0.4}\right)-0.2 y=0.8\)

⇒ \(\frac{1.19}{0.4}-\frac{0.21 y}{0.4}-0.2 y=0.8\)

2.975-0.525y-0-2y=0.8

-0-725y = 0. = 0 .525y-2.975

– 0.725y = 2175

⇒ \(0.7\left(\frac{1.7-0.3 y}{0.4}\right)-0.2 y=0.8\)

⇒ \(\frac{1.19}{0.4}-\frac{0.21 y}{0.4}-0.2 y=0.8\)

2.975y – 0.525y – 0.2y = 0.8

-0.725y  = 0.8 – 2.975

-0.725y  = -2.175

\(y=\frac{-2.175}{-0.725}\)

Substituting the value of ‘y’ from the equation

⇒ \(x=\frac{1.7-0.3(3)}{0.4}\)

⇒ \(x=\frac{1.7-0.9}{0.4}\)

⇒ \(x=\frac{0.8}{0.4}\)

x = 2

The solution is

y =3

x =2

6. y = 2x – 6

y= 0

Solution:

y = 2x-6 (1)

y= 0 (2)

Substituting y value in Equation (1)

0=2x-6

-2x = -6

⇒ \(x=\frac{6}{2} \Rightarrow x=3\)

Solution is x = 3

y = 0

7. 65x -33y = 97

33x-65y = 1

Solution: Given equations are 65x-33y=97 (1)

33x-65y = 10 (2)

on adding equation ( & we get

⇒ 98x-98y=98 x-y=1 → (3)

On Subtracting equation (2)from (1) we get

– 32x-32y=-96

– 32(x+y)=+96

(x + Y) = \(\frac{96}{32}\)

x + y = 3 → (4)

Adding equation (3) and (4)

putting x = 2 in equation (3)

2-y=1 =) -y= 1-2 =) -y=-1 =) y=1

Hence, the Solution is x=2

4 = 1

8. 217x+13ly=913

131x+217y=827

Solution: Given equations 217x+131y=9130 (1)

131x +2174=827 (2)

on adding (1) and (2) we get

348x+3484 = 1740

348(x+y)=1740

⇒ \(x+y=\frac{1740}{348}\) → (3)

On Subtracting the equation we get

-86x+86y=-86

-86(x-4)=-86

x-y=1 → (4)

Adding equation (3) and (4)

2x = 6

⇒ \(x=\frac{6}{2} \Rightarrow x=3\)

putting x = 3 in equation (3)

3 + y = 5

y = 5 – 3

y = 2

Hence the solution is x = 3

y = 2

CBSE Solutions For Class 10 Mathematics Chapter 3 Linear Equation In Two Variables

Question 2. Solve the following System of equations by using the cross-multiplication method:

1. 2x+y=5

3x+2y=8

Solution: The given equation can be written as

2x+y-5=0

3x+2y-8=0

By Gross multiplication method, we get

x y 1

1 -5 2 1

2 -8 3 1

⇒ \(\frac{x}{-8+10}=\frac{y}{-15+16}=\frac{1}{4-3}\)

⇒ \(\frac{x}{2}=\frac{y}{1}=\frac{1}{1}\)

⇒ \(\frac{x}{2}=\frac{1}{1} \text { or } x=2\)

and \(\frac{y}{1}=\frac{1}{1} \text { or } y=1\)

Hence, x=2 and y=1 is the required Solution.

2.  8x+13y-29=0

12x-74-17=0

Solution:

The given equations are

8x+13y-29=0

12x-7y-17=0

By Cross multiplication method, we get

x y 1

13 -29 8 13

-7 -17 12 -7

⇒ \(\frac{x}{-221-203}=\frac{y}{-348+136}=\frac{1}{-56-156}\)

⇒ \(\frac{x}{-424}=\frac{y}{-212}=\frac{1}{-212}\)

⇒ \(\frac{x}{-424}=\frac{1}{-212}\)

⇒ \(x=\frac{-424}{-212}\)

x = 2

and \(\frac{y}{-212}=\frac{1}{-212}\)

y = 1

Hence, x=2 and y=1 is the required Solution.

3. 2y+2x=0

4y+3x=5

Solution: The given equation Can be written as 2x +3y=0 3x+4y-5

By Cross multiplication method, we get

x y 1

3 0 2 3

4 -5 3 4

⇒ \(\frac{x}{-15-0}=\frac{y}{0+10}=\frac{1}{8-9}\)

⇒ \(\frac{x}{-15}=\frac{y}{10}=\frac{1}{-1}\)

when \(\frac{x}{-15}=\frac{1}{-1} \Rightarrow x=15\)

and\(\frac{y}{10}=\frac{1}{-1} \Rightarrow y=-10\)

Hence, x=15 and y=-10 is the required solution.

4. \(\frac{x}{6}+\frac{4}{15}=4\)

\(\frac{x}{3}-\frac{4}{12}=\frac{19}{4}\)

Solution: The given equation can be written as

⇒ \(\frac{x}{6}+\frac{y}{15}-4=0\)

⇒ \(\frac{5 x+2 y-120}{30}=0\)

5 x+2 y-120 = 0  → (1)

⇒ \(\frac{x}{3}-\frac{y}{12}-\frac{19}{4}=0\)

⇒ \(\frac{8 x-2 y-114}{24}=0\)

8x – 2y – 114 = 0→ (2)

By Cross multiplication method & we get

x         y            1

2    – 120     5       2

-2  -114      8        2

⇒ \(\frac{x}{-225-240}=\frac{y}{-960+570}=\frac{1}{-10-16}\)

⇒ \(\frac{x}{-468}=-\frac{y}{-390}=\frac{1}{-26}\)

when \(\frac{x}{-4 6 8}=\frac{1}{-26}\)

⇒ \(x=\frac{-468}{-26} \Rightarrow x=18\)

⇒ \(\text { and } \frac{y y}{-390}=\frac{1}{-26} \Rightarrow y=\frac{-390}{- 20} \Rightarrow y=15\)

Hence X-18 and y = 15 is the required solution.

5. x+y=a+b

ax-by=a2=62

Solution: The given equations Can be written as

x+y=(a-6)=0

ax-by-(a2+63)=0

By Cross multiplication method, we get

x            y               1

1      -(a-b)        1       1

-b     -(a2+b2)   a    -b

⇒ \(\frac{x}{-\left(a^2+b^2\right)-b(a-b)}=\frac{y}{-a(a-b)+\left(a^2+b^2\right)}=\frac{1}{-b-a}\)

⇒ \(\frac{x}{-a^2-b^2-a b+b^2}=\frac{y}{-a^2+a b+a^2+b^2}=\frac{1}{-b-a}\)

⇒ \(\frac{x}{-a^2-a b}=\frac{y}{a b+b^2}=\frac{1}{-b-a}\)

when \(\frac{x}{-a^2-a b}=\frac{1}{-b-a}\)

⇒ \(x=\frac{-b(a+b)}{-(a+b)}\)

x = a

and \(\frac{y}{a b+b^2}= \frac{1}{-b-a}\)

⇒ \(\frac{y}{-b(-a-b)}=\frac{1}{-b-a}\)

⇒ \(y=\frac{-b(-b-a)}{+b-a}\)

Hence x=a and y = b are the required Solution

6. ax-by=a2+b2

x+y=20

Solution: The given equations Can be written as,

ax-by-(a2=+b2)=0

x+y-2a=0

By Cross multiplication method, we get

x                y             1

-b         -(a2+b2)      a      -b

1               -2a           1        1

⇒ \(\frac{x}{2 a b+\left(a^2+b^2\right)}=\frac{y}{-\left(a^2+b^2\right)+2 a^2}=\frac{1}{a+b}\)

⇒ \(\frac{x}{2 a b+\left(a^2+b^2\right)}=\frac{y}{-\left(a^2+b^2\right)+2 a^2}=\frac{1}{a+b}\)

⇒ \(\frac{x}{(a+b)^2}=\frac{y}{a^2-b^2}=\frac{1}{a+b}\)

⇒ \(\frac{x}{(a+b)^2}=\frac{1}{a+b}\)

⇒ \(x=\frac{(a+b)^2}{(a+b)}\)

x = a+b

and \(\frac{y^2}{a^2-b^2}=\frac{1}{a+b}\)

⇒ \(y=\frac{(a+b)(a-b)}{(a+b)}\)

y = a-b

Hence x= a+b an is the required solution.

7. ax+by = c

bx+ay=1+c

Solution: The given equations can be written as

ax+by = c

bx+ay= (C+1)=0

B Cross multiplication method, we get

x                  y                  1

b                -c                 a             b

a             -(c+1)            b               a

⇒ \(\Rightarrow \frac{x}{-b(c+1)+a c}=\frac{y}{-b c+a(c+1)}=\frac{1}{a^2-b^2}\)

⇒ \(\frac{x}{-b c-b+a c}=\frac{y}{-b c+a c+a}=\frac{1}{a^2-b^2}\)

⇒ \(x=\frac{ac-b-b c}{a^2-b^2}\)

and \(\frac{y}{-b c+a c+a}=\frac{1}{a^2-b^2}\)

⇒ \(y=\frac{-(b c-a-a c)}{-\left(-b^2-a^2\right)} \Rightarrow y=\frac{b c-a-a c}{b^2-a^2}\)

Hence x \(=\frac{a c-b-b c}{a^2-b^2} \text { and } y=\frac{b c-a-a c}{b^2-a^2}\).

8.  (a-b)x + (a+b)y = a2- 2ab-b2

(a+b) (x+y)= a2+b2

Solution:

the given equation can be written as

(a-b)x + (a+b)y – (a2+2a+b2) = 0

(a-b)x + (a+b)y – (a+b)2 = 0 (1)

(a+b)(x+y) = a2+b2

ax+ay+bx+by – (a2-b2) = 0

(a+b)x + (a+b)y – (a-b)2 = 0 (2)

By cross-multiplication methods (1)and(2)

x              y                 1

(a+b)      -(a+b)2      (a-b)     (a+b)

(a+b)      -(a2-b2)     (a+b)    (a+b)

⇒ \(\frac{x}{-(a+b)\left(a^2-b^2\right)+(a+b)^3}=\frac{y}{-(a+b)^3+\left(a^2-b^2\right)(a-b)}=\frac{1}{(a-b)(a+b)-(a+b)^2}\)

⇒ \(\frac{x}{-a^3+a b^2-a^2 b+b^3+a^3+b^3+3 a^2 b+3 a b^2}\)

⇒ \(=\frac{y}{-p^2-p^2-3 a^2 b-3 a b^2+a^3-a^2 b-a b^2+b^3}\)

⇒ \(=\frac{1}{a\not^2-b^2-\not a^2-b^2-2 a b}\)

⇒ \(\frac{x}{2 b^3+4 a b^2+2 a^2 b}=\frac{y}{-4 a^2 b-4 a b^2}=\frac{1}{-2 b^2-2 a b}\)

⇒ \(\frac{x}{2\left(b^3+2 a b^2+a^2 b\right)}=\frac{y}{-4\left(a^2 b-a b^2\right)}=\frac{1}{-2 b(b+a)}\)

when \(\frac{x}{2\left(b^3+2 a b^2+1 a^2 b\right)}=\frac{1}{-2 b(b+a)}\)

⇒ \(x=\frac{2\not\left(b^3+2 a b^2+a^2 b\right)}{-2 b(b+a)}\)

⇒ \(x=\frac{\not b\left(b^2+2 a b+a^2\right)}{\not b(b+a)}\)

⇒ \(x=\frac{(a+b)^2}{(a+b)}\)

x = a+b

and\(\frac{y}{-4\left(a^2 b-a b^2\right)}=\frac{1}{-2 b(b+a)}\)

⇒ \(y=\frac{-4 b\left(d^2-a b\right)}{-2 b(b+a)}\)

⇒ \(y=\frac{2 a^2-2 a b}{a+b} \times \frac{1}{2 a^2}\)

⇒ \(y=\frac{-2 a b}{a+b}\)

Hence x=a+b and \(y=\frac{-2 a b}{a+b}\) is the required solution.

Question 3. Each of the following System of equations determines whether the System has a unique solution, no Solution, or infinitely many Solutions. In Case there is a unique solution, find it:

1. 2x-3y=17

4x+9=13

Solution: Given equations are

2x-3y=170 (1)

4x+y=13 (2)

Here a1 =2, b1 =-3, C1 = 17

a2 =4, b2 =1 and C2 =13

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2} \text { and } \frac{b_1}{b_2}=\frac{-3}{1}\)

Since\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} .\) Hence, the given system has a unique solution.

By Cross multiplication method, we have

x            y         1

3      -17         2         -3

1      -13        4           1

⇒ \(\frac{x}{39+17}=\frac{y}{-68+26}=\frac{1}{2+12}\)

⇒ \(\frac{x}{56}=\frac{y}{-42}=\frac{1}{14}\)

When \(\frac{x}{56}=\frac{1}{14}\)

⇒ \(x=\frac{56}{14} \Rightarrow x=4\)

And \(\frac{y}{-42}=\frac{1}{14}\)

⇒ \(y=\frac{-42}{14} \Rightarrow y=-3\)

Hence x=4 and y=-3 is the required solution.

2. 5x+2y=16

3x+ \(\frac{6}{5}\) y = 2

Solution: Given equation are 5x+2y=16

3x+\(\frac{6}{5}\)

Here a1=5, b1=2, c1=16

a2=3, b2 = \(\frac{6}{5}\), c2=2

Now,

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

⇒ \(\frac{5}{3}=\frac{2}{6 / 5} \neq \frac{16}{2}\)

⇒ \(\frac{5}{3}=\frac{10}{6} \neq 8\)

⇒ \(\frac{5}{3}=\frac{5}{3} \neq 8\)

Since,\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2} .\) hence the given system has no solution.

3. 3x+4=2
6x+2y=4

Solution: Given equations are 3x+y=2

6x+2y=4

Here a1 = 3, b1 = 1, c1=2

a2=6, b2=2, c2=4

Now,

⇒ \(\frac{a_1}{a_2}=\frac{3}{6} \text { and } \frac{b_1}{b_2}=\frac{1}{2} \text { and } \frac{c_1}{c_2}=\frac{2}{4}\)

Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} .\) Hence the given system has an infinite solution.

4. \(\frac{x}{3}+\frac{y}{2}=3\)

x-2y = 2

Solution: Given equations are \(\frac{x}{3}+\frac{y}{2}=3\)

x-2y = 2

Here a1 = \(\frac{1}{3}\), b1 = \(=\frac{1}{2}\)
, c1=2

a2=6, b2=2, c2=4

Now

⇒ \(\frac{a_1}{a_2}=\frac{1 / 3}{1}=\frac{1}{3} \text { and } \frac{b_1}{b_2}=\frac{1 / 2}{-2}=\frac{-1}{4}\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} .\) Here the given system has unique solutions

By Cross multiplication method, we get

x           y             1

1/2        -3          1/3        1/2

-2          -2          1           -0

⇒ \(\frac{x}{-\left(\frac{1}{2}\right)(2)-6}=\frac{y}{-3+2\left(\frac{1}{3}\right)}=\frac{1}{-2\left(\frac{1}{3}\right)-1\left(\frac{1}{2}\right)}\)

⇒ \(\frac{x}{-1-6}=\frac{y}{-3+\frac{2}{3}}=\frac{1}{-\frac{2}{3}-\frac{1}{2}}\)

⇒ \(\frac{x}{-7}=\frac{y}{\frac{-9+2}{3}}=\frac{y}{\frac{-4-3}{6}}\)

⇒ \(\frac{x}{-7}=\frac{y}{-7 / 3}=\frac{1}{-7 / 6}\)

⇒ \(\frac{x}{-7}=\frac{-3 y}{7}=\frac{-6}{7}\)

⇒ \(\frac{x}{-7}=\frac{-3 y}{7}=\frac{-6}{7}\)

⇒ \(\text { when } \frac{x}{-7}=\frac{-6}{7}\)

⇒ \(x=\frac{-6(-7)}{7}\)

⇒ \(x=\frac{42}{7}\)

x = 6

and \(\frac{-3 y}{7}=\frac{-6}{7}\)

⇒ \(-3 y=\frac{-6(7)}{7}\)

7(-3)=-6(7)

-21y = -42y

⇒ \(y=\frac{42}{21}\)

y = 2

Hence x = 6 and y=2 is a required solution.

5. Kx+2y=5

3x+9=1

Solution: The given system of equations is

kx+2y=5

3x+y=1

Here, a1=k, b1 =2 and c1=5

a2=3, b2=1 and c2=1

The System has unique Solution \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{k}{3} \neq \frac{2}{1}\)

⇒ \(\Rightarrow \quad k \neq 6\)

So, k can take any real value except 6.

6. 4x+ky +1=0

2x+2y+2=0

Solution: The given System of equations is 4x+ky+8=0

2x+2y+2=0

Here a1 = 4, b1 = k and c1 =8

a2=2, b2 =2 and C2 =2

The System has unique solution if \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\begin{aligned}
& \Rightarrow \frac{2 y}{x} \neq \frac{k}{2} \\
& \Rightarrow k \neq 4
\end{aligned}\)

So k can take any real value except 4.

7. 2x-37-5=0

kx-by-8=0

Solution: The given System of equations are 2x-34-5=0

kx-6x-8=0

Here, a1 =2, b1=3 and c1 =-5

a2=k, b2=6 and C2 =-8

The System has unique solution if \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{2}{k}+\frac{-3}{-6}\)

⇒ \(\frac{2}{k}+\frac{1}{2}\)

⇒ \(\begin{aligned}
& \frac{4}{k} \neq 1 \\
&\quad k \neq 4
\end{aligned}\)

So k can take any real value except 4.

8. 8x+5y=9

kx+10y=18

Solution: The given System of equations is 8x+5y=9

Here a1 =8, b1=5 and c1 =9

a2 =k, b2 = 10 and c2 = 18

kx+10y=18

The System has infinitely many solutions if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{8}{k}=\frac{5}{10}=\frac{9}{18}\)

⇒ \(\frac{8}{k}=\frac{1}{2}=\frac{1}{2}\)

when \(\frac{8}{k}=\frac{1}{2} \Rightarrow k=16\)

Hence, for k = 16, the given System equations will have infinitely many solutions.

Question 4. The Sum of the two numbers is 85. Suppose the langer number exceeds four times the Smaller one bys. Find the numbers.
Solution: let the two numbers x and y, and x>y.

According to question x+y=85 → (1)

and X-4y=5 → (2)

Subtracting equation (2) from (1), we get

5y=80 =) y=16

Putting y=16 in equation (, we get

x+16=85

2=85-16

– x = 69

CBSE Solutions For Class 10 Mathematics Chapter 1 Real Numbers

Real Numbers Exercise – 1.1

CBSE Solutions For Class 10 Mathematics Chapter 10 Circles

Question .1 Use Euclid division algorithm, to find the H.C.F of the following:

  1. 70 and 40
  2. 180 and 45
  3. 165 and 225
  4. 155 and 1385
  5. 105 and 135
  6. 272 and 1032

Solution:

1. 70 and 40

Here 40 > 70

7o = 4O * 1 + 30

4o = 30 * 1 + 10

3o = 10 * 1 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F (40,70)

2. 180 and 45

Hare 18 > 45

Real Numbers Euclid division algorithm

45 = 18 * 2 + 9

18 = 9 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 18,45) = 9

3. 165 and 225

Here 165 > 225

Real Numbers Euclid division algorithm

225 = 165 * 1 + 90

165 = 90 * 1 + 75

90 = 75 * 1 + 15

75 = 15 * 5 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 165,225) = 15

4. 155 and 1385

Hare 155 and 1385

Real Numbers Real Numbers Euclid division algorithm

1385 = 155 * 8 + 145

155 = 145 * 1 + 0

145 = 10 * 14 + 5

10 = 5 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 155,1385) = 15

5. 105 and 135

Here 105 and 135

Real Numbers Euclid division algorithm

135 = 105 * 1 + 30

105 = 30 * 30 + 15

30 = 15 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H, C.F ( 105, 135)

6. 272 and 1032

Here 272 and 1032

Real Numbers Euclid division algorithm

1032 = 272 * 3 + 216

272 = 216 * 1 + 56

216 = 56 * 3 + 48

48 = 8 * 6 + 0

Since remainder = 0

The recent divisor is the H.C.

H, C.F ( 272, 1032)

Question: 2. The H.C.F of 408 and 1032 is expressible in the form of 1032×2-408xy, then find the Value of y.
Solution:

First, we will find the H-CF of 408 and 1032

Here 408 >1032

Real numbers The H.C.F Expressible

1032 = 408×2+216

408 = 216×1+192

216= 192X1+24

192=24×8+0

Since remainder = 0

The recent divisor is the H-C.F

H.C.F. of (408, 1032) = 24

Now, 1032×2-408xy = 24

= -408xy = 24-1032X 2

-y = \(\frac{24-2064}{408}\)

-y = \(\frac{-2040}{408}\)

y = 5

Question 3. If the H.C.F of 56 and 72 is expressible in the form of 56x+72×53, then find the Value of x.
Solution:

First, we will find the H.C.F of 56 and 72

Here 56 > 372

Real numbers The H.C.F Expressibles

72 = 56 x 146

56= 16 x 3 + 8

16 = 8 x 2 + 0

Since remainder = 0

The recent divisor is the H·C.F

H.C.F. of (56,72)=8

Now, 56x+72×53 = 8

56x = 8 – 72 x 53

56x= 8-3,816

x = \(\frac{-3,808}{56}\)

x = – 68

Question 4. Express the H.C.F of 18 and 24 in the form
solution:

Here 18>24

Real numbers Express the H.C.F

24 = 18× 1+6

18=6×2+6

6=6×1+0

Since remainder =0

The recent divisor is the H.C.F

H.C.F. (18,24)=6

Now,

6=18-6×2

6=18-(24-18X1)

= 18-24 +18 x |

18×2-24 = 18x +244

where x = 2, y=-1

Question 5. Express the H.CF of 30 and 36 in the form of 30x + 36y.
Solution:

Here 30> 36

Real numbers Express the H.C.F Of Number

30= 200 6×4+6

6= 6X1+0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F. (30,36) = 6

Now, 6=30-6×4

6=30-(36-30×1)

= 30-36+30X1

= 30X2-36

= 30x+364

where, x=2 and y=-1

Question 6. Find the largest number that divides 189 and 249 9 in each case.
Solution: We have to find a number, which divides the other numbers

Means H.C.F.

It is then that the required Number, when divided between 189 and 249 leaves the remainder 9; 9 is extra in each number. It means that if these numbers are 6 less, then there is no remainder in each case.

89-9=180 and 249-9=240 are completely divisible by the required number.

Real Numbers The Largest Number That Divides

H.C.F. (180,240) = 60

Hence, the required number =  60.

Question 7. Find the largest number that divides 280 and 1248 u and 6 respectively. leaving the remainder solution.
Solution: We have to find a number, which divides the other numbers

means→H.C.F.

It is given that the required number, when divided between 280 and 1248, leaves the remaining 4 and 6 respectively. It means that if 280 is 4 less than, 1248 is 6 less, then on division, gives no remainder.

280-4=276 and 1248-6=1242 are Completely divisible to the required number.

First, we will find the H.C.F of 276 and 1242

Real Numbers Completely Divisible

H.C.F. (2.76, 1242) = 138

Hence, the required number =138.

Question 8. Find the greatest number that divides 699, 572, and 442 leaving remainders 6, 5, and 1 respectively.
Solution: We have to find a number, which divides the other numbers means→H.C·F.

It is given that the required numbers when divided into 699, 572, and 442, leave the remaining 6,5 and I respectively. It means that if 699 is 6 less, 572 is s less, and 442 is I less, then on division, gives no remainder.

699-6=693, 572-5=567 and 442-1=441 are and 442-1=441 are completely by the required number .

First, we will find the H.C.F of 693 and 567.

Real Numbers Completely By The Required Number

693 = 567×1 +126

567 = 126×4 +63

126 = 63X2  + o

Now, we filled the H.C.F of 63 and 441.

Real Numbers Completely By The Required Numbers

441=63×7 +0

H.C.F. (63, 441) = 63

Required Number = 63

Question 9. A sweet Seller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray what is the number of Sweets that Can be Placed In each stack for this purpose? Also, find the number of stacks.
Solution: Maximum number of burfis in each stack = H.C.F of 420 and 130

420 = 2X2 X3 X5 X 7

130 = 2X5X 13

H·CF = 2×5 = 10

Maximum number of burfis in each stack = 10

Also, number of stacks = \(=\frac{420}{10}+\frac{130}{10}=42+13=55 .\)

Question 10. Three sets of English, Hindi, and Mathematics books have to be stacked In such a way that all the books are stored topicwise and the height of each stack is the Same. The number of English books is 96, the number of Hindi books is 240 and the number of mathematics books is 336. Assuming that the books are of the Same thickness, determine the number of Stacks of English, Hindi, and Mathematics books and hence the total number of Stacks.
Solution: Maximum number of books in each stack H.C.F. of

96,240 and 336

96= 2x2x2x 2 x 2 x 3

240 = 2x2x2x2x3x5

336 = 2x2x2 × 2 × 3 × 7

H·C.F = 2x2x2x2x3

Maximum number of books in each stack = 48
Also, number of stacks = 96

Also number Stacks = \(\frac{96}{48}+\frac{240}{48}+\frac{336}{48}\)

Real Numbers Exercise 1. 2

Question 1. Express each of the following as a product of prime factors:

  1. 96
  2. 48
  3. 150
  4. 3072

Solution:

Real Numbers Product Of Prime Factors

96 = 2 * 2 * 2 * 2 * 2 * 3

96 = 25 * 3

Real Numbers Product Of Prime Factors

84 = 2 * 2 * 3 * 7

84 = 22 * 3 * 7

Real Numbers Product Of Prime Factors

150=2 x 3 x 5 x 5

150 = 2 x 3 x 52

Real Numbers Product Of Prime Factors

3072 = 2× 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

3072= 210 X 3

Question 2. Find the H.C.F and L.C.M of the following pairs using the prime factorization method:

  1. 12 and 25
  2. 20 and 25
  3. 96 and 404
  4. 336 and 56

Solution:

1. 12 and 25

Real Numbers The Prime Factorization Method

Now

Real Numbers The Prime Factorization Methods

H.C.F = 3

and L.C.M= 2 x 2 x 3 x 5

= 60

2. 20 and 25

Real Numbers The Prime Factorization Methods

Now

Real Numbers The Prime Factorization Methods

H.C.F = 5

and L.C.M = 2 x 2 x 5 x 5

= 100

3. 96 and 404

Real Numbers The Prime Factorization Methods

Now

Real Numbers The Prime Factorization Methods

H.C.F = 2 x 2= 4

and L.C.M = 2 x 2 x 2 x 2 x 2 x 3 x 101

= 9696

4. 336 and 56

Real Numbers The Prime Factorization Methods

Now

Real Numbers The Prime Factorization Methods

H.C.F = 2 x 2 x 2 x 7 = 56

and L.c.M = 2 x 2 x 2 x 2 x 3 x 7 = 336

Question 3. Using the prime factorization Method, find the HC.F and L.C.M of the following Pairs. Hence Verify H-C.F. XL.C.M= Product of two numbers.

  1. 96 and 120
  2. 16 and 20
  3. 396 and 1080
  4. 144 and 192

Solution:

1. 96 and 120

Real Numbers The Prime Factorization

96=2x2x2 x 2 x 2 x 3

120 = 2x 2 x 2 × 3 × 5

Now, H.C.F. = 2×2×2×3=24

and LCM2 = 2 X 2 X 2 X 2 X 2 X 3 X 5 = 480

Now H.C.F X L.C. M = 12×144=24×480 = 11,520

and Roduct of two numbers = 96X120 = 11,520

Hence, H-C.F X L.C.M = Product of two numbers

2. 16 and 20

Real Numbers The Prime Factorization

16 = 2 X 2 X 2 x 2

L.C.M = 2 * 2 * 2 * 5 = 80

Now H.C.F x L .C.F = 4 * 80 = 320

and product of two numbers = 16 * 20

Hence H.C.F= product of two numbers

3. 396 and 1080

Real Numbers The Prime Factorization

396 = 2×2 × 3 × 3 × 11

1080 = 2x2x2 × 3 × 3 × 3 × 5

Now, H.C.F. = 2 x2 x3x3 =36

and L.C.M. = 2X2 X2 X3 X3X3 X5x11

=11880

Now H.C.FXL.CM = 36X11880 = 4,27,680 and Product of two numbers = 4,27,680

Hence, H.C.F. XL.C.M = Product of two numbers

4. 144 and 192

Real Numbers The Prime Factorization

144 = 2x2x 2×2×3×3

192 = 2x2x2 × 2 × 2 × 2 × 3

Now, H.C.F=2x2x2x2 x3 = 48

and L.C.M= 2×2 X2 X2 X2 X2 X3 X3

= 576

Now H.C.FXL.CM = 48X576 = 27,648 and Product of two numbers = 144 X 192

= 27,648

Hence, H. C. F. XL.C.M.= Roduct of two numbers

Question 4. The H.C.F and LCM of the two numbers are 145 and 2175 respectively. If the first number is 435, find the second number.
Solution:

Here, H.GF = 145

L.C.M = 2175

Now, First no. x Second no. = H.C.F. X L.C.M.

Second no = \(\frac{\text { H.C.F. XLC.M }}{\text { Firstno. }}\)

Second no = \(\frac{145 \times 2175}{435}\)

Second no = \(\frac{315375}{435}\)

Question 5. check whether 18 n can end with the digit o for the natural number n.
Solution:

Real Numbers The Prime Factorization

18= 2 x 3 x 3 = 22 x 32

18n = (2×32) n = 2 n x32 n

It has no term containing

No Value of MEN for which 18 n ends with digit 0.

Question 6. On a morning walk, three persons step off together and their steps are 40cm, 42cm, and us cm respectively, what is the difference minimum distance each should walk so that each Can Cover the Same distance In Complete Steps?
Solution: We have to find a number (distance) that is divided by each number Completely, which means →→L.CM

we have to find the L.C.M. of 400m, 42cm, and our cm to get the required distance.

Real Numbers On a morning walk

Now, L. C.M = 2 x 2 x 2 x 5×3×7 × 3 = 2520

Minimum distance each should walk = 2520 Cm

Question 7. Write the missing numbers in the following factor tree:

Real Numbers Factor Tree

Solution:

Real Numbers Factor Trees

  1. The upper box, on 7 and 13 is filled by the Product of 7 and 13, 9.
  2. The upper next box, on Sand 91 is filled by the product of 5 and 91, 1.e., 455
  3. The upper next box, on 3 and USS is filled by the product of 3 and USS, i.e., 1365
  4. The topmost box, on 3 and 1365 will be filled by the product of 3 and 1365 i.e, 4095

Question 8. State whether the given statements are true or false:
Solution:

  1. The Sum of two rationals is always rational. (True)
  2. The Sum of two irrationals is always irrational. (False)
  3. The product of two votionals is always rational. (True)
  4. The product of two irrationals is always irrational, (False)
  5. The Sum of a rational and an irrational is always rational. (False)

The product of a rational and an irrational is always rational. (True)

CBSE Solutions For Class 10 Mathematics Chapter 2 Polynomials

CBSE Solutions For Class 10 Mathematics Chapter 2

Question 1. x2+9x+20
Solution:

Let P(x) = = x2+9x+20

= x2+5x+4x+20

= x(x+5)+4(x+5)

(x+4)(x+5)

P(x)=0

(x+4)(x+5)=0

x+4=0 or x+5=0

x=-4 x=-5

Zeros of P(x) are -4 and -5 \(=\frac{-9}{1}=\frac{- \text { Coefficient of } x}{\text { coefficient of } x^2}\)

Now, Sum of Zeros = -4+(-5) \(\frac{20}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

Question 2. x2 – 9
Solution:

let P(x) = x2 = 9 = x2=-32 = (x-3)(x+3)

P(x)=0

(x-3)(x+3)=0

x-3=0 Or x+3=0

x=3 x=-3

Zeros of p(x) are 3 and -3

Now, Sum of Zeros = 3+(-3)=0 = \(\frac{-0}{1}=-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)

and Product of Zeros = (3)(-3) = −9 = \(\frac{-9}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

Question 3. Find the quadratic polynomials, the Sum of whose Zeroes is 17 and the product is 60. Hence, find the Zeroes of the polynomial.
Solution:

Let and B be the Zeroes of the polynomial P(X).

Given that X+B=17 and αβ=60

Now, P(x) = x2= (α+β)γ + αβ

= x2= 17x+60

= x2=12x-5x+60

= x(x-12)-5(x-12)

= (x-5)(x-12)

There may be So many different polynomials that satisfy the given Condition. The general equation quadratic polynomial will be k(x2=-17x+60), where k = 0

P(x)=0

(x-5) (x-12) = 0

(x-5)=0 or (x-2)=0

2=5 Or x=12

Zeros are 12 and 5.

Question 4. Find a quadratic polynomial, the Sum of whose Zeros is 7 and the product is -60. Hence, verify the relation between Zeros and Coefficients of the polynomial.
Solution:

Let and B be the Zeros of the polynomial P(x).

Given that α+β=ϒ and αβ = -60

Now, P(x) = x2 – (α+β) γ+ αβ

= x2-7x-60

= x2-12x+5x-60

= x(x-2)+5(x-12)

= (x-12)+(x-5)

There may be so many different polynomials which satisfy the given Condition. The general quadratic polynomial will be k (x2-7x-60), where k = 0.

P(x)=0

(x-12) (x+5)=0

(x-12)=0

x=12 or (x+5)=0

x=-5

Zeros are 12 and -5.

Question 5. If the product of Zeroes of the polynomial 30+ 5x+k is 6, find the value of k.
Solution:

Given polynomial = 31751+k

Product of Zeroes = \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

6= \(\frac{k}{3}\)

= 6×3=k

⇒ k = 18

CBSE Solutions For Class 10 Mathematics Chapter 2 Polynomials

Question 6. If the Sum of Zeroes of the polynomial x2+2x-12 is 1, find the value of k.
Solution:

Given polynomial = x2+2kx-12

Sum of Zeroes = \(-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)

1 = \(\frac{-2 k}{1}\)

1 = -2K

⇒ \(k=\frac{-1}{2}\)

Question 7. If x = \(\frac{5}{3}\) and x = \(\frac{-1}{2}\) are the Zeroes of the polynomial ax=7x+b, then find the values of a and b,
Solution: Let P(x) = ax27x+b

⇒ \(x=\frac{5}{3} \text { and } x=\frac{-1}{2}\) are zeroes of p(x)

⇒ \(P\left(\frac{5}{3}\right)=0\)

⇒ \(a\left(\frac{5}{3}\right)^2-7\left(\frac{5}{3}\right)+b\)

⇒ \(\frac{25 a}{9}-\frac{35}{3}+b\)

⇒ \(b=\frac{-25 a}{9}+\frac{35}{3}\)

⇒ \(P\left(\frac{-1}{2}\right)=a\left(\frac{-1}{2}\right)^2-7\left(\frac{-1}{2}\right)+b\)

⇒ \(\frac{a}{4}+\frac{7}{2}+b\)

⇒ \(\frac{a}{4}+\frac{7}{2}-\frac{25 a}{9}+\frac{35}{3}=0\)

⇒ \(\frac{a}{4}-\frac{25 a}{9}=-\frac{7}{2}-\frac{35}{3}\)

⇒ \(\frac{9 a-100 a}{36}=\frac{-21-70}{6}\)

⇒ \(-\frac{91 a}{36}=\frac{-91}{6}\)

⇒ \(\frac{91 a}{6}=91\)

91a = 546

⇒ \(a=\frac{546}{91}\)

a = 6

⇒ \(b=\frac{-25(6)}{9}+\frac{35}{3}\)

⇒ \(b=\frac{-25(6)}{9}+\frac{35}{3}\)

⇒ \(b=\frac{-150+105}{9}\)

⇒ \(b=\frac{-45}{9}\)

b = -5

Question 8. Verify that 1,-2, 4 are Zeros of the Cubic polynomial x3-3x2-6x+8. Also, Verify the relation between Zeroes and Coefficients of the polynomial.
Solution:

Here, P(x) = x3-3x2-6x+8

P(1) = (1)2 = 3(1)2 – 6(1) +8 = 1-3-6+ 8 = −9+9=0

P(-2) = (2) 3 -3(-2)=6(-2)+8=-8-12+12+8 = 0

P(4) = (4)3-3(4) -6(4) +8 = 64-48-24+8 = 0

1-2 and 4 are Zeroes of P(x).

Now, α+B+= 1-2+4=3 = \(\frac{-3}{1}=-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)

XB+B++√α = (1)(-2)+(-2) (4)+(4)(1) = −2-8+4

= -10+4

⇒\(\frac{-6}{1}=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^3}\)

and LB7 = (1)(-2) (4) =\(-\frac{8}{1}=\frac{8}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^3}\)

Question 9. Verify that 2-4. and are zeroes of the Cubic polynomial 3x 3 +5x2 = 26x+8. Also, verify the relation between Zeroes and Coefficients of the polynomial.
Solution:

Here, P(x)=3x3 +5 x 2 – 26x+8

P(2) = 3(2)3 +5(2)2 = 26(2)+8=24+20-52+8 = 52-52 = 0

P(-4)=3(-4)3 + 5(-4)=26(-4)+8=-(92+80+104+8=-192+192=0

⇒ \(P\left(\frac{1}{3}\right)=3\left(\frac{1}{3}\right)^3+5\left(\frac{1}{3}\right)^2-26\left(\frac{1}{3}\right)+8=\frac{3}{27}+\frac{5}{9}-\frac{26}{3}+8=\frac{3+15-234+216}{27}\)

⇒ \(=\frac{234-234}{27}=0\)

2,-4. and — are Zeroes of P(x),

Now \(\text { , } \alpha+\beta+\gamma=2-4+\frac{1}{3}=-2+\frac{1}{3}=\frac{-6+1}{3}=-\frac{5}{3}=\frac{5}{3}=-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)

⇒ \(\alpha \beta+\beta \gamma+\gamma \alpha=(2)(-4)+(-4)\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)(2)=-8-\frac{4}{3}+\frac{2}{3}=\frac{-28+2}{3}=\frac{-26}{3}\)

⇒ \(=\frac{- \text { Coefficient of } x^2}{\text { coefficient of } x^3}\)

and \(\alpha_\beta \beta=(2)(-4)\left(\frac{1}{3}\right)=-\frac{-8}{3}=\frac{8}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^3}\)

Question 10. Find a Cubic polynomial whose Zeroes are 5, 6, and -4.
solution:

Let α= 5, β=6 and γ=-4

α+β+ γ = 5+6-4 =) ||- 4 = 7

αB+ βγ + γα = 5(6)+6(-4)+(-4) (5)

= 30-24-20

= 30-44

= -14

XB = 5(6)(u)

= -120

Cubic Polynomial = x3-(x+B++) x2 + (αB+ βγ + γα) x-px

= x2 – (7) x2 + (-14)x-(-120)

= x3-7x2 = 14x+120

Question 11. Find a Cubic polynomial whose Zeroes are 11 and -1.
Solution:

let α= \(\frac{1}{2},\),B = 1 and 2 =-1

⇒ \(\alpha+\beta+\gamma=\frac{1}{2}+1-1=\frac{1}{2}\)

⇒ \(\alpha_\beta+\beta^1+\gamma \alpha=\frac{1}{2}(1)+(1)(-1)+(-1)\left(\frac{1}{2}\right) \Rightarrow \frac{1}{2}-1-\frac{1}{2} \Rightarrow-1\)

⇒ \(\alpha \beta \gamma=\frac{1}{2}(1)(-1) \Rightarrow \frac{-1}{2}\)

Cubic polynomial = x3= (α+β+γ) x2 + (αβ+ βγ + γα)x-αßγ

⇒ \(x^3-\frac{1}{2} x^2+(-1) x-\left(-\frac{1}{3}\right)\)

⇒ \(x^3-\frac{1}{2} x^2-x+\frac{1}{2}\)

⇒ \(2 x^3-x^2-2 x+1\)

Question 12. Find the quotient and remainder in each of the following and verify the division algorithm:

1. P(x) = x2=14x2+2x-1 is divided by g(x)=x+2
Solution:

Polynomial verify The Division Algorithm

Now, quotient = x2=6x+14, remainder = -29

dividend = x3-4x2+2x-1 and divisor = x+ 2

and quotient x divisor + remainder = (x2-6x+(4)(x+2)-29

= x2-6x2+14x+2x2=1271+28-29

= x2-4x2 +2x-1

= dividend

2. P(x) = x2 + 2x2= x + 1 is divided by g(x) = x2+1
solution:

Polynomial Quotient Divisor

Now, quotient = x2+1, remainder = -x, dividend = x2 + 2x = x +1 and

divisor = x+1

and quotient divisor + remainder = (x2 + 1)(x2 + 1) – α = ) x + x2 + x2 + 1-x

x 4 + 2x = x + 1 =) dividend

Question 13. Actual division shows that x+2 is a factor of x3 + 4x2+3x-2.
Solution:

Polynomial Actual division

Question 14. If I am a zero of the polynomial x2-4x2=7x+10, find its other two Zeroes
solution:

let P(x)=x3=_472-70+10

x= 1 is a zero of P(x)

(x-1) is a factor of P(x)

Polynomial Other Two Zeroes

P(x) = x3- 4x2-7x+ 10 =) (x-1) (x=3x-10)

(x-1) (x2+2x-5x-10)

(x-1) ((x(x+2)-5(x+2))

(x-1)(x+2)(x-5)

Now, P(x)=0

⇒ (x-1)(x+2)(x-5)=0

⇒ α-1=0 or x+2=0 or x-5=0

x=1 or x=-2 or x=5

Hence, other Zeroes are -2 and 5

Question 15. If Land -2 are two Zeroes of the polynomial x4+x3=11x=9x+18, find the other two Zeroes.
Solution:

Let P(x) = x2 + x2 – 1172=9x+18

x=1, x=-2 is a zero of p(x)

(x-1) (x+2) is a factor of P(x)

x2+2x-x-2 =) x2+x-2 is a factor of P(x)

Polynomial Zeroes of the polynomial

p(x0 = x4 = x3 – 11×2 – 9x + 18 = (x-1)(x+2)(x2-9)

= (x-1)(x+2)(x2-32)

= (x-1)(x+2)(x+3)(x-3)

Now P(x)

(x-1)(x+2)(x+3)(x-3)=0

= (x-1)(x+2)(x+3)(x-3)

⇒ x-1=0 x+2=0

⇒ x+3=0 X-3=0

⇒ x=1 x=-2 x=-3 X=3

Hence, other Zeroes are 3 and -3.

Question 16. Find all Zeroes of x + x3-23x=-3x+60, if it is given that two of its Zeroes are √3 and -√3.
Solution:

Let P(x) = x2+x3-28x=-3x+60

√3 and -√3 are Zeroes of P(x).

(x−√3)(x+√3) = x2= 3 is a factor of P(x).

Polynomial Zeroes

P(x) = x2+x3-x2=3x+60 = (x2=3)(x2+11-20)

= (x2-3) [x2+5x-4x-20]

=(x2-3)(x(x+5)-4(x+5))

= (x2-3)(x-4)(x+5)

The other Zeroes are given by

x-4=0 or x+5=0

⇒ x=4 Οr x=-5

Hence, other Zeroes are 4 and -5.

Question 17. Find Zeroes of the polynomial f(x) = x2 – 13x2 + 32x-60, if it is given that the Product of its two Zeroes is 10.

Solution: Let α, B, be Zeroes of the given polynomial p(x), Such that &p=10-)(1)

⇒ \(\alpha+\beta+\gamma=\frac{-(-13)}{1}=13 \longrightarrow(2)\)

⇒ \(\alpha \beta+\beta 1+\alpha \gamma=\frac{32}{1}=32 \longrightarrow(3)\)

⇒ \(\alpha \beta \gamma=-\frac{(-60)}{1}=60 \longrightarrow(4)\)

From (1) and (4)

10s=60

⇒ \(\gamma=\frac{60}{10} \Rightarrow \gamma=6 \rightarrow(5)\)

Put 7=6 in (2), we get α+3 +6 = 13

α+B=7

Now, (α-B)2 = (x+3)=4xß

= (7)=4(10)

= 49-40

= 9

α-B = ± 3 —–(6)

Solving (5) and (6), we get

α=2,B=5 or α=2, B=5 and 1=6.

So, Zeroes are 2,5 and 6.

Question 18. What must be added to P(x) = 4x2 – 5x = 39x = 46x=-2, so that the resulting Polynomial is divisible by g(x)=4x2+7x+2?
Solution:

P(x) = 4×4 -5x3-39x2– 46x-2

9(x)= 4x2+7x+2

Polynomial Polynomial

Biodiversity And Conservation Class 12 Important Questions And Answers Biology Chapter 13

Biodiversity And Conservation Important Questions And Answers

Question 1. Name the three important components of biodiversity.
Answer:

CBSE Class 12 BioIogy Chapter 13 Biodiversity And Conservation Components Of Biodiversity

Question 2. How do ecologists estimate the total number of species present in the world?
Answer:

There are two methods to estimate the number of species in the world :

  1. By estimating the rate of discovery of new species. ,
  2. By statistical comparison of the temperate-tropical species richness of an exhaustively studied group of insects and extrapolate this ratio to other groups of animals and plants to come up with a gross estimate of the number of species on earth.

Question 3. Give three hypotheses for explaining why tropics show the greatest levels of species richness.
Answer:

The three hypotheses to explain species richness in the tropics are:

  1. The constant environment in the tropics promotes niche specialization and increased species diversity.
  2. There is longer exposure to solar radiation in tropical regions that contributes directly to higher productivity and indirectly to greater species diversity.
  3. There occurred no glaciation in the tropical region and it remained undisturbed. Thus organisms living in the tropics continued to flourish and evolved more species diversity.

Question 4. What is the significance of the slope of regression in a species-area relationship?
Answer:

  • The slope of regression in a species-area relationship indicates that species richness decreases with the decrease in area.
  • The regression coefficient (Z) is 0,1 – 0.2 regardless of the taxonomic group or the region example plants in Britain or birds in California.
  • However, when very large areas like the entire continent are analyzed, it was found that the slope of the line is much steeper with Z values in the range of 0.6 to 1.2.

Question 5. What are the major causes of species losses in a geographical region?
Answer:

There are four major causes (The Evil Quartet):

CBSE Class 12 BioIogy Chapter 13 Biodiversity And Conservation Geographical Region

Question 6. How is biodiversity important for ecosystem functioning?
Answer:

Read and Learn More Class 12 Biology Chapter Wise

Importance of biodiversity for ecosystem functioning –

  1. Stability: Biodiversity is an important aspect of the stability of an ecosystem. Ecologists believe that communities with more species tend to be more stable than those with less species.
  2. Productivity: Ecosystems with higher biodiversity show more productivity than ecosystems with lower biodiversity. David Tilman’s long-term ecosystem experiments using outdoor plots provide confirmation.
  3. Ecosystem health: Rich biodiversity is not only essential for ecosystem health but also imperative for the survival of the human race on Earth. Species are interlinked and so, killing or disappearance of one would affect the others also.
  4. Resilience: Increased biodiversity provides resilience of the ecosystem against natural or man-made disturbances.

Question 7. What are sacred groves? What is their role in conservation?
Answer:

  • Sacred groves are forest patches for worship in several parts of India All the trees and wildlife in them are venerated and given total protection.
  • They are found in Khasi and Jaintia Hills in Meghalaya, Western Ghat regions of Karnataka Maharashtra, etc.
  • Tribes do not allow anyone to cut even a single branch of a tree in these sacred groves thus sacred groves have been free from all types of exploitations.

Question 8. Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Answer:

  1. Control of soil erosion: Plant roots hold the soil particles tightly and do not allow the topsoil to drift away by winds or moving water. Plants increase die porosity and fertility of die soil.
  2. Control of floods: It is earned out by retaining water and preventing runoff of rainwater. Litter and humus of plants function as sponges thus retaining the water which percolates down and gets stored as underground water. Hence, the flood is controlled.

CBSE Class 12 BioIogy Chapter 13 Biodiversity And Conservation Important Questions And Answers

Question 9. The species diversity of plants (22%) is much less than that of animals (72%). What could be the explanations for how animals achieved greater diversification? [IMP.]
Answer:

Animals have achieved greater diversification than plants due to the following reasons:

  1. They are mobile and thus can move away from their predators or unfavorable environments. On the other hand, plants are fixed and have fewer adaptations to obtain the optimum amount of raw materials and sunlight therefore they show lesser diversity.
  2. Animals have a well-developed nervous system to receive stimuli against external factors and thus can respond to them. On the other hand, plants do not exhibit any such mechanism thus they show lesser diversity than animals.

Question 10. Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:

  • Species that are harmful to human beings can be made extinct example  HIV, poliovirus, etc.
  • Such microorganisms are not part of any food chain and tints, their extinction would not affect die ecosystem.

Ecosystem Class 12 Important Questions and Answers Biology Chapter 12

Ecosystem Question And Answers

Question 1. Difference Between the Grazing food chain and the Detritus food chain
Answer:

Ecosystem Difference Between Grazing Food Chain And Detritus Food Chain

Question 2. Difference Between Litter And Detritus
Answer:

Ecosystem Difference Between Litter And Detritus

Question 3. Difference Between  Food chain and Food web
Answer:

Ecosystem Difference Between Food Chain And Food Wevb

Question 4. Difference Between  Production And Decomposition
Answer:

Ecosystem Difference Between Production And Decomposition

Question 5. Difference Between  Upright and Inverted Pyramid
Answer:

Ecosystem Difference Between Upright pyramid And Inverted pyrmaid

Question 6. Difference Between Primary Productivity And Secondary Productivity
Answer:

Ecosystem Difference Between Primary Productivity And Secondary Productivity

Question 7. Describe the components of an ecosystem.
Answer:

An ecosystem consists of two types of components, ie., biotic or living and abiotic or non-living. There are three main types of biotic components based on the mode of obtaining their food- producers, consumers, and decomposers,

CBSE Class 12 BioIogy Chapter 12 Ecosystem Components Of An Ecosystem

  1. Producers (autotrophs): are photosynthetic or autotrophic plants that synthesize their organic food from inorganic raw materials with the help of solar radiation. Common producers are algae, plants, and photosynthetic bacteria. Phytoplanktons are the producers of aquatic ecosystems.
  2. Consumers (heterotrophs): They are animals that feed on other organisms or producers to obtain their nourishment. Common consumers are deer, goats, etc.
  3. Decomposers: They are saprotrophs that obtain nourishment from organic remains. They release digestive enzymes to digest the organic matter. Common decomposers are detritivores,
    1. For Example:  Earthworm.

Abiotic components of the ecosystem consist of non-living substances and factors that are as follows:

  1. Temperature
  2. Light
  3. Wind
  4. Humidity
  5. Precipitation
  6. Water

CBSE Class 12 BioIogy Chapter 12 Ecosystem Important Questions And Answers

Question 8. Define ecological pyramids and describe with examples, pyramids of number and biomass.
Answer:

The relationship between producers and consumers at different trophic levels in an ecosystem can be graphically represented in the form of an ecological pyramid.

Structure: The base always represents tire producers or the first trophic level and the apex represents top-level consumers or the last.

Ecological pyramids are of three types;

  1. Pyramid of number
  2. Pyramid of biomass
  3. Pyramid of Energy

Pyramid of numbers: The relationship between producers and consumers in an ecosystem can be represented in the form of a pyramid in terms of several organisms at different trophic levels called the pyramid of numbers.

Note: It is inverted when you count several insects feeding on a big tree.

Ecosystem Pyramid Of Numbers In A grassland Ecosystem

Pyramid of biomass: The relationship between producers and consumers in an ecosystem can be represented in the form of a pyramid in terms of biomass called a pyramid of biomass. It can be

  1. Upright, Example. in case of grassland ecosystem;
  2. Inverted, Example. in case of pond ecosystem as the biomass of fishes exceeds that of phytoplankton

Ecosystem Pyramid Of Biomass Shows A SharpDecrease In Biomass At Higher Trophic Level

Ecosystem Inverted Pyramids Of Biomass Of Pond ecosystem

Question 9. What is primary productivity? Give a brief description of factors that affect primary productivity.
Answer:

Primary productivity is the rate of synthesis of biomass by producers, per unit time, per unit area through the process of photosynthesis.

The factors affecting primary productivity are the following

  • Plant species inhabiting a particular area.
  • Environmental factors:
  1. Sunlight: The sunlight directly regulates the primary productivity because the plants perform photosynthesis with the help of sunlight As the tropical region receives maximum sunlight it exhibits higher productivity.
  2. Temperature: Temperature regulates the activity of enzymes. So, the optimum temperature is required for the proper functioning of the enzyme.
  3. Moisture: Rain (humidity) is required for higher primary productivity. Deserts have the lowest primary productivity as dead soil is deficient in moisture.

Question 10. Define decomposition and describe the processes and products of decomposition.
Answer:

The process of breaking down complex organic matter into inorganic substances like CO, water, and nutrients is called decomposition. The raw materials for decomposition including dead plant remains like leaves, bark, flowers, and animal remains and their fecal matter are called detritus.

Steps of Decomposition

  1. Fragmentation: The process of breaking down detritus into smaller particles is called fragmentation, for example., as done by Earthworm (fanner’s friend).
  2. Leaching: The process by which water-soluble inorganic nutrients go down into the dead soil horizon and get precipitated as unavailable salts is called leaching.
  3. Catabolism: The enzymatic process by which bacterial and fungal enzymes degrade detritus to simpler inorganic substances is called catabolism.
  4. Humification: The process of accumulation of a dark-colored amorphous substance, called humus, that is highly resistant to microbial action and undergoes decomposition at an extremely slow rate is called humification. Humus being colloidal is the reservoir of nutrients.
  5. Mineralization: The process by which humus is further degraded by some microbes to release inorganic nutrients is called mineralization.

Ecosystem Digrammatic representation Of Decomposition Cycle In A Terrestrial Ecosyste

Question 11. Give an account of energy flow in an ecosystem.
Answer:

The sun is the only source of energy for all ecosystems on earth. Out of the total incident solar radiation, only 50% of it is photosynthetic active radiation (PAR).

  • Plants capture only 2-10 % of the PAR and this small amount of energy sustains the entire living world. So, there is unidirectional flow of energy from the sun to producers and then to consumers
  • The energy is transferred in an ecosystem, in the form of food which is degraded and loses a major part of food energy as heat during metabolic activities and only a very small fraction becomes stored as biomass
  • This is correlated to the second law of thermodynamics, i.e., ecosystems need a constant supply of energy to synthesize molecules they require, to counteract the universal tendency towards increasing disorderliness. –
  • The green plants in the ecosystem that can trap solar energy to convert it into chemical bond energy are called producers.
  • All the animals that depend on food on plants are called consumers or heterotrophs.
  • Consumers are divided into the following categories:
  1. Primary consumers: Animals that feed directly on plants, i.e., herbivores.
  2. Secondary consumers: Consumers that feed on primary consumers, i.e., carnivores.
  3. Tertiary consumers: Consumers that feed on secondary consumers
  • Lindeman’s 10 percent law: At each step of the food chain, when food energy is transferred from one trophic level to the next higher trophic level, only about 10 percent of energy is passed on to the next trophic level. This is known as Lindeman’s 10 percent law given by Lindeman in 1942.

Ecosystem Multiple-Choice Questions

Question 1. Which one of the following has the largest population in a food chain?

  1. Producers
  2. Primary consumers
  3. Secondary consumer
  4. Decomposers

Answer: 1. Producers (decomposers can be maximum but they are excluded from the food chain)

Question 2. The second trophic level in a lake is

  1. Phytoplankton
  2. Zooplankton
  3. Benthos
  4. Fishes

Answer: 2. Zooplankton

Question 3. Secondary producers are

  1. Herbivores
  2. Producers
  3. Carnivores
  4. None of the above

Answer: 1. Herbivores

Read and Learn More Class 12 Biology Chapter Wise

Question 4. What is the percentage of photosynthetically active radiation (PAR) in the incident solar radiation?

  1. 100%
  2. 50%
  3. 1-5%
  4. 2-10%

Answer: 2. 50%

Ecosystem Fill In The Blanks

1. Plants are called Producers because they fix carbon dioxide.

2. In an ecosystem dominated by trees, the pyramid (of numbers) is an inverted type.

3. In aquatic ecosystems, the limiting factor for the productivity is Light.

4. Common detritivores in our ecosystem are Earthworms, ants, and mites

5. The major reservoir of carbon on earth is Oceans (71% dissolved carbon)