CBSE Solutions For Class 10 Mathematics Chapter 15 Probability

Class 10 Maths Probability

Question 1. In any situation that has only two possible outcomes, each outcome will have Probability. Find whether it is true or false.
Solution: False.

The probability of each outcome will be \(\frac{1}{2}\), only when the two outcomes are equally likely.

Question 2. A Marble is chosen at random from 6 marbles numbered 1 to 6. Find the Probability of getting a marble having number 2 and 6 on it.
solution:

The favourable Case is to get a marble on which both numbers 2 and are written. But there is no such marble.

So, N(E) = 0 and n(S) = 6

∴ Required probability = \(\frac{n(E)}{n(S)}=\frac{0}{6}=0\)

Question 3. A marble is chosen at random from 6 marbles numbered Ito 6. Find the Probability of getting a marble having number 2 or 6 on it.
Solution:

Here N(E) = 2

and n(S) = 6

∴ Required Probability = \(\frac{2}{6}=\frac{1}{3}\)

CBSE Class 10 Maths Solutions Probability

Question 4. It is given that in a group of 3 Students, the probability of 2 students not having the Same birthday is 0.992. What is the probability that the 2 students have the Same birthday.
Solution:

The probability of 2 students not having the Same birthday = 0.992

∴ Probability of 2 students having the same birthday

= 1-0.992

= 0.008

Question 5. A bag Contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

  1. Red?
  2. Not ved?

Solution:

Total balls = 3+5 = -8

Total possible outcomes of drawing a ball at random from the bag = 8

1. Favorable outcomes of drawing a batt at random red ball =3

∴ The probability of drawing a red ball

⇒ \(\frac{\text { Favourable outcomes of drawing a red ball }}{\text { Total possible outcomes }}\)

⇒ \(\frac{3}{8}\)

2. Probability that the ball drawn is not red = 1- probability that the ball drawn

⇒ \(1-\frac{3}{8}=\frac{5}{8}\)

6. A die is thrown once. Find the probability of getting:

  1. A prime number
  2. A number lying between 2 and 6
  3. An odd number.

Solution:

Possible outcomes in one throw of a die = {1,2,3,4,5,6}

Total possible outcomes = 6

1. Prime numbers = {2,3,5}=3

∴ Probability of getting prime number = \(\frac{3}{6}=\frac{1}{2}\)

2. Numbers lying between 2 to 6 = {3, 4, 5} = 3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

3. Odd numbers = {1,3,5}=3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

Question 7. 12 defective pens are accidentally mixed with 132 good to just look at a pen and tell whether or not it is taken out at random from this lot. Determine the probability Out is a good one.
Solution:

No. of good pens = 132

No. of defective pens = 12

Total pens = 132 +12 = 144

Total Favourable outcomes of drawing a pen = 144

Favourable outcomes of drawing a good pen = 132

∴ Probability of drawing a good pen = \(\frac{132}{144}=\frac{11}{12}\)

Question 8. A child has a die whose Six faces show the letters as

CBSE School For Class 10 Maths Chapter 15 Probability A Child Has A Die Whose Six Fases Show The Letters

given below; The die is thrown once. what is the probability of getting

  1. A?
  2. D?

Solution:

Total possible outcomes in a throw of die = 6

1. Favourable outcome of getting A = 2

∴ Probability of getting \(A=\frac{2}{6}=\frac{1}{3}\)

2. Favourable outcomes of getting D = 1

∴ Probability of getting D = \(\frac{1}{6}\)

Question 9. A box Contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

  1. A two-digit number,
  2. A number divisible by 5.

Solution:

We have, n(S) = 90

1. Let A be the event of getting “a two-digit number”.

∴ Favourable cases are 10,11,12,13,14, …….., 90

∴ n(A) = 81

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}\)

2. let B be the event of getting ་་ a number divisible by 5″.

∴ Favourable Cases are 10, 15, 20, 25, 30, …… 90.

Let there be n in numbers.

∴ Tn = 90

∴ 10+ (n-1)5 = 90

⇒ (n-1)5=80

⇒ n-1 = 16

⇒ n = 17

∴ n(B) = 17

P(B) = \(\frac{n(B)}{n(S)}=\frac{17}{90}\)

Question 10. It is known that a box of 600 electric bulbs Contains 12 defective bulbs. One bulb is taken out at random from this box, what is the probability that it is a non-defective bulb?
Solution:

The number of non-defective bulbs in the box = 600-12=588

So, probability of taking out a non-defective bulb = \(\frac{588}{600}=\frac{49}{50}\)

=0.98

CBSE Solutions For Class 10 Mathematics Chapter 14 Statistics

Class 10 Maths Statistics

Question 1. Find the mean by the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean By Direct Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean By Direct Method

Now,

mean \(\bar{x}=\frac{\sum f_{i x_i}}{\sum f_i}=\frac{1100}{50}=22\)

Question 2. Find the mean using the Direct Method:

CBSE School For Class 10 Maths Chapter 14 Statistics CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean Using Direct Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean Using Direct Method

Now, mean \(\bar{x}=\frac{\sum f_{i x i}}{\sum f_i}=\frac{13,200}{50}=264\)

Question 3. The mean of the following frequency distribution is 25. Find the value of P using the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 25.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 25

Now, Mean

⇒ \(\bar{x}=\frac{\sum f_{i x i}}{\sum f_i}\)

⇒ \(\bar{x}=\frac{1230+15 p}{42+p}\)

⇒ \(25=\frac{1230+15 P}{42+P}\)

⇒ 1050 + 25P = 1230 + 15P

⇒ 25P – 15P = 1230 + 15P

⇒ 10P = 180

⇒ P = \(\frac{180}{10}\)

P = 18

CBSE Class 10 Maths Solutions Statistics

Question 4. The mean of the following distribution is 54. Find the value of P using the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 54.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 54

Now, Mean \(\bar{x}=54\)

⇒ \(\bar{x}=\frac{\sum f_{i x_i}}{\sum f_i}\)

54 = \(\frac{2070+70 p}{41+P}\)

2214+54p=2070+70P

70P-54P = 2214-2070

16P = 144

P = \(\frac{144}{16}\)

P = 9

Question 5. Find the mean from the following table using the Short Cut Method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Short Cut Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Short Cut Method

Now, Mean a = 22.5

Mean \(\bar{x}=a+\frac{\sum f_{i d i}}{\sum f_i}\)

⇒ \(\bar{x}=22.5+\frac{0}{67}\)

⇒ \(\bar{x}=22.5\)

Question 6. Find the mean from the following table using the step. deviation method:

CBSE School For Class 10 Maths Chapter 14 Statistics Step Deviation Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Step Deviation Method

Now, a = 37.5, h = 5

Mean \(\bar{x}=a+\frac{\sum f_{i u_i}}{\sum f_i} \times h\)

⇒ \(\bar{x}=37.5+\frac{-46}{50} \times 5\)

⇒ \(\bar{x}=37.5-\frac{46}{10}\)

⇒ \(\bar{x}=\frac{375-46}{10}\)

⇒ \(\bar{x}=\frac{329}{10}\)

⇒ \(\bar{x}=32.9\)

Question 7. Find the mean from the following table using the step deviation method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Step Deviation Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Step Deviation Method

Now, a = 27.5, h = 5

mean \(\bar{x}=a+\frac{\sum f_i u_i}{\sum f_i} \times h\)

⇒ \(\bar{x}=27.5+\frac{-20}{40} \times 5\)

⇒ \(\bar{x}=\frac{220-20}{8}\)

⇒ \(\bar{x}=\frac{200}{8}\)

⇒ \(\bar{x}=25\)

Question 8. Find the median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 8.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 8

Here, N = 58

∴ For Median class \(\frac{N}{2}=\frac{58}{2}=29\)

∴ Median class = 10-13

Here l1 = 10, l2 = 13

⇒ i = 13-10 = 3

f = 0, C.f = 29

∴ Median M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

M = \(10+\frac{29-29}{0} \times 3\)

M = \(10+\frac{0}{0} \times 3\)

M = 10

Question 9. Find the Median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 9.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 9

Here, N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

=25

∴ Median Class = 20-30

Now, l1 = 20, l2 = 30

i = 30-20 = 10

f = 12, C = 15

and Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(20+\frac{(25-15)}{12} \times 10\)

M = \(20+\frac{10}{6} \times 5\)

M = \(\frac{120+50}{6}\)

M = \(\frac{170}{6}\)

M = 28.33

Question 10. Find the median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

∴ The median class is 15-20

Now, l1 = 15, l2 = 20, i = 20-15 = 5, f = 15, C = 20

and Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(15+\frac{25-20}{15} \times 5\)

M = \(\frac{45+5}{3}\)

M = \(\frac{50}{3}\)

M = 16.67

Question 11. Find the Median for the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median For The Following Frequency Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median For The Following Frequency Distribution

Now, N=340

⇒ \(\frac{N}{2}=\frac{340}{2}=170\)

∴ Median class = 39.5-46.5

∴ l1 = 39.5, l2 = 46.5, i= 46.5-39.5 = 7, f = 102, C = 199

Median M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

M = \(39.5+\frac{(170-199)}{102} \times 7\)

M = \(39.5-\frac{29}{102} \times 7\)

M = \(39.5-\frac{203}{102}\)

M = \(\frac{4029-203}{102}\)

M = \(\frac{3826}{102}\)

M = 36.5

Question 12. Find the Median for the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution

Here N = 123

⇒ \(\frac{N}{2}=\frac{123}{2}=61.5\)

∴ Median Class = 20,5-25.5

l1 = 20.5, l2 = 25.5, i = 25.5-20.5 = 5, f = 24, C = 65

Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{F} \times i\)

M = \(20.5+\frac{(61.5-65)}{24} \times 5\)

M = \(20.5-\frac{3.5}{24} \times 5\)

M = \(\frac{492-17.5}{24}\)

M = \(\frac{474.5}{24}\)

M = 20.1

Question 13. If the Median of the following frequency distribution is 32.5. Find the value of F.

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution Is 325.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution Is 325

Here N = 34+ P

⇒ \(\frac{N}{2}=\frac{34+P}{2}\)

Median = 325 ⇒ Median class=30-40

∴ 11 = 30, l2 =40, i = 40-30=10

f = 12, C = 17

M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

32.5 = \(30+\frac{\left(\frac{34+p}{2}-17\right)}{12} \times 10\)

32.5-30 = \(\frac{10}{12}\left(\frac{34+P}{2}-17\right)\)

2.5 = \(\frac{5}{6}\left(\frac{34+p-34}{2}\right)\)

2.5 = \(\frac{5 P}{12}\)

30 = 5P

P = \(\frac{30}{5}\)

P = 6

Question 14. Determine the median for the following income distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean For The Following Income Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean For The Following Income Distribution

Here N = 100

⇒ \(\frac{N}{2}=\frac{100}{2}=50\)

∴ Medion class = 300-400

∴ l1 = 300, l2 = 400, i= 400-300 = 100, f = 30, C = 33

M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(300+\frac{50-33}{30} \times 100\)

M = \(300+\frac{17}{30} \times 100\)

M = \(\frac{9000+1700}{30}\)

M = \(\frac{10700}{30}\)

M = 356.67

Question 15. Find the mode of the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Median And Mode Of The Following Data.

Solution:

Clearly, the modal class is 60-80 as it has the maximum frequency

∴ l = 60, f1 = 12, f0 = 10, f2 = 6, h = 20

Mode M = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(60+\frac{12-10}{2(12)-10-6} \times 20\)

M = \(60+\frac{2}{24-16} \times 20\)

M = \(60+\frac{2}{8} \times 20\)

M = \(\frac{480+40}{8}\)

M = \(\frac{520}{8}\)

M = 65

Question 16. Given below is the frequency distribution of the heights of players in a school;

CBSE School For Class 10 Maths Chapter 14 Statistics The Frequency Distribution Of The Heights Of Players In A School.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Frequency Distribution Of The Heights Of Players In A School

Clearly, the modal class is 165.5-1685 as it has a maximum frequency

∴ l = 165.5, f1 = 142, f0 =118, f2 = 127, h=3

Mode = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(165.5+\frac{142-118}{2(142)-118-127} \times 3\)

M = \(165.5+\frac{24}{284-245} \times 3\)

M = \(165.5+\frac{24}{39} \times 3\)

M = \(\frac{6454.5+72}{39}\)

M = \(\frac{6526.5}{39}\)

M = 167.35

Question 17. Find the mode of the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Frequency Distribution

Solution:

Clearly, the modal class is 50-60, as it has the maximum frequency

∴ l = 50, f1 = 11, f0 = 9, f2 = 6, h = 10

Mode = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(50+\frac{11-9}{2(11)-9-6} \times 10\)

M = \(50+\frac{2}{22-15} \times 10\)

M = \(50+\frac{2}{7} \times 10\)

M = \(\frac{350+20}{7}\)

M = \(\frac{370}{7}\)

M = 52.86

Question 18. The following distribution represents the height of 160 students in a class;

CBSE School For Class 10 Maths Chapter 14 Statistics The Height Of 160 Students Of A Class

Solution:

Clearly, the modal class is 155-160 as it has the maximum

∴ l = 155, f1 = 38, f0 =30, f2 =24, h=5

Mode = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(155+\frac{38-30}{2(38)-30-24} \times 5\)

M = \(155+\frac{8}{76-54} \times 5\)

M = \(155+\frac{40}{22}\)

M = \(\frac{3410+40}{22}\)

M = \(\frac{3450}{22}\)

M = 156.82

Question 19. The following table gives the weekly wage of workers in a factory:

CBSE School For Class 10 Maths Chapter 14 Statistics The Weekly Wage Of Workers In A Factory.

Find (1) the mean (2) the modal class (3) the mode

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Weekly Wage Of Workers In A Factory

1. Mean:

Mean = \(\frac{\sum f_{i x i}}{\sum f_i}=\frac{5520}{80}\)

= 69

2. Modal class = 55-60

3. Mode: clearly, the Modal class is 55-60 it has the maximum frequency

l = 55, f1 = 20, f0 = 5, f2 = 10, h = 5

M = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(55+\frac{20-5}{2(20)-5-10} \times 5\)

M = \(55+\frac{15}{40-15} \times 5\)

M = \(55+\frac{15}{25} \times 5\)

M = \(\frac{1375+75}{25}\)

M = \(\frac{1450}{25}\)

M = 58

Question 20. The mode of the following Series is 36. Find the missing frequency in it:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Following Series Is 36 Find The Missing Frequency

Solution:

Clearly, 30-40 is the modal class as mode 36 lies in this class

Here l = 30, f1 = 16, f0 = x (Say), f2 = 12 and h = 10 and mode 36

Mode (M) = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

36 = \(30+\frac{16-x}{2(16)-x-12} \times 10\)

36 = \(30+\frac{16-x}{32-x-12} \times 10\)

36 = \(30+\frac{16-x}{20-x} \times 10\)

36 – 30 = \(\frac{16-x}{20-x} \times 10\)

6(20-x) = 10(16-x)

120-6x = 160-10x

10x-6x = 160-120

4x = 40

x = \(\frac{40}{4}\)

x = 10

Question 21. Compute the mode of the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Data

Clearly, Modal class 29.5 – 29.5 as it has the maximum frequency

∴ l = 19.5, f1 = 23, f0 = 16, f2 = 15, h = 10

M = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(19.5+\frac{23-16}{2(23)-16-15} \times 10\)

M = \(19.5+\frac{7}{46-31} \times 10\)

M = \(\frac{292.5+70}{15}\)

M = 24.17

Question 22. Find the mean, Median, and mode of the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Following Frequency Distribution

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Median And Mode Of The Following Data

let assumed mean A = 70, h = 20, Ef = 50, and Σfu = -19

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_u}{\sum f}\right]\)

⇒ \(\bar{x}=70+\left[20 \times \frac{-19}{50}\right]\)

⇒ \(\bar{x}=70+[20 x-0.38]\)

⇒ \(\bar{x}=70-7.6\)

⇒ \(\bar{x}=62.4\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}\) = 25

Cumulative frequency just greater than 25 is 36 and the Corresponding class is 60-80.

∴ l1 = 60, f = 12, l2 = 80, C = 24, i = 80-60 = 20

Now, median (M) = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(60+\frac{(25-24)}{12} \times 20\)

M = \(60+\frac{1}{12} \times 20\)

M = \(\frac{720+20}{12}\)

M = \(\frac{740}{12}\)

M = 61.66

Mode = 3(Median)-2(Mean)

Mode = 3(61.66)-2(62.4)

= 184.98-124.8

M = 60.18

Question 23. 100 Surnames were randomly picked from a local directory and the distribution of a number of letters of the English alphabet in the Surname was obtained as follows:

CBSE School For Class 10 Maths Chapter 14 Statistics 100 Surnames Were Randomly Picked From A Local Directony And Letter Of The English Alphabet.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics 100 Surnames Were Randomly Picked From A Local Directony And Letter Of The English Alphabet

let assumed mean A = 11.5, h = 3, Σf = 100, Σfu = -106

Mean \(\bar{x}=A+\left[h \times \frac{\sum(f u)}{\sum f}\right]\)

⇒ \(\bar{x}=11.5+\left[3 \times \frac{-106}{100}\right]\)

⇒ \(\bar{x}=11.5+[3 \times-1.06]\)

⇒ \(\bar{x}=11.5-3.18\)

⇒ \(\bar{x}=8.32\)

Here, N = 100

⇒ \(\frac{N}{2}=\frac{100}{2}\)

= 50

Cumulative frequency just greater than 50 is 76 and the corresponding class is 10-13

∴ l1 = 7, f = 40, l2 = 10, C = 36, i = 10-7 = 3

Median (M) = \(\ell+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(7+\frac{50-36}{40} \times 3\)

M = \(7+\frac{14}{40} \times 3\)

M = \(\frac{280+42}{40}\)

M = \(\frac{322}{40}\)

M = 8.05

Mode = 3(Median)-2(Mean)

= 3(8.05)-2(8-32)

= 24.15-16.64

M = 7.51

Question 24. The following table gives the daily income of such workers of a factory:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Income Of 50 Workers Of A Factory.

Find the mean, mode, and median of the above data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Income Of 50 Workers Of A Factory

let assumed mean A = 150, h = 20, Σf = 50, Σfu = -12

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_4}{\sum P}\right]\)

⇒ \(\bar{x}=150+\left[20 \times \frac{-12}{50}\right]\)

⇒ \(\bar{x}=150+[20 x-0.24]\)

⇒ \(\bar{x}=150-4.8\)

⇒ \(\bar{x}=145.2\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

Cumulative frequency just greater than 25 is 36 and the Corresponding is 120-140

∴ l1 = 120, f = 14, l2 = 140, C = 12, i = 140-120 = 20

Now, Median = \(l_1+\frac{\left(\frac{N}{2}- C\right)}{f} \times i\)

⇒ \(120+\frac{(25-12)}{14} \times 20\)

⇒ \(120+\frac{13}{14} \times 20\)

⇒ \(\frac{1680+260}{14}\)

⇒ \(\frac{1940}{14}\)

M = 138.57

Mode = 3(Median) – 2 (Mean)

M = 3(138.57)-2((45.2)

= 415.71-290.4

= 125.31

Question 25. A Survey regarding the heights (in cm) of so girls in a class was conducted and the following data was obtained:

CBSE School For Class 10 Maths Chapter 14 Statistics A Survey Regarding The Height In Cm Of 50 Girls Of A Class Was Conducted And The Following Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics A Survey Regarding The Height In Cm Of 50 Girls Of A Class Was Conducted And The Following Data

Let assumed Mean A=145, h=10, Σf=50, and Σfu=24

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_u}{\sum f}\right]\)

⇒ \(\bar{x}=145+\left[10 \times \frac{24}{50}\right]\)

⇒ \(\bar{x}=145+[10 \times 0.48]\)

⇒ \(\bar{x}=145+4.8\)

⇒ \(\bar{x}=149.8 \mathrm{~cm}\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

Cumulative frequency just greater than 25 is 42 and the Corresponding is 150-160

l1 = 150, f = 20, l2 = 160, C = 22, i = 160-150 = 10

Now, Median (M) = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{F} \times i\)

⇒ \(150+\frac{(25-22)}{20} \times 10\)

⇒ \(150+\frac{30}{20}\)

⇒ \(\frac{3000+30}{20}\)

⇒ \(\frac{3030}{20}\)

= 151.5 cm

Mode = 3(Median)-2(Mean)

Mode = 3(151.5)-2(149.8)

= 454.5-299.6

= 154.9

Question 26. The table below shows the daily expenditure on food of 30 households in a locality:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Expenditure On Food Of 30 House Holds In A Locality

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Expenditure On Food Of 30 House Holds In A Locality.

Let assumed mean A=225, h = 50, Σf=30, Σfu = -12

Mean \(\bar{x}=A+\left[h \times \frac{\sum f u}{\sum f}\right]\)

⇒ \(\bar{x}=225+\left[50 \times \frac{-12}{30}\right]\)

⇒ \(\bar{x}=225+[50 \times -0.4]\)

⇒ \(\bar{x}=225-20\)

⇒ \(\bar{x}=205\)

Here, N = 30

⇒ \(\frac{N}{2}=\frac{30}{2}=15\)

Median Class = 200-250

l1 = 200, f = 12, l2 = 250, C = 13, i = 250-200 = 50.

Median (M) = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{F} \times i\)

M = \(200+\frac{15-13}{12} \times 50\)

M = \(\frac{2400+100}{12}\)

M = \(\frac{2500}{12}\)

M = 208.33

CBSE Solutions For Class 10 Mathematics Chapter 13 Volume And Surface Area Of Solids

Class 10 Maths Volume And Surface Area Of Solids

Question 1. A tent of cloth is Cylindrical upto I’m height and Conical above it of the Same radius of base. If the diameter of the tent is 6m and the slant height of the Conical part is 5m, find the cloth required to make this tent.
Solution:

Diameter of base 2r = 6m

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Tent Of Cloth Is Cylindrical Up To Height And Conical Of The Sum Radius Of Base

r = \(\frac{6}{2}\) = 3m

Height of Cylindrical path h = 1m

The slant height of Conical part l = 5m

Cloth required in tent = 2πrh +πrl

= πr(2h+1)

⇒ \(\frac{22}{7}\) x 3 (2×1 +5)

⇒ \(\frac{22}{7}\) x 3(10)

= 66m2

CBSE Solutions For Class 10 Mathematics Chapter 13 Volume And Surface Area Of Solids

Question 2. The volume and Surface area of a Solid hemisphere are numerically equal. What is the diameter of the hemisphere?
Solution:

We have,

Volume of hemisphere = Surface area of hemisphere

⇒ \(\frac{2}{3}\) = 3πr2

⇒ \(\frac{2}{3}\) r = 3

2r = 9

Hence, the diameter of the hemisphere = 9 units.

Question 3. 2 Cubes each of volume 64 cm3 are joined end to end. Find the Surface area of the resulting Cuboid.
Solution:

Given,

volume of cube = 64 cm3

(Side)3 = 64

(Side)3 = 43

Side = 4cm

Side of cube = 4cm

A Cuboid is formed by joining two Cubes together.

∴ For Cuboid

length l = 4+4=8cm,

breadth b = 4cm

height b = 4cm

Now, the total surface area of the cuboid.

= 2(l.b+b.h+l.h)

2(8×4+ 4×4+8×4)

= 2(32+16+32)

2(80)

= 160 cm2

Question 4. From a Solid Cylinder whose height is 2.4cm and diameter is 1.4 cm, a Conical Cavity of the Same height and diameter is hollowed out. Find the total Surface area of the remaining Solid to the nearest Cm2.
Solution:

Diameter of Cylinder 2r = 1.4cm

r = 0.7cm

∴ Radius of Cylinder = radius of cone = r= 0.70m

Height of Cylinder = height of cone

h = 2.40m

If the slant height of a cone is l, then

l2 = h2+r2 = (2.4)2 + (0.7)2

5.76 +0.49 = 6.25

1 = \(\sqrt{6.25}\) = 2.5cm

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Solid Cylinder Whose Height And Diameter The Total Surface Area Of The Remaining Solid To The Nearest Cm

The surface area of the remaining solid = area of the base of the cylinder + curved Surface of the cylinder + curved surface of the cone

πr2 + 2πrh + πrl

=πr (r+2h+1)

= \(\frac{22}{7}\) × 0.7 (0.7+2×2.4+2.5)

= \(\frac{22}{7}\) × 0.7 × (5.86) 223×0.7

= \(\frac{22}{7}\): × 4.102

= 17.6cm2

Question 5. The radius and height of a solid right Circular Cone are in the ratio of 5:12. If its volume is 314 cm3, find its total Surface area. [Take π = 3.14]
Solution:

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids The Radius And Height Of A Solid Right circular Cone

Let the radius of Cone = 5x

∴ Height of Cone =12x

l2 = (5x)2+ (12x)2

l2 = 25x2 +144x2

l2 = 169x2

l = \(\sqrt{169 x^2}\)

l = 13x

It is given that volume = 314 cm2

∴ \(\frac{1}{3}\)π(5x)2 (12x) = 314

⇒ \(\frac{1}{3}\) × 3.14 x 25×12 × x3 = 314

⇒  x3 = \(\frac{314 \times 3}{3.14 \times 25 \times 12}\) = 1

∴ x = 1cm

∴ Radius r = 5×1 = 5cm

Height h = 12×1 = 12cm

and slant height 1 = 13×1 = 13cm

Now, total surface area of Cone = πr(l+r) = 3.14 × 5(13+5) = 3.14×5×18

= 282.60cm2

Hence, the total surface area of cone is 282.60cm2

Question 6. The Curved Surface area of a Cone of height 8m is 188.4m2. Find the volume of Cone.
Solution:

πrl = 188.4

⇒ rl = \(\frac{188.4}{3.14}\) = 60

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids The Curved Surface Area Of A Cone

r2l2 = 3600

r2 (h2 +r2)=3600

r2 (64+r2)=3600

r4+64r2 = 3600 = 0

(r2 +100) (r2 – 36)=0

∴ r2 = -100 or r2=36

r = 6

(r2 = -100 is not possible)

∴ Volume of a cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times 3.14 \times 36 \times 8=301.44 \mathrm{~m}^3\)

Question 7. A Conical tent is required to accommodate 157 persons, each person must have 2m2 of space on the ground and 15m3 of air to breathe. Find the height of the tent” Also Calculate the slant height.
Solution:

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Conical Tent Is Required To Accommodate Persons And Also Calculate The Slant Height

1 person needs 2m2 of Space.

∴ 157 Persons needs 2 × 157 m2 of Space on the ground.

πr2 = 2 × 157

∴ r2 = \(\frac{2 \times 157}{3.14}\)

r2 = 100

r = 10m

Also, 1 person needs 15m3 of air.

∴ 157 Persons need 15×157 m2 of air.

∴ \(\frac{1}{3}\) πr2h = 15×157

⇒ h = \(\frac{15 \times 157 \times 3}{3.14 \times 100}\) = 22.5m

∴ l2 = h2 + r2 = (225)2 + (10)2 = 606.25

∴ l = \(\sqrt{606.25}\) = 24.62m

Question 8. Three Cubes of metal whose edges are in the ratio 3:4:5 are melted down into a single cube whose diagonal is \(12 \sqrt{3}\) cm. Find the edges of the three cubes.
Solution:

The ratio in the edges = 3:4:5

Let edges be 3x, 4x and 5x respectively.

∴ Volumes of three cubes will be 27x3, 64x3, and 125x3 in cm3 respectively.

Now, the Sum of the Volumes of these three Cubes = 27x3 + 64x3+125x3

216x3 cm3

Let the edge of the new cube be a cm.

∴ Diagonal of new Cube = \(a \sqrt{3}\) cm

∴ \(a \sqrt{3}=12 \sqrt{3}\)

⇒ a = 12

∴ Volume of new Cube = (12)3 = 1728 cm3

Now by the given Condition

216x3 = 1728

x3 = 8

x = 2

∴ Edge of 1 Cube = 3×2 = 6cm

Edge of 2 Cube = 4×2 = 8 Cm

Edge of 3 Cube = 5×2 = 10cm

Question 9. A Solid is in the shape of a Cone Standing on a hemisphere with both their radii being equal to Icm and the height of the cone is equal to its radius. Find the volume of the Solid in terms of π.
Solution:

Rodius of hemisphere = radius of Cone = r = 1 cm

Height of Cone h = radius of Cone = 1 cm

Volume of hemisphere = \(\frac{2}{3}\) πr3

volume of Come = \(\frac{1}{3}\) πr2h

∴ Volume of Solid = Volume of hemisphere + volume of Come

⇒ \(\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h\)

⇒ \(\frac{2}{3} \pi(1)^3+\frac{1}{3} \pi(1)^2(1)\)

⇒ \(\frac{2}{3} \pi+\frac{1}{3} \pi\)

= π cm3

Question 10. A granary is in the shape of a Cuboid of a size 8m x 6m x 3m. If a bag of grain occupies a space of 0.65 m3, how many bags Can be stored in the granary?
Solution:

The Size of the granary is 8m x 6m x 3m.

∴ volume of granary =8×6×3 = 144m3

the volume of one bag of grain = 0.65m3

∴ The number of bags that can be stored in the granary

= \(\frac{\text { volume of granary }}{\text { volume of each bag }}\)

= \(\frac{144}{0.65}\)

= 221.54 or 221 bags.

Question 11. A cylindrical bucket 28cm in diameter and 72cm high is full of water. the water is emptied into a rectangular tank 66 cm long and 28cm wide. Find the height of the water level in the tank.
solution:

Let the height of the water level in the tank = xm, then according to problem

πr2h = l×b×x

Or \(\frac{22}{7} \times 14 \times 14 \times 72=66 \times 28 \times x\)

Or \(x=\frac{\frac{22}{7} \times 14 \times 14 \times 72}{66 \times 28}\)

x = 24 Cm

Question 12. A Solid Spherical ball of Iron with a radius of 6cm is melted and recast into three Solid Spherical balls. The radii of the two balls are 3cm and 4cm respectively, determining the diameter of the third ball.
Solution:

Let the radius of the third ball = r cm

∴ The volume of three balls formed = volume of the ball melted

⇒ \(\frac{4}{3} \pi(3)^3+\frac{4}{3} \pi(4)^3+\frac{4}{3} \pi(r)^3=\frac{4}{3} \pi(6)^3\)

⇒ 27 +64 +r3 = 216

⇒ r3 = 125, i.e., r = 5cm

The diameter of the third ball = 2×5cm = 10cm

Question 13. A Semicircle of radius 17.5cm is rotated about its diameter. Find the Curved Surface of So generated Solid.
Solution:

The Solid generated by a circle rotated about its diameter is a Sphere

Now, a radius of Sphere r = 17.5 cm

and its Curved Surface = 4π2

= \(4 \times \frac{22}{7} \times 17.5 \times 17.5\)

= 3850 cm2

Question 14. A sphere of radius 6cm is melted and recast into a Cone of height 6cm. Find the radius of the Cone.
Solution:

Radius of Sphere = 6cm

∴ Volume of Sphere = \(\frac{4}{3} \pi(6)^3\) = 288π cm3

Let radius of Cone = r

Height of Cone = 6cm

∴ Volume of Cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 \times 6=2 \pi r^2\)

Given that, Volume of Cone = volume of a sphere

2πr2 = 288π

r2 = 144

r = 12

Therefore, a radius of Cone = 12cm

Question 15. A metallic Cylinder of diameter 16 cm and height 9cm is melted and recast Into Sphere of diameter 6cm. How many such Spheres can be formed?
Solution:

For the Cylinder,

Radius = \(\frac{16}{2}\) = 8cm

Height = 9cm

∴ Volume of Cylinder = π (8)2(9) = 576 π cm3

Diameter of Sphere = 6cm

∴ Radius of Sphere = \(\frac{6}{2}\) = 3 cm

Now, Volume of one Sphere = \(\frac{4}{3} \pi(3)^3\) = 36π cm3

∴ Number of Spheres formed = \(=\frac{\text { volume of Cylinder }}{\text { volume of one Sphere }}\)

⇒ \(\frac{576 \pi}{36 \pi}\)

= 16

Question 16. The volume of a sphere is 288π Cm3, 27 Small Spheres Can be formed with this Sphere. Find the radius of the Small Sphere.
Solution:

Volume of 27 Small Spheres = Volume of One big Sphere = 288π

⇒ Volume of 1 Small Sphere = \(\frac{288}{27} \pi\)

⇒ \(\frac{4}{3} \pi r^3=\frac{22}{3} \pi\)

⇒ r3 = 8

⇒ r = 2cm

Question 17. A metallic Sphere of radius 4.2cm is melted and recast into the Shape of a Cylinder of radius 6cm, Find the height of the cylinder.
Solution:

Radius of Sphere R = 4.2cm

Radius of Cylinder r = 6cm

Let the height of the Cylinder = h

Now, the volume of the cylinder = Volume of the Sphere

⇒ \(\pi r^2 h=\frac{4}{3} \pi R^3\)

⇒ h = \(\frac{4 R^3}{3 r^2}\)

⇒ \(\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \mathrm{~cm}\)

= 2.744 сm

∴ Height of Cylinder = 2.744cm

Question 18. Metallic Spheres of radii 6cm, 8cm, and 10 cm, respectively, are melted to form a Single Solid Sphere. Find the radius of the resulting Sphere.
Solution:

Let r1 = 6cm r2 =8cm and r3 = 10cm

let the radius of a bigger Solid Sphere = R

The volume of bigger Solid Volume = Sum of volumes of three given Spheres

⇒ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3+\frac{4}{3} \pi r_3^3\)

⇒ \(R^3=r_1^3+r_2^3+r_3^3\)

⇒ \(R^3=6^3+8^3+10^3\)

⇒ R3 = 216 +512 +1000

⇒ R3 = 1728

⇒ R2 = 123

⇒ R = 12cm

∴ Radius of new Solid Sphere = 12 cm

Question 18. A 20m deep well with a diameter 7m is dug and the earth from digging is evenly Spread out to form a platform 22m of the platform.
Solution:

Radius of well, r = \(\frac{7}{2} m\)

and depth h = 20m

Let the height of the platform be H meter.

∴ The volume of platform = Volume of well

⇒ 22 x 14 x H = πr2h

⇒ \(22 \times 14 \times H=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)

⇒ H = \(\frac{7 \times 5}{14}\)

⇒ H = \(\frac{35}{14}\)

⇒ H = 2.5mn

Height of platform = 2.5m

Question 20. A Cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base, Compare the volume of the two parts.
Solution:

We Can Solve this using Similarity

Let r and h be the radius and height of a Cone OAB

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Cone Is Divided Into Two Parts By Drawing A Plane Through The Mid Point Of Its Axis

Let OE = \(\frac{h}{2}\)

As OED and OFB are Similar

∴ \(\frac{OE}{O F}=\frac{ED}{F B}\)

⇒ \(\frac{h / 2}{h}=\frac{ED}{r}\)

⇒ ED = \(\frac{r}{2}\)

Now volume of Cone OCD = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \pi \times\left(\frac{r}{2}\right)^2 \times \frac{h}{2}\)

⇒ \(\frac{\pi r^2 h}{24}\)

and volume of Cone DAB = \(\frac{1}{3} \times \pi x r^2 \times h\) = \(\frac{\pi r^2 h}{3}\)

∴ \(\frac{\text { volume of Part } O C D}{\text { volume of Part } C D A B}\)

⇒ \(\frac{\frac{\pi r^2 h}{24}}{\frac{\pi r^2 h}{3}-\frac{\pi r^2 h}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{8-1}{24}}\)

⇒ \(\frac{1}{7}\)

Question 21. A well of diameter 3m is dug lum deep. The earth taken out of it has been Spread evenly all around it in the Shape of a Circular ring of width um to form an embankment. Find the height of the embankment.
Solution:

Diameter of well 2r = 3m

r = \(\frac{2}{3}\) = 1.5m

and depth h = 14m

∴ Volume of earth taken out from well = πr2h

⇒ \(\frac{22}{7} \times 1.5 \times 1.5 \times 14=99 \mathrm{~m}^3\)

Now, the outer radius of the well, R = 1.5+4 = 5.5m

∴ Area of the ring of platform = π(R2– r2)

⇒ \(\frac{22}{7}\left[(5.5)^2-(1.5)^2\right]\)

⇒ \(\frac{22}{7}[30.25-2.25]\)

⇒ \(\frac{22}{7} \times 7 \times 4=88 \mathrm{~m}^2\)

let the height of the embankment = H

∴ 88 x H=99

⇒ \(H=\frac{99}{88}=\frac{9}{8}\)

= 1.125m

∴ Height of embankment = 1.125m

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