S Chand Class 10 CHAPTER 18 Arithmetic Mean, Median, Mode and Quartiles Exercise 18 A

 Exercise 18 A 

Question 1

Find the number half way between $0.2$ and $0.02$,

Ans: Halfway number  (average)$\frac{0.2+0.02}{2}=\frac{0.22}{2}=0.11$

Question 2

Find the mean of the following sets of number

(i) $4,5,7,8$ (ii) $3,5,0,2,8$ (iii) $2.5,2.4,3.5,2.8,2.9,3.5$ and 3.6 (iv) $-6,-2,-1,0,1,2,5,9$

(v) first five prime number  (vi) First eight even natural number (ix) x, x+1, x+2,x+3,x+4,x+5 and x+6

 

Sol: (i) here n= 4

$\therefore$ Mean $=\frac{4+5+7+8}{4}=\frac{24}{4}=6$

(ii) $n=5$

$\therefore \text { Mean }=\frac{3+5+0+2 \pi 8}{5}=\frac{18}{5}=3.6$

(iii) $n=7$

$\therefore$ Mean $=\frac{2.5+2.4+3.5+2.8+2.9+3.3+3.6}{7}$

$=\frac{21.0}{7}=3.0$

(iv) n=8

$\begin{aligned} \therefore \text { Mean } &=\frac{-6+(-2)+(-1)+0+1+2+5+9}{8} \\ &=\frac{-6-2-1+0+1+2+5+9}{8}=\frac{8}{8}=1 \end{aligned}$

(v) First 5 prime number are 2,3,,4,711

Mean = $\frac{2+3+5+7+11}{5}=\frac{28}{5}=5.6$

(vi) First 8 even natural number are 2,3,6,8,10,12,14 and 16

Mean = $\frac{2+4+6+8+10+12+14+16}{8}=\frac{72}{8}=9$

(vii) all factors of 30 are 1,2,3,5,6,10,15,30

Here n- 8

Mean = $\frac{1+2+3+5+6+10+15+30}{8}=\frac{72}{8}=9$

(viii) First 5 multiple of 8[ are 8,16,24,32,40 here n = 5

$\therefore$ Mean $=\frac{8+16+24+32+40}{5}=\frac{120}{5}=24$

(ix) Here $n=7$

$\begin{aligned}\therefore \text { Mean } &=\frac{x+x+1+x+2+x+3+x+4+x+5+x+6}{7} \\&=\frac{7 x+21}{7}=\frac{7(x+3)}{7}=x+3\end{aligned}$

Question 3

The mean of the number 6,y,4 , x ,14 is 8 express y in terms of x 

Sol: Mean $=\frac{6+y+7+x+14}{5}$

$8=\frac{x+y+27}{5}$

$40=x+y+27$

$y+x=40-27 \Rightarrow y+x=13$

$\therefore y=13-x$

Question 4

Nisha secured 73,86 ,78 and 75 marks in four tests what is the least number of points she should secure in her test if she has to have an average of 80.

Sol: Number of total tests = 5 

Average of 5 tests -80 

Suppose that the next test secure is x than 

$\begin{aligned} \text { Mean } &=\frac{73+86+78+75+x}{5} \\ \Rightarrow 80 &=\frac{312+x}{5} \\ \Rightarrow 400 &=312+x \\ \therefore x &=400-312=88 \end{aligned}$

Question 5

A class of 10 students was given a test in math . The marks out of 50 secured by the students were as follows  $31,36,27,38,45,39,32,29,41,38$. find the mean Score

Sol: Here n= 10 and scores are $31,36,27,38,45,39,32,29,41,38$

$\begin{aligned} \therefore \text { Mean } &=\frac{31+36+27+38+45+39+32+29+41+38}{10} \\ &=\frac{356}{10}=35.6 \end{aligned}$

Question 6

Find the mean of the following frequency distribution: 

$\begin{aligned}&\text { (a) }\\&\begin{array}{|l|c|c|c|c|c|}\hline \text { Weight } & 30 & 31 & 32 & 33 & 34 \\\hline \text { Noogslucenk } & 8 & 10 & 15 & 8 & 9 \\\hline\end{array}\end{aligned}$

(b) 

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Marks } & 20 & 25 & 30 & 35 & 40 \\\hline \text { Student } & 5 & 10 & 12 & 8 & 5 \\\hline\end{array}$

(c)

 $\begin{array}{|c|c|c|c|c|}\hline x & 2 & 5 & 7 & 8 \\\hline y & 2 & 4 & 6 & 3 \\\hline\end{array}$

(d)

 $\begin{array}{|c|c|c|c|c|c|c|}\hline x & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 \\\hline y & 30 & 60 & 20 & 40 & 10 & 50 \\\hline\end{array}$

Sol: (a)

$ \begin{array}{|c|c|c|}\hline \text { Weight } & \text { No of students } & f \times x \\\hline 30 & 8 & 240 \\31 & 10 & 310 \\32 & 15 & 480 \\33 & 8 & 264 \\34 & 9 & 306 \\\hline \text { Total } & 50 & 1600 \\\hline\end{array}$

$\begin{aligned} \therefore \text { Mean } &=\frac{\sum f x}{\sum f} \\ &=\frac{1600}{50} \\ &=32 \mathrm{~kg} \end{aligned}$

(b)

$\begin{array}{|c|c|c|}\hline \text { Marks }(x) & \text { Students }(t) & f \times x \\\hline 20 & 5 & 100 \\25 & 10 & 250 \\30 & 12 & 360 \\35 & 8 & 280 \\40 & 5 & 200 \\\hline \text { total} & 40 & 1190 \\\hline\end{array}$

$\begin{aligned} \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\ &=\frac{1190}{40} \\ &=\frac{119}{4}=29.75 \end{aligned}$

(c) 

 $\begin{array}{|c|c|c|}\hline x & f & f \times x \\\hline 2 & 2 & 4 \\5 & 4 & 20 \\7 & 6 & 42 \\8 & 3 & 24 \\\hline \text { total } & 15 & 90 \\\hline\end{array}$

$\begin{aligned} \therefore \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\ &=\frac{90}{15} \\ &=6 \end{aligned}$

(d)

$\begin{array}{|c|c|c|}\hline x & f & f \times x \\\hline 0.1 & 30 & 3 \\0.2 & 60 & 12 \\0.3 & 20 & 6 \\04 & 40 & 16 \\0.5 & 10 & 5 \\0.6 & 50 & 30 \\\hline \text { Tolat } & 210 & 72 \\\hline\end{array}$

$\begin{aligned}\therefore \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\&=\frac{72}{210}=\frac{12}{35} \\&=0.34 .\end{aligned}$

Question 7

Fill in the blanks 

While calculating the mean of a the grouped data. We make the assumption that frequency is any class is entered at its.......

Sol: Mean of grouped data in any class is called its class mark 

Question 8

The frequency distribution of marks obtained by 40 students of a class is as under calculate the arithmetic mean 

$\begin{array}{|c|c|c|c|c|c|c|}\hline \text { Marks } & 0-8 & 8-16 & 16-24 & 24-32 & 32-40 & 40-48 \\\hline \text { Students } & 5 & 3 & 10 & 16 & 4 & 2 \\\hline\end{array}$

Sol:  

$\begin{array}{|c|c|c|c|}\hline \text { marks } & 8 & x & f x \\\hline 0-8 & 5 & 4 & 20 \\8-16 & 3 & 12 & 36 \\16-24 & 10 & 20 & 200 \\24-32 & 16 & 28 & 448 \\32-40 & 4 & 36 & 144 \\40-48 & 2 & 42 & 88 \\\hline \text { total } & 40 & & 936 \\\hline\end{array}$

$\begin{aligned} \text { Meam } &=\frac{\varepsilon f x}{\varepsilon t} \\ &=\frac{936}{40} \\ &=23.4 \text { Marks } \end{aligned}$

Question 9

Fill the mean of the following data: 

(a)

$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Marks } & 10-14 & 15-19 & 20-24 & 25-29 & 30-34 \\\hline \text { Student } & 4 & 6 & 12 & 5 & 3 \\\hline\end{array}$

(b) 

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Class } & 0-10 & 11-20 & 21-30 & 31-40 & 41-50 \\\hline \text { frequency } & 3 & 4 & 2 & 5 & 6 \\\hline\end{array}$

Sol: (a) 

$\begin{array}{|c|c|c|c|}\hline \text { Marks } & \text { Student }(\text { f) } & x & f \times x \\\hline 10-14 & 4 & 12 & 48 \\15-19 & 6 & 17 & 102 \\20-24 & 12 & 22 & 264 \\25-29 & 5 & 27 & 135 \\30-34 & 3 & 32 & 96 \\\hline \text { Tolel } & 30 & & 645 \\\hline\end{array}$

Mean$=\frac{\varepsilon f x}{\varepsilon f}$= $\frac{578.5}{20}$

$=28.925$

Question 10

In a class of 60 boys the marks obtained in a monthly test were as under : 

$\begin{array}{|c|c|c|c|c|c|}\hline \text { Marks } & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\\hline \text { Students } & 10 & 25 & 12 & 00 & 05 \\\hline\end{array}$

Find the mean marks of class 

Sol: $\begin{array}{|l|l|l|l|}\hline \text { Marks } & \text { students (f) } & x & f \times x \\\hline 10-20 & 10 & 15 & 150 \\20-30 & 25 & 25 & 625 \\30-40 & 12 & 35 & 420 \\40-50 & 08 & 360 \\56-60 & 05 & 55 & 275 \\\hline & 60 & 1830 & 183 \\\hline\end{array}$

Mean $=\frac{\sum f x}{\sum f}$= $\frac{1830}{60}=\frac{183}{6}=30.5$ 

Question 11

(i) Direct method and (ii) short cut method 

$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}\hline \text { Class } & 5-10 & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 & 40-45 & 45-50 \\\hline \text { frequency } & 10 & 6 & 4 & 12 & 8 & 4 & 2 & 1 & 3 \\\hline\end{array}$

Sol: (image to be added)

(i) Direct method

$\text { Mean }=\frac{\sum \cdot d}{\sum f}=\frac{1100}{50}=22$

(ii) Short cut method 

Mean = A+ =\frac{\sum \cdot d}{\sum f}

$=27.5+\frac{-275}{50}$

$=27.5-5.5=22$

Question 12

The following table givens the classification of 100 cows of a dairy form . According to the amount of milk given by each in a dairy 

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline \begin{array}{c}\text { Amount of milk in kg} \\\text {no. of cows} 1 \mathrm{~kg}\end{array} & 0-2 & 2-4 & 4-6 & 6-61 & 8-10 & 10-12 & 12-74 & 14-16 & 16-18 \\\hline \text { No.0fectss } & 4 & 14 & 17 & 20 & 10 & 13 & 12 & 10 & 10 \\\hline\end{array}$

Calculate the mean correct to first place of decimal 

Sol: 

$\begin{array}{|c|c|c|c|}\hline \text { Amount of milk} & \text { cows }(f) & x & f \times x \\\hline 0-2 & 4 & 1 & 4 \\2-4 & 14 & 3 & 42 \\4-6 & 17 & 5 & 85 \\6-8 & 20 & 7 & 140 \\8-10 & 10 & 9 & 90 \\10-12 & 13 & 11 & 143 \\12-14 & 12 & 13 & 156 \\14-16 & 10 & 15 & 150 \\16-18 & 10 & 17 & 170 \\\hline \text { Jotel } & 150 & & 980 \\\hline\end{array}$

$\therefore$ Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{980}{110}=8.9=8.9$

Question 13

The weight of 50 apples picked out at random from a are given below:-$82,118,80,110,104,84,106,107,76,82,109,107,115$ $93,187,95,123,125,111,92,86,70,126,70,130,129,139,119,115,128,100,186$

$84,99,113,204,111,141,136,123,90,115,98,110,78,90,107,81,131,75 .$

(i) What the range of weights 

(ii) Form a frequency distribution with class- interval $70-89,90-109$ and So on 

(iii) Use your frequency distribution to calculate the mean 

Sol: (i) Maximum weight = 204g

Minimum weight = 70g 

Range= 204-70=134gm

(ii)Frequency distribution:-

(IMAGE TO BE ADDED)

(iii) $\begin{aligned} \text { Mean } &=A+\frac{\varepsilon f d}{\varepsilon f} \\ &=139.5+\frac{(-1460)}{50} \\ &=139.5-\frac{146}{5} \\ &=139.5-29.2 \end{aligned}$

=110.3gm

Question 14

The following table given the weekly wages of workers in a factory 

$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text { Weekly Wages (Rs) } & 50-55 & 55-60 & 60-65 & 65-70 & 70-75 & 75-80 & 80-85 &85-90 \\\hline \text { No. of worker} & 5 & 20 & 10 & 10 & 9 & 6 & 12 & 8 \\\hline\end{array}$

Calculate (i) The mean(ii) The number of workers getting weekly wages below Rs 80 and (iii) the number of workers getting Rs 65 and more but tea than 85 as weekly wages 

Sol:

$\begin{array}{|l|c|c|c|c|}\hline 50-55 & 5 & 52.5 & -15 & -75 \\55-60 & 20 & 57.5 & -10 & -200 \\60-65 & 10 & 62.5 & -5 & -50 \\65-70 & 10 & 67.5=A & 0 & 0 \\70-75 & 9 & 72.5 & 5 . & 65 \\75-80 & 6 & 77.5 & 10 & 15 \\80-85 & 12 & 82.5 & 15 & 160 \\85-90 & 8 & 97.5 & 20 & 120 \\\hline \text { Total } & 80 & & & 120 \\\hline\end{array}$

(i) $\begin{aligned} \text { Mean }=A+\frac{\varepsilon f d}{\varepsilon t} &=67.5+\frac{120}{80} \\ &=67.5+1.5 \\ &=69.0 \end{aligned}$

(ii) Number of workers getting wages below 80 = 60

(iii) Number of workers getting more than Rs 65 but less then Rs 85 = $10+9+6+12=37$

Question 15

Find the mean of following data

(IMAGE TO BE ADDED)

Sol:  

$\begin{array}{|c|c|c|c|c|}\hline \text { Marks obtained } & \text { Class-mark No } & \text { No of workers } & f & f \times x \\\hline 0-10 & 5 & 7 & 7 & 35 \\10-20 & 15 & 19 & 12 & 100 \\20-30 & 25 & 32 & 13 & 325 \\30-40 & 35 & 42 & 10 & 350 \\40-50 & 45 & 50 & 8 & 360 \\\hline & & & 50 & 1250 \\\hline\end{array}$

Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{1250}{50}=\frac{125}{5}=25$

Question 16

The following table into the form of an ordinary frequency distribution and determine the value of mean 

(IMAGE TO BE ADDED)

Sol: 

(IMAGE TO BE ADDED)

$\begin{aligned} \text { Mean } &=A+\frac{\varepsilon f f}{\varepsilon f} . \\ &=27.5+\frac{-800}{150} \\ &=27.5-\frac{80}{15}=27.5-\frac{16}{3} \\ &=27.5-5.3=22.2 \end{aligned}$

Question 17

Find the missing frequency in the following 

$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}\hline \text { class } & 4-8 & 8-12 & 12-16 & 16-20 & 20-24 & 24-28 & 28-32 & 32-36 & 36-40 \\\hline \text { Frequency } & 11 & 13 & 16 & 14 & - & 9 & 17 & 6 & 4 \\\hline\end{array}$

Sol: Suppose that the missing frequency is p

$\begin{array}{|c|c|c|c|}\hline \text { Class } & \text { frequency } f & x & f \times x \\\hline 4-8 & 11 & 6 & 66 \\8-12 & 13 & 10 & 130 \\12-16 & 16 & 14 & 224 \\16-20 & 14 & 18 & 252 \\20+24 & p & 22 & 22 p \\24-28 & 9 & 26 & 234 \\22-32 & 17 & 30 & 510 \\32-36 & 6 & 34 & 204 \\36-40 & 4 & 38 & 152 \\\hline \text { Toral } & 90+p & \varepsilon f x=1772+22 p \\\hline\end{array}$

mean $=\frac{\varepsilon f x}{\varepsilon t}$

$19.92=\frac{1772+22 p}{90+p} \Rightarrow 1772+22 p=1792.8+19.92 p$

$\begin{aligned} \Rightarrow 22 p-19.92 p &=1792.8-1772 \\ 2.08 p &=20.8 \Rightarrow p=\frac{20.8}{2.08}=10 \end{aligned}$

Question 18

Calculate the A.M correct to one decimal place for the following frequency distribution of marks obtained in an arithmetic test

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Marks } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\\hline \begin{array}{c}\text { No.of students) } \\\text { }\end{array} & 2 & 5 & 20 & 8 & 7 \\\hline\end{array}$

Sol: 

$\begin{array}{|c|c|c|c|}\hline \text { Marks } & \text { no of students } & x & f \times x \\\hline 0-10 & 2 & 5 & 10 \\10-20 & 5 & 15 & 75 \\20-30 & 20 & 25 & 500 \\30-40 & 8 & 35 & 280 \\40-50 & 7 & 45 & 315 \\\hline & \varepsilon f=42 &\varepsilon x=1180 \\\hline\end{array}$

Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{1180}{42}=28.095=28.1$

Question 19

Following table gives marks scored by student in an examination 

4\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text { Marks } & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 \\\hline \text { No.of } & 3 & T & 15 & 24 & 16 & 8 & 5 & 2 \\\hline \text { Student } & & & & & & \\\hline\end{array}$

Ans: 

$\begin{array}{|l|c|l|c|c|c|}\hline \text { Marks } & \text { No of student(f)} & x & A & d=x-A & f \times d \\\hline 0-5 & 3 & 2.5 & & -15 & -45 \\5-10 & 7 & 7.5 & & -10 & -70 \\10-15 & 15 & 12.5 & & -5 & -75 \\15-20 & 24 & 17.5 & 17.5 & 0 & 0 \\20-25 & 16 & 22.5 & & 5 & 80 \\25-30 & 8 & 27.5 & & 10 & 80 \\30-35 & 5 & 32.5 & & 15 & 75 \\35-40 & 2 & 37.5 & & 20 & 40 \\\hline total & \sum f=80 & & & & \varepsilon f d=85 \\\hline\end{array}$

$\begin{aligned} \text { Mean } &=A+\frac{\varepsilon t d}{\varepsilon f} \\ &=17.5+\frac{85}{80} \\ &=17.5+1.06 \\ &=18.56 \\ &=18.56 \end{aligned}$

S Chand Class 10 CHAPTER 17 Heights and Distances Exercise 17

 Exercise 17 

Question 1

The angle of elevation of the top a tower from a point at a distance of 100m from its  a Horizontal plane is found to 60 degree. Find the height of tower

Sol: Suppose that AB = h is tower and O' is  point a horizontal plane is found that angle of elevation $60^{\circ}$

(IMAGE TO BE ADDED)

In $\triangle O A B$

$\begin{aligned}&\tan 60^{\circ}=\frac{A B}{O A} \\&\sqrt{3}=\frac{h}{100} \\&h=100 \sqrt{3}\end{aligned}$

$=100 \times 1.732=173.2 \mathrm{~m}$

Question 2

A vertical flags staff stand on a horizontal plane . From a point distant 150m from it foot the angle of elevation of its top is found to be 30 degree find the height of the flag staff

Sol: Suppose  that $A B=h$ is a flag staff and $O$ is a point at a distance.

$150 \mathrm{~m}$ from the foot of flag staffs makes an angle $30^{\circ}$

(IMAGE TO BE ADDED)

In  $\triangle O A B$ $\tan 30^{\circ}=\frac{A B}{O A}$

$\frac{1}{\sqrt{3}}=\frac{h}{150}$

$h=\frac{150}{\sqrt{3}}=\frac{150 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{150 \sqrt{3}}{3}=50 \mathrm{\sqrt{3}}=50 \times 1.732=86.6 \mathrm{~m}$

Question 3

The string of a kite is 150m and it makes an angle 60 with horizontal. find the height of the kite from the ground 

Sol:  (IMAGE TO BE ADDED)

Suppose that the height of kite h and makes an angle 

Elevation is 60 and length of string 150m 

In $\triangle O B A, \quad \sin 60^{\circ}=\frac{A B}{O A}$ =$\frac{\sqrt{3}}{2}=\frac{R}{150}$

$2 h=150 \sqrt{3}$

$h=\frac{150 \sqrt{3}}{2}=75 \mathrm {\sqrt{3}}=75 \times 1.732=129.9 \mathrm{~m}$

Question 4

If the shadow of a tower is 30 meter when the sum is 30 what is length of shadow when the sun's elevation of 60 

Sol: In first case $\theta=30^{\circ} \quad d=30 \mathrm{~m}$

Suppose that the height of tower is AB= h

$\triangle O A B$

$\begin{aligned}\operatorname{tom} 30^{0} &=\frac{h}{30} \\\frac{1}{\sqrt{3}} &=\frac{h}{30}\end{aligned}$ $\Rightarrow h=\frac{30}{\sqrt{3}}=10 \sqrt{3}$

Now second case the length of shadow is x then 

In $\triangle O A B$. $\tan 60^{\circ}=\frac{h}{x}$

$v \overline{3}=\frac{10 \sqrt{3}}{x}$

$x=\frac{10 \sqrt{3}}{\sqrt{3}}=10 \mathrm{~m}$

 (IMAGE TO BE ADDED)

Question 5

S Chand Class 10 CHAPTER 16 TRIGONOMETRY Exercise 16 B

 Exercise 16 B

Question 1 

Ans: Using the since, cosine and tangent tables 

(a) $\quad 15^{\circ} 27^{\prime}$

$=\sin 15^{\circ} 24^{\prime}+3^{\prime}$ (mean difference of 3)

$=0.26556+84$

$=0.26640=0.2664$

$\cos 15^{\prime} 27^{\prime}=\cos 15^{\circ} 24^{\prime}+3^{\prime}$

$=0.96410-23$ (Mean difference of 3 )

$=0.96387=0.9639$

$\tan 15^{\circ} 27^{\prime}=\tan 15^{\circ} 24^{\prime}+3^{\prime}$

$=0.27545+94=0.27639=0.2764$

(b) $\sin 3748^{\prime}=$ $0.6129$

$\operatorname{Cos} 37^{\circ} 48^{\prime}=0.79015=0.7902$

$\tan 37.48^{\prime}=0.77568=0.7757$

(c) $\sin 55^{\circ} 17^{\prime}=\sin 5555^{\circ} 12^{\prime}+5^{\prime}$

$=0.82115+82$ (Mean dillerence of $5^{\prime}$ )

$=0.82917=0.8219$

$\cos 55^{\circ} 17^{\prime}=\cos 55^{\circ} 12^{\prime}+5^{\prime}$

$=0.57071-120=0.56951=0.5695$

$\tan 55^{\circ} 17^{\prime}=\tan 55^{\circ} 12^{\prime}+5^{\prime}$

$=1.4388+453=1.44334=0.4433 .$

(d) $\sin 83^{\circ} 37^{\prime}=$ $\sin 83^{\circ} 36^{\prime}+1^{\prime}$

$0.99377+3=0.99380=0.9938$

$\operatorname{Cos} 83^{\circ} 37^{\prime}=\operatorname{Cos} 83^{\circ} 36^{\prime}+1$

$0.11147-29=0.1118=0.1112$

$\tan 8337^{\prime}=8.91520=8.9152$

Question 2

Ans: Find the acute angle A, given 

(a) $\quad \sin A=0.4919$

$=0.4919=0.49090+$ difference 

$=100$

$=\sin 2924^{\prime}+4=\sin 29.28^{\prime}$

$\therefore A=29.281$

(b) $\tan A=2.7775$

$=2.7775=2.77761$ (it is nearest to 2.777 So)

$\begin{aligned} \therefore \tan A &=\tan 70^{\circ} 12^{\prime} \\ \therefore \quad A &=70^{\circ} 12^{\prime} \end{aligned}$

(c) $\tan A \quad 3.412$

=3.41973

$=\tan 73^{\circ} 42'$ (∵ 3.91973 is nearest to 3.412)

$\therefore \quad A=73^{\circ} 42^{\prime}$

(d) $\cos A=0.4651$

$=0.46484+16$

$=\cos 62^{\prime} 18-1^{\prime}=\operatorname{Cos} 62^{\circ} 17^{\prime}$

$\therefore A=62.17^{\prime}$

(e)

 $\begin{aligned} \sin A &=0.95190=095150+31 \\ &=\sin 72^{\circ} 6^{\prime}+3^{\prime}=\sin 72^{\circ} 9^{\prime} \\ \therefore A &=72^{\circ} 9^{\prime} \end{aligned}$

 

(f) $\operatorname{Cos} A=$ $0.57570=.57501+69$

$=\operatorname{Cos} 54^{\circ} 54^{\prime}-3^{\prime}=$

$\operatorname{Cos} 54^{\prime} 51$

$\therefore A=54.5^{\prime}$

Question 3

Ans:  Using tables , find the value of $(2sin\theta- Cos \theta)$

(i) when $\theta=35^{\circ}$

(a) $2 \sin \theta-\cos \theta=2 \sin 35^{\circ}-\cos 35^{\circ}$

$=2(0.57358)-0.81915$ (from table)

$=1.14716-0.81915$

$=0.32801=0.3280$

(b) When tan $\theta=0.2679$

tan 19.56

$\begin{aligned} \therefore & 2 \sin \theta-\cos \theta \\=& 2 \sin 1456^{\prime}-\cos 1456^{\prime} \\=& 2(0.25769)-0.9662 e \\ & 0.51538-0.96622=-0.45084 \end{aligned}$

Question 4

Ans: State for any acute angle $\theta$

 (1) Whether sin $\theta$ increase or decrease as increase 

(i) We know that sin $\theta$=0 and sin 90= 1

ஃ It is clear that sin$\theta$ increase as $\theta$ increase 

(2) Whether cos $\theta$ increase or decrease as $\theta$ decrease 

(ii) We know that cos $\theta$ and cos 90 = 0

ஃ it is clear that as  $\theta$  decrease , cos $\theta$ increase  

Question 5

Ans: $\begin{aligned} \sin x^{\circ} &=0.67 \\ &=0.67043 \\=& \sin 42^{\circ} .6^{\prime}-2^{\prime} \\=& \sin 42^{\circ} 4^{\prime} \end{aligned}$

(a) 

$\begin{aligned} \cos x^{\circ} &=\cos 42^{\circ} 4^{\prime} \\ &=0.74314-77 \\ &=0.742 .37 . \\ &=0.7423 \end{aligned}$

(b) $\cos x^{\circ}+\tan x^{\circ}$

$=\cos 42^{\circ} 4^{\prime}+42^{\circ} 4^{\prime}$

$=0.7423+(0.90040+214)$

$=0.7423+0.90254$

$=0.7423+0.90 .25$

$=1.6448$

Question 6

Ans: $\sin A=0.1822$

$\sin A=0.18224$

$A=\sin 10^{\circ} 30$

$A=10^{\circ} 30$

Question 7

Ans: Given, 

Rectangle ABCD , AC is its diagonal 

AB = 23CM

$\angle C A B=35^{\circ}$

Let $B C=x$

In Right $\triangle A B C$

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

$\tan \theta=\frac{B C}{A B}$

$\tan 35^{\circ}=\frac{x}{23}$

$0.70021=\frac{x}{23}$

$\begin{aligned} x &=23 \times 0.70021 \\ &=16.10483 \\ &=16.1048 \\ &=16.11 \\ BC &=16.11 \mathrm{~cm} \end{aligned}$

Question 9

Ans: Given,

$\begin{aligned} B C &=12 \mathrm{~cm} \\ A B &=4 \mathrm{~cm} ; \\ \angle A E B &=50^{\circ}, \\ \angle B &=50^{\circ} \text { and } \\ \angle C &=30^{\circ} \end{aligned}$

(i) In right angle $\triangle A E B$;

$\operatorname{Cos} 50^{\circ}=\frac{B E}{A B}$

$.6428=\frac{8 E}{4}$

$B E=: 6428 \times 4 .$

$B E=2.5712 \mathrm{~cm}$

(ii) $\begin{aligned} \operatorname{Sin} 50^{\circ} &=\frac{A E}{A B} \\ \cdot 7660 &=\frac{A E}{4} \\ \cdot 7660 \times 4 &=A E \\ 3.0640 &=A E \\ A E &=3.064 \mathrm{~cm} \end{aligned}$

In $\triangle A E C$

$\begin{aligned}\sin 30^{\circ} &=\frac{A E}{A C} \\.5000 &=\frac{3.064}{A C}\end{aligned}$

$A C=\frac{3.0640}{.5000}$

$A C=\frac{3064}{500}$

$A C=6.128 \mathrm{~cm} .$

Question 10

Ans: Given, 

From $\triangle A B C$

$\angle B=90^{\circ}$

$\angle C=30^{\circ}$

$\begin{aligned} \therefore \angle A &=180^{\circ}-\left(90^{\circ}+30^{\circ}\right) \\ \angle A &=180^{\circ}-920^{\circ} \\ \angle A &=60^{\circ} \end{aligned}$

(i) In right angle $\triangle A B C$,

$\tan 30^{\circ}=\frac{A B}{B C} .$

$\frac{1}{\sqrt{3}}=\frac{12}{B C}$

$B C=12 \sqrt{3} \mathrm{~cm}$.

(ii) In right angle $\triangle B D A$

$\cos 60^{\circ}=\frac{A D}{A B} .$

$\frac{1}{2}=\frac{A D}{12}$

$\frac{12}{2}=A D$

$6=A D$

$A D=6 \mathrm{~cm} .$

(iii) In right angle $\triangle A B C$

$\sin 30^{\circ}=\frac{A B}{A C}$

$\frac{1}{2}=\frac{12}{A C}$

$A C=12 \times 2$

$A C=24 \mathrm{~cm} .$

Question 11

Ans: Radius of the circle with center C is 15cm

(IMAGE TO BE ADDED)

ex $A C=13 C=15 \mathrm{~cm}$

$\angle A C B=131^{\circ}$

From C, Draw CL $\perp A B$ Now in $D A B C$, $\angle C=131^{\circ} \mathrm{C}$ $A C=B C$

$\begin{aligned} \therefore \angle A=\angle B &=\frac{180^{\circ}-131^{\circ}}{2} \\ &=\frac{49^{\circ}}{2} \\ &=24. 5^{\circ}=24^{\circ} 30^{\prime} \end{aligned}$

(i) Now in right triangle ACL, LA =  $20^{\circ} 36^{\circ}$

 $\begin{aligned} \therefore \quad \cos o=\frac{A L}{A C} &=\cos 24^{\circ} 30^{\circ} \\ &=\frac{A L}{15} \end{aligned}$

$0.90996=\frac{A L}{15}$ $A L=15 \times 0.90996$

$=\quad A L=13.6494$

and $\begin{aligned} A B &=2 A L=2 \times 13.6494 \\ &=27.2988=27.3 \mathrm{~cm} \end{aligned}$

(ii) Sinθ $=\frac{C L}{A C}$ So, $\sin 24^{\circ} 30^{\circ}=\frac{C L}{15}$

 $=0.41469=\frac{C L}{15}$ $C L=15 \times 0.41469$

$\Rightarrow C L=6.22035=6.22 \mathrm{CM}$

Hence , the distance of AB from the center C= 6.22cm

Question 12

Ans: (IMAGE TO BE ADDED)

Given, 

$A P=20 \mathrm{~km}$

$A B=80 \mathrm{~km} .$

$A B$ making an angle of $30^{\circ}$

$\begin{aligned}\angle B A D &=90^{\circ}-30^{\circ} \\&=60^{\circ}\end{aligned}$

(i) In right angle $\triangle A D B$,

$\sin 60^{\circ}=\frac{B D}{A B}$

$\frac{\sqrt{3}}{2}=\frac{B D}{80}$

$B D=\frac{80}{2} \sqrt{3}$

$B D=40 \sqrt{3}$

$B D=40 \times 1.732 .$

$B D=69.280 \mathrm{~km}$

$\begin{aligned} \therefore B C &=B D+D C . \\ &=69.280420 \\ &=89.280 \mathrm{~km} \end{aligned}$

(ii) In right angle  $\triangle A D B$

$\operatorname{Cos} 60^{\circ}=\frac{A D}{A B}$

$\frac{1}{2}=\frac{A D}{80}$

$A D=\frac{80}{2}$

$A D=40 \mathrm{~km}$

So, $A D=P C=40 \mathrm{~km}$. 

Hence the horizontal distance of point C from point P is 40km

Question 13

Ans: BCDE is a rectangle in which ED = 3.88cm 

BC = 3.88CM

A is a point such that AD = 10cm and A lie 

On CB on producing AE is joined 

Let angle AEB = $\theta$

(Image to be added)

(i) In right triangle ACD 

$\begin{aligned} & \sin 0=\frac{C D}{A D} \\ & \sin 23^{\circ} 35^{\circ}=\frac{C D}{10} \\ \therefore & 0.40008=\frac{C D}{10} \\ \text { so, } C D &=4.00 \mathrm{⊥} \mathrm{CM} \end{aligned}$

(ii) $\cos \theta=\frac{A C}{A D}=$ $\cos 23^{\circ} 35^{\prime}=\frac{A C}{L O}$

$\begin{aligned} \therefore \quad 0.91648=\frac{A C}{10}=A C &=9.1648 \\ &=9.165 \end{aligned}$

$A C=2.165 \mathrm{~cm}$

(iii) Now $A B=A C-B C=9.165-3.880=5.285$

and $E B=C D=4.00 \mathrm{~L}$

$\therefore \tan \theta=\frac{A B}{E B}=\frac{5.285}{4.001}$

$=\frac{5.285}{4001}=1.32092$

$=1.31745+347$

$=\tan 52^{\circ} .48^{\circ}+5^{\circ}$

$=\tan 52^{\circ} 53^{\prime}$

$\theta=52^{\circ} 53^{\circ}$

$\angle A E B=52^{\circ} 53^{'}$

Question 14

Ans: In right angle triangle, ABC, 

$\begin{aligned} \angle B &=90^{\circ} \\ B C &=3 \mathrm{~cm}, \\ A B &=4 \mathrm{~cm} . \end{aligned}$

$\begin{aligned} \therefore A C^{2} &=B C^{2}+A B^{2} . \\ A C^{2} &=(3)^{2}+(4)^{2} \\ A C &=\sqrt{9+16} \\ A C &=\sqrt{25} \\ A C &=5 \mathrm{Cm} . \end{aligned}$

From $\triangle A B C$ and $\triangle D B C$,

$\begin{aligned}&\angle A B C=\angle B D C . \\&\angle C=\angle C\end{aligned}$

$\therefore \triangle A B C \sim \triangle D B C .$  (By AA)

$\triangle A B C \sim \triangle A B D .$

$\frac{A C}{B C}=\frac{A B}{B D}=\frac{B C}{C D} .$

$\frac{A B}{B C}=\frac{B D}{C D} .$ (By alternate)

$\frac{B D}{C D}=\frac{4}{3}$

$\frac{C D}{B D}=\frac{3}{4} .$

(i) $\therefore \tan \angle D B C=\frac{C D}{B D}=\frac{3}{4}$.

 (ii) In right angle$\triangle A B D$

sin $\angle D E A=\frac{A D}{A B}=\frac{A B}{A C}=\frac{4}{5}$

Question 15

Ans: Let BC be the building and 

AB be the flag pole on the building 

So BC = x and 

AB= y

Angle of elevation $\angle B D C=63^{\circ}$

and angle ADC= $63^{\circ}+3^{\circ}$

$=66^{\circ}$

From right angle  $\triangle B C D$

$\tan 63^{\circ}=\frac{B C}{D C}$

$1.9626=\frac{x}{50}$

$x=1.9626 \times 50$

$x=98.1300$

$x=98 \mathrm{~m}$

From right angle  $\triangle A C D$

$\begin{aligned} \tan 66^{\circ} &=\frac{A B+B C}{D C} \\ 2.2460 &=\frac{x+y}{50} \\ 2.2460 &=\frac{98+y}{50} \\ 2-2460 \times 50 &=98+y \\ 112 \cdot 3000 &=98+y \\ 112 \cdot 3000-98 &=y \\ 14 \cdot 3000 &=y \\ y &=14 \mathrm{~m} \end{aligned}$

Question 16

Ans: (IMAGE TO BE ADDED)

TR is the tree which was broken from Q and its top T touched the ground at S. So that SR= 25m

and angle QSR = 30

In the figure TQ= QS 

$\tan \theta=\frac{Q R}{S R}=\tan 30^{\circ}=\frac{Q R}{25}$

$=\frac{1}{\sqrt{3}}=\frac{Q R}{25}=Q R=\frac{25}{\sqrt{3}} .$

$=Q \cdot R=\frac{25}{1.732}=14.43$

and $\cos \theta=\frac{S R}{S Q} \Rightarrow \cos 30^{\circ}=\frac{25}{S Q}$

$=\frac{\sqrt{3}}{2}=\frac{25}{SQ}$

$=5Q=\frac{25 \times 2}{\sqrt{3}}=\frac{50}{\sqrt{3}}

Height of tree = TQ +QR 

$=Q S+Q R=\frac{25}{\sqrt{3}}+\frac{50}{\sqrt{3}}=\frac{75}{\sqrt{3}}$

$=\frac{75 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{75 \sqrt{3}}{3}$

$=25 \sqrt{3} \mathrm{~m}$

$=25(1.732)$

$=43.3 \mathrm{~m}$

$=43 \mathrm{~m}$

Question 17

Ans: In equilateral triangle ABC with each side 6cm

If D is a point on BC such that BD = 2cm 

E is the mid point of BC 

DE = DE - BD = 3-1 =2cm

(if  E is mid point of BC )

 (IMAGE TO BE ADDED)

(i) if $E$ is mid point of $B C$

$\text { so } A E \perp B C$

and $A D=\frac{\sqrt{3}}{2}$ side $=\frac{\sqrt{3}}{2} \times 6=3 \sqrt{3} \mathrm{~cm}$

(ii) In right triangle ADE 

 $\begin{aligned} \tan \angle A D C &=\tan \angle A D E=\frac{A E}{D E}=\frac{3 \sqrt{3}}{2} \mathrm{CM} \\ &=\frac{3(1.732)}{2}=3 \times 0.866= 2.598 \end{aligned}$

(iii)  $\begin{aligned} \tan \angle A D C &=2.59156 \\ &=68^{\circ} 54^{\circ}=69^{\circ} \end{aligned}$

(iv) $\tan \angle D A E=\frac{D E}{A E}$

$=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{3 \times \sqrt{3} \times \sqrt{3}}$

$\begin{aligned}=\frac{2 \sqrt{3}}{9} &=\frac{2(1.732)}{9}=\frac{3.464}{9} \\ &=0.3 .85 \end{aligned}$

$\begin{aligned}=0.38587 &=\tan 21^{\circ} 6^{\prime} \\ &=\tan 21^{\circ} . \end{aligned}$

So $\angle D A E=21^{\circ}$

But $\angle B A C=\angle B A E-\angle D A E$ 

$=30^{\circ}-21^{\circ}=9^{\circ}$(If AE also bisects angle A)

Question 18

Ans: K be the kite which is 75 m above the ground and its string makes angle of 60 with the ground 

(IMAGE TO BE ADDED)

so In $\triangle K B T$

$\begin{aligned}&K T=75 \mathrm{~m} \\&\angle B=60^{\circ} \\&\angle T=90^{\circ} \\&\text { Let } K B=x \mathrm{~m}\end{aligned}$

$\begin{aligned} \therefore \quad & \sin \theta=\frac{K T}{K B} \\ & \sin 60^{\circ}=\frac{75}{x} \\ & \frac{\sqrt{3}}{2}=\frac{75}{x} \end{aligned}$

$x=\frac{75$x=\frac{75 \times 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{150 \sqrt{3}}{3}=50 \sqrt{3}$ \times 2}{\sqrt{3}}$

$=50(1.732)=86.6=87$

 So length of string of the kite = 87m 

S Chand Class 10 CHAPTER 16 TRIGONOMETRY Exercise 16 A

 Exercise 16 A

Question 1

Ans: $\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}=1$

$L \cdot H \cdot S=\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}$ $\left[\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right]$

$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta}=1=R \cdot \mathrm{H} \cdot \mathrm{S}$

Question 2

Ans: $\frac{1-\sin ^{2} \theta}{\cos ^{2} \theta}$=1

L.H.S $=\frac{1-\sin ^{2} \theta}{\cos ^{2} \theta} \quad\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\frac{\cos ^{2} \theta}{\cos ^{2} \theta}=1=R \cdot H \cdot S$

Question 3

Ans: $\sin A \cdot \cot A=\cos A$

L.H.S=$\sin A \cdot \cot A$

$=\sin A \cdot \frac{\cos A}{\sin A}$  $\left[\cot A=\frac{\cos A}{\sin A}\right]$

$=\cos A=R . H .S$

Question 4

Ans: $\frac{1}{\cos ^{2} \theta}-\tan ^{2} \theta=1$

$L . H . S=\frac{1}{\cos ^{2} \theta}-\tan ^{2} \theta \quad\left[\frac{1}{\cos ^{2} \theta}=\sec ^{2} \theta\right]$

$=\sec ^{2} \theta-\tan ^{2} \theta$

$=1=R \cdot H \cdot S$

Question 5

Ans: $\tan ^{2} \cos ^{2} A=1-\cos ^{2} A$

$L \cdot H \cdot S=\tan ^{2} A \cos ^{2} A$ $\left[\tan ^{2} A=\frac{\sin ^{2} A}{\cos ^{2} A}\right]$

$=\frac{\sin ^{2} A}{\cos ^{2} A} \cos ^{2} A$

$=\sin ^{2} A=1-\cos ^{2} A=R \cdot H \cdot S$

Question 6

Ans: $\tan \theta=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$

 $\Rightarrow R . H . S=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$ $\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\frac{\sin \theta}{\sqrt{\cos ^{2} \theta}}$

$=\frac{\sin \theta}{\cos \theta}=\tan \theta=$ L.H.S

Question 7

Ans: $\frac{1+\cos \theta}{\sin ^{2} \theta}=\frac{1}{1-\cos \theta}$

L.H.S $=\frac{1+\cos \theta}{\sin ^{2} \theta}=\frac{1+\cos \theta}{1-\cos ^{2} \theta}\left[\sin ^{2} \theta=1-\cos ^{2} \theta\right]$

$=\frac{1+\cos \theta}{(1-\cos \theta)(1-\cos \theta)}\left[a^{2}-b^{2}=(a-b)\left(a+
 b\right)\right.$

$=\frac{1}{1-\cos \theta}=$ R.H.S

Question 8

Ans: $\cot ^{2} \theta\left(1-\cos ^{2} \theta\right)=\cos ^{2} \theta \quad\left[\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

$L \cdot H \cdot S=\cot ^{2} \theta\left(1-\cos ^{2} \theta\right)$ $\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right]$

$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta} \times \sin ^{2} \theta$

$=\cos ^{2} \theta$ = R.H.S

Question 9

Ans: $\tan ^{2} \theta\left(1-\sin ^{2} \theta\right)=\sin ^{2} \theta$

L.H.S $=\tan ^{2} \theta\left(1-\sin ^{2} \theta\right)$

$\begin{array}{ll}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \cos ^{2} \theta & {\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]} \\ & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]}\end{array}$

$=\sin ^{2} \theta=R \cdot H \cdot S$

Question 10

Ans: $\left(1-\sin ^{2} \theta\right) \operatorname{sic}^{2} \theta=1$

L.H.S $=\left(1-\sin ^{2} \theta\right) \sec ^{2} \theta \quad\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\cos ^{2} \theta \cdot \sec ^{2} \theta$

$=1=$ R.H.S

Question 11

Ans: 

$\begin{aligned}\left(1-\cos ^{2} \theta\right) \operatorname{cosec}^{2} \theta &=1 \\ L \cdot H \cdot S=\left(1-\cos ^{2} \theta\right) & \operatorname{cosec}^{2} \theta \\ 1-\cos ^{2} \theta &=\sin ^{2} \theta \\ \operatorname{cosec} \theta &=\frac{1}{\sin \theta} \\ &=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta} \\ 1^{-} &=R \cdot H \cdot S \end{aligned}$

Question 12

Ans: $\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}=1$ 

$L \cdot H \cdot S=\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}$

$=\sin ^{2} \theta+\frac{1}{\sec ^{2} \theta} \quad\left[1+\tan ^{2} \theta=\sec ^{2} \theta\right]$

$\doteq \sin ^{2} \theta+\cos ^{2} \theta$ $\left[\frac{1}{\sec ^{2} \theta}=\cos ^{2} \theta\right]$

$R \cdot H .S=1$

Question 13

Ans:  $\cos ^{2} \theta+\frac{1}{1+\cot ^{2} \theta}=1$

$L \cdot H S=\cos ^{2} \theta+\frac{1}{1+\cot ^{2} \theta}$

$\left[1+\cot ^{2} \theta=\cos \operatorname{sc}^{2} \theta\right]$

$=\cos ^{2} \theta+\frac{1}{\operatorname{cosec}^{2} \theta}$

$\left\{\frac{1}{\operatorname{cosec}}=\sin \theta\right.$

$=\cos ^{2} \theta+\sin ^{2} \theta$

$R \cdot H \cdot S=1$

Question 14

Ans: $\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sec ^{2} \theta-\tan ^{2} \theta}=1$

$L \cdot H \cdot S=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sec ^{2} \theta-\tan ^{2} \theta}=1$

$\sin ^{2} \theta+\cos ^{2} \theta=1$

$\sec ^{2} \theta-\tan ^{2} \theta=1$

$R \cdot H \cdot S=1$

Question 15

Ans: $\left[\frac{\cos ^{2} A}{\sin ^{2} A}+1\right] \tan ^{2} A=\frac{1}{\cos ^{2} A}$

$L \cdot H \cdot S=\left[\frac{\cos ^{2} \cdot A}{\sin ^{2} A}+1\right] \tan ^{2} \theta$

$\frac{\cos A}{\sin A}=\cot A$

$=\left(\cot ^{2} A+1\right) \tan ^{2} \theta$

$\quad \cos ^{2} A+1=\operatorname{cosec}^{2} A$

$=\operatorname{cosec}^{2} A \times \tan ^{2} A$

$\quad \operatorname{cosec} A=\frac{1}{\sin A}$

$=\frac{1}{\sin ^{2} A} \times \frac{\sin ^{2} A}{\cos ^{2} A}$

$R \cdot H \cdot S=\frac{1}{\cos ^{2} A}$

Question 16

Ans: $\sin ^{2} \theta+\sin ^{2} \theta \cos ^{2} \theta=\sin ^{2} \theta .$

$\begin{aligned} L \cdot H \cdot S=& \sin ^{4} \theta+\sin ^{2} \theta \cos ^{2} \theta \\=& \sin ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right.\\ & \sin ^{2} \theta+\cos ^{2} \theta=1 \\=& \sin ^{2} \theta \times 1 \\ R \cdot H \cdot S=& \sin ^{2} \theta \end{aligned}$

Question 17

Ans: $\sin ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta+\cos 4 \theta=1$

$\begin{aligned} \text { L.H.S }=& \sin ^{2} \theta+2 \sin ^{2} \theta(a)^{2} \theta+\cos ^{4} \theta \\ & a^{2}+2 a b+b^{2}=(a+b)^{2} \\=&\left(\sin ^{2} \theta\right)^{2}+2 \sin ^{2} \theta \cos ^{2} \theta+\left(\cos ^{2} \theta\right)^{2} \\=&\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2} \\ & \sin ^{2} \theta+\cos ^{2} \theta=1 \\=&(1)^{2} \end{aligned}$

R.H.S= 1

Question 18

Ans: $\sin 4 A \operatorname{cosec}^{2} A+\cos 4 A \sec ^{2} A=1$

L.H.S $=\sin 4 A \operatorname{cosec}^{2} A+\cos ^{4} A \sec ^{2} A$

$=\sin 4 A \times \frac{1}{\sin ^{2} A}+\cos ^{4} A \times \frac{1}{\cos ^{2} A}$  $\left[\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

$\therefore \sin ^{2} \theta+\cos ^{2} \theta=1$ $\left[\sec \theta=\frac{1}{\cos \theta}\right]$

R.H.S =1

Question 19

Ans: $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=1$

$\begin{aligned} L \cdot H \cdot S=& \sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A \\=& \sin ^{2} A \times \frac{\cos ^{2} A}{\sin ^{2} A}+\cos ^{2} A \times \frac{\sin ^{2} A}{\cos ^{2} A} \\ & \frac{\cos A}{\sin A}=\cot \theta, \frac{\sin \theta}{\cos \theta}=\tan \theta \\=& \cos ^{2} A+\sin ^{2} A \cdot=1 \\ \text { R.H.S }=& 1 \end{aligned}$

Question 20

Ans: $\tan \theta+\cot \theta=\sec \theta \cdot \operatorname{cosec} \theta \cdot .$

$L \cdot H \cdot S=\tan \theta+\cot \theta$

${\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right] }$

$=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \sin \theta}$

$=\left[\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta}\right]$

$=\frac{1}{\cos \theta \sin \theta}$

${\left[\frac{1}{\cos \theta}=\sec \theta, \frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] }$

$R \cdot H \cdot S=\operatorname{Sec} \theta \cdot \operatorname{cosec} \theta$

Question 21

Ans:  $(\tan A+\cot A) \sin A \cos A=1$

$L \cdot H \cdot S=(\tan A+\cot A)(\sin A \cos A)$

$\begin{aligned} & {\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta \doteq \frac{\cos \theta}{\sin \theta}\right] } \\=& {\left[\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right] \sin A \cos A . } \\=& \frac{\sin ^{2} A+\cos ^{2} A}{\cos A \sin A} \times \sin A \cos A \\ & {\left[\sin ^{2} \theta+\operatorname{Cos}^{2} \theta=1\right] }\end{aligned}$

$=\frac{1}{\cos A \sin A} \times \sin A \cos A$

R.H.S = 1

Question 22

Ans:   $\begin{aligned} \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta & \\ L \cdot H \cdot S=& \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)} \\=& \frac{1-\sin ^{2} \theta+\cos \theta}{\sin \theta(1+\cos \theta)} \\ & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right] } \\=& \frac{\cos \theta+\cos ^{2} \theta}{\sin \theta(1+\cos \theta} \end{aligned}$

$=\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$

$=\frac{\cos \theta}{\sin \theta}$

${\left[\frac{\cos \theta}{\sin \theta}=\cot \theta\right] }$

$R \cdot H \cdot S=\cot \theta$

Question 23

Ans: 

 $\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=2 \operatorname{cosec}^{2} \theta$

$\begin{aligned} \text { L.H.S } &=\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta} \\ &=(a+b)(a-b)=a^{2}-b^{2} \\ &=\frac{1+\cos \theta+1-\cos \theta}{(1-\cos \theta)(1+\cos \theta)}=\frac{2}{1-\cos ^{2} \theta} \end{aligned}$

$\begin{aligned} & {\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right] } \\=& \frac{2}{\sin ^{2} \theta} . \\ & {\left[\frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] }\end{aligned}$

$R \cdot H \cdot S=2 \operatorname{cosec}^{2} \theta$

Question 24

Ans:  $\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\tan ^{2} \theta .$

$L \cdot H \cdot S=\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\frac{1-\tan ^{2} \theta}{\frac{1}{\tan ^{2} \theta}-1}$ 

$=\frac{1-\tan ^{2} θ}{\frac{1-\tan ^{2} θ }{\tan ^{2} \theta}}$

$=\tan ^{2}\theta=R.H.S$

Question 25

Ans:  $\frac{1}{\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$

$L \cdot H \cdot S=\frac{1}{\operatorname{Sec} \theta+\tan \theta}$

$=\frac{\sec \theta-\tan \theta}{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$

$=\frac{\sec \theta-\tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta} \quad\left[\sec ^{2} \theta-\tan ^{2} \theta=1\right]$

$=\frac{\sec \theta-\tan \theta}{1}$

$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$

R.H.S=$ \frac{1-\sin \theta}{\cos \theta}$

Question 26

Ans: $(\operatorname{cosec} A-\sin A) \cdot(\sec A-\cos A)(\tan A+\cot A)=1$

L.H.S $=(\operatorname{cosec} A-\sin A) \cdot(\sec A-\cos A)(\tan A+\cot A)$

$\left[\operatorname{cosec} A=\frac{1}{\sin A}, \sec A=\frac{1}{\cos A}, \tan A=\frac{\sin A}{\cos A}\right.$

$\left.\cot A=\frac{\cos A}{\sin A}\right]$

$=\left[\frac{1}{\sin A}-\sin A\right]\left[\frac{1}{\cos A}-\cos A\right]\left[\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right]$

$=\frac{1-\sin ^{2} A}{\sin A} \times \frac{1-\cos ^{2} A}{\cos A} \times \frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}$

$=\frac{\left[\sin ^{2} A+\cos ^{2} A=1\right]}{\sin A} \times \frac{\sin ^{2} A}{\cos A} \times \frac{1}{\sin A \cos A}$

$=\frac{\cos ^{2} A \sin ^{2} A}{\sin ^{2} A \cos ^{2} A}$

R.H.S $=1$

Question 27

Ans: $\frac{1+\sin \theta}{1-\sin \theta}=(\sec \theta+\tan \theta)^{2}$

$\begin{aligned} \text { L.H.S. } &=\frac{1+\sin \theta}{1-\sin \theta} \\ &=\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)} \\ &=\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta} \end{aligned}$

$\begin{aligned} & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right] } \\=& \frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta} \\=& {\left[\frac{1+\sin \theta}{\cos \theta}\right]^{2} } \\=& {\left[\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right]^{2} }\end{aligned}$

$\frac{1}{\cos \theta}=\sec \theta, \frac{\sin \theta}{\operatorname{con} \theta}=\tan \theta$

$R \cdot H \cdot S=(\sec \theta+\tan \theta)^{2}$

Question 28

Ans: $\frac{1+\cos \theta}{1-\cos \theta}=(\operatorname{cosec} \theta+\cot \theta)^{2}$

$\begin{aligned} L \cdot H \cdot S &=\frac{1+\cos \theta}{1-\cos \theta} \\ &=\frac{(1+\cos \theta)(1+\cos \theta)}{(1-\cos \theta)(1+\cos \theta)} \cdot\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right] \end{aligned}$

$=\frac{\left(1+\cos (\theta)^{2}\right.}{1-\cos ^{2} \theta}$

$=\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}$

$=\left[\frac{1+\cos \theta}{\sin \theta}\right]^{2}$

$=\left[\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right]^{2}$

$\left(\frac{1}{\sin \theta}=\operatorname{cosec} \theta, \frac{\cos \theta}{\sin \theta}=\cot \theta\right)$

$R \cdot H \cdot S=(\cos \theta+\theta+\cot \theta)^{2}$

Question 29

Ans: $\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\operatorname{cosec} \theta+\cot \theta-\frac{1+\cos \theta}{\sin \theta}$

$L \cdot H \cdot S=\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{\cot A+\operatorname{cosec} \theta-\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{\cot \theta+\operatorname{cosec} \theta+\left(\cot ^{2} \theta-\operatorname{cosec} 2 \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\cot \theta+\operatorname{cosec} \theta)+\cot \theta+\operatorname{cosec} \theta)(\cot \theta-\operatorname{cosec} \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\cot \theta+\operatorname{cosec} \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\operatorname{cotc}+\operatorname{cosec} \theta)(\cot \theta-\operatorname{cosec} \theta+1)}{(\cot \theta-\operatorname{cosec} \theta+1}$

$=\cot \theta+\operatorname{cosec} \theta$

$=\operatorname{cosec} \theta+\cot \theta$

$=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta} .$

$R \cdot H \cdot S=\frac{1+\cos \theta}{\sin \theta}$

Question 30

Ans: $\tan \theta+\cot \theta=2$

$(\tan \theta+\cot \theta)^{2}=4 \quad$ [Square both side]

$\tan ^{2} \theta+\cot ^{2} \theta+2 \tan \theta \cot \theta=4$

${[\tan \theta \cdot \cot \theta=1] }$

$\begin{aligned} \tan ^{2} \theta+\cot ^{2} \theta+2 \times 1 &=4 \\ \tan ^{2} \theta+\cot ^{2} \theta+2 &=4 \\ \tan ^{2} \theta+\cot ^{2} \theta &=4-2 \\ &=2 \\ \tan ^{2} \theta+\cot ^{2} \theta &=2 \end{aligned}$

Question 31

Ans:  $a^{2}+b^{2}=(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{2}$

$=-\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta+\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \epsilon \cos \theta$

$=2 \sin ^{2} \theta+2 \cos ^{2} \theta$

$=2\left(\sin ^{2} \theta+\left(\cos ^{2} \theta\right)\right.$

$=2 \times 1$

$=2$

S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 C

 Exercise 15 C

Question 1

Ans: (i) Given, 

Radius = 3cm

Height = 4cm

Slant height (l) =$\sqrt{r^{2}+h^{2}}$

$=\sqrt{(3)^{2}+(4)^{2}}$

$=\sqrt{3+16}$

$=\sqrt{25}$

$=5 \mathrm{~cm}$

Curved surface = $\pi r l$

$=\pi \times 3 \times 5$

$=15 \pi \mathrm{cm}^{2}$

Area of base = $\pi r^{2}$

$\pi \times 3 \times 3$

$=9 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=15 \pi+9 \pi$

$=24 \pi \mathrm{cm}^{2}$

Volume  =  $=\frac{1}{3} \pi r^{2} h$

$-\frac{1}{3} \pi \times 3 \times 3 \times 4 $

$=12 \pi \mathrm{cm}^{3}$

(ii) Given, 

Radius = 20cm

Slant height = 25 cm

$l^{2}=r^{2}+1^{2}$

$\left(25^{2}-r^{2}=h^{2}\right.$

$\sqrt{625}-(20)^{2}=1^{2}$

$\sqrt{225}=h$

$15=h$

$h=15 \mathrm{~cm}$

Curved surface = πrl

$=\pi \times 20 \times 25$

$=500 \pi \mathrm{cm}^{2}$

area of base =  π$r^{2}$

$=\pi \times 200 \times 20$

$=400 \pi \mathrm{cm}^{2} .$

Total surface area = $\pi r l+\pi r^{2}$

$\begin{aligned}=& 500 \pi+400 \pi \\=& 900 \pi \mathrm{cm}^{2} \end{aligned}$

Volume $=\frac{1}{3} \pi{r}^{2} h$

$=2000 \pi \mathrm{cm}^{3}$

(iii) Given, 

Height = 18cm

Slant height = 30 cm

$\therefore \quad l^{2}=h^{2}+r^{2} .$

$(30)^{2}=(18)^{2}+r^{2}$

$900-324=r^{2} .$

$\sqrt{576}=r$

$24=r$

Curved surface = πrl 

$=\pi \times 24 \times 30$

$=720 π c m^{2}$

Area of base = $\pi r^{2}$

$=\pi \times 24 \times 24$

$=576 \pi \mathrm{m}^{2}$

Total surface area= $\pi r l+\pi r^{2}$

729π + 576π

$=1296 \mathrm π{cm}^{2}$

Volume = $\frac{1}{3} \pi r^{2} h$

$=576 \times 6 \pi .$

$=3456 \pi \mathrm{cm}^{3}$

(iv) Given, 

Radius = 27cm

Height = 36cm

$\begin{aligned} \because \quad l &=\sqrt{h^{2}+r^{2}} \\ &=\sqrt{(36)^{2}+(27)^{2}} \\ &=\sqrt{1296+729} \\ &=\sqrt{2025} \\ &=45 \mathrm{~cm}\end{aligned}$

Curved surface =  $\pi$ rl

= $\pi$ $\times 27 \times 45$

$=1215 \pi \mathrm{cm}^{2} .$

Area of base =  $\pi r^{2}$ 

$=\pi \times 27 \times 27$

$=729 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=1215 \pi+729 \pi$

$=1944 \pi \mathrm{cm}^{2}$

Volume = $\frac{1}{3}$ $\pi r l+\pi r^{2}h$

$=8748 \pi \mathrm{cm}^{3}$

(v) Given, 

Radius = 5cm,

Curved surface = 65π $\left(cm^{2}\right)$

  

∴ Curved surface =πrl 

65π = π $5\times l$

$\frac{65} π { π  \times 5}=l$

$l=13 \mathrm{~cm}$

$\because \quad l^{2}=h^{2}+r  ^{2}$

$(13)^{2}=h^{2}+(5)^{2}$

$16 y-25=h^{2}$

$144=h^{2}$

$\sqrt{144}=h$

$12=h$

$h=12 \mathrm{~cm} .$

Area of base = $πr^{2}$

$=\pi \times 5 \times 5$

$=25 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=65 \pi+25 \pi$

$=90 \pi \mathrm{m}^{2} .$

Volume $=\frac{1}{3}\pi r^{2} h$

$=\frac{1}{3} \pi \times(5)^{2} \times 12$

$=25 \times 4 \pi$

$=100 \pi \mathrm{cm}^{3}$

(vi) Given,

Radius = 35cm

Total surface area  = $13860 \mathrm{~cm}^{2}$

Total surface area = πrl+π $r^{2}$

$13860:=\pi r(l+r) .$

$4410 \pi=\pi \times 35(l+35) .$

$4410 \pi=35 \pi(l+75)$

$126=l+35$

$126-35=l$

$91=l$

$\ell=91 \mathrm{~cm} .$

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\(91)^{2} &=h^{2}+(35)^{2} \end{aligned}$

$8281-1225=h^{2}$

$\sqrt{8281-1225}=h$

$\sqrt{7056}=h$

$84=h$

$h=84 \mathrm{~cm}$

Curved surface = $\pi rl$

$=\pi \times 35 \times 91$

$=3185 \pi \mathrm{cm}^{2}$

Area of base $=\pi r^{2}$

$=\pi \times 35 \times 35$

$=1225 \pi c m^{2}$

Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi \times 35 \times 35 \times 84$

$=34300 \pi \mathrm{cm}^{3}$

Question 2

Ans: (i) Given, 

Height = 8m, 

Area of base = $156 \mathrm{~m}^{2}$

$\begin{aligned} \therefore \text { Area of base } &=\pi r^{2} \\ 156 &=\frac{22}{7} \times \r^{2} \\ \frac{156 \times 7}{22} &=r^{2} \end{aligned}$

$\frac{546}{11}=r^{2}$

$r^{2}=\frac{546}{11} \mathrm{~m}$

ஃ Volume = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{546}{11} \times 8$

$=2 \times 26 \times 8$

$=416 \mathrm{~m}^{3}$

(ii) Given,

Slant height (l)= 17cm,

Radius (r)= 8cm

$\therefore l^{2}=h^{2}+r^{2}$

$l^{2}-r^{2}=h^{2} .$

$(17)^{2}-(8)^{2}=h^{2}$

$\sqrt{289-64}=h .$

$\sqrt{225}=h$

$15=h$

$h=15 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 15$

$=176 \times 40$

$=\frac{7.040}{7}$

$=1005.71 \mathrm{~cm}^{3}$

(iii) Given, 

Height = 8cm,

Slant length = 10cm,

$\therefore \quad l^{2} = h^{2}+r^{2}$

$l^{2}-h^{2}=r^{2}$

$(10)^{2}-(8)^{2}=r^{2}$

$100-64=r^{2}$

$\sqrt{36}=r$

$6=r$

$r=6 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 8$

$=\frac{44 \times 48}{7}$

=301.71$cm^{3}$

(iv) Given, 

Height=5cm 

Perimeter of base = 8cm

Perimeter of base= $2 \pi r$

$8=2 \times \frac{22}{7} \times r$

$\frac{{8} \times 7}{2 \times 22}=6$

$\frac{14}{11}=r$

$r=\frac{14}{11} \mathrm{~cm}$

∴ Volume = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{14}{11} \times \frac{14}{11} \times 5$

$=\frac{4 \times 70}{33}$

$=\frac{280}{33}$

$=8.48 . \mathrm{cm}^{3}$

Question 3

Ans: (i) Given, 

Height = 8m,

Slant height = 10m

$\therefore l^{2}=h^{2}+r^{2}$

$l^{2}-h^{2}=r^{2} .$

$(10)^{2}-(8)^{2}=r^{2}$

$\sqrt{100-64}=r$

$\sqrt{36}=r$

$6=r$

r=6 m

∴ Curved surface area = πrl

$\begin{aligned} &=\frac{22}{7} \times 6 \times 10 \\=& \frac{1320}{7} \end{aligned}$

=199.6 $m^{2}$

(ii) Given, 

Perimeter of base = 88cm,

slant height = 2dm = $2 \times 10cm=20cm

∴ Perimeter of base $=2 \pi r$

88= $=2 \times \frac{22}{7} \times r$

$\frac{88 \times 7}{2 \times 22}=r$

r=14cm

Curved surface area $=\pi r l$

$\begin{aligned} & \frac{22}{7} \times 14 \times 20 \\=& 44 \times 20 \\=& 880 \mathrm{~cm}^{2} \end{aligned}$

(iii) Given,

Area of base= $154 \mathrm{~cm}^{2}$

Height 24cm

Area of base= $\pi r^{2}$

$154=\frac{22}{7} \times r^{2}$

$\frac{154 \times 22}{7}=r^{2}$

$49=r^{2}$

$\sqrt{49}=r$

$7=r$

$r=7 \mathrm{~cm}$

$\therefore \ell=\sqrt{h^{2}+r^{2}}$

l=$\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=25 \mathrm{~cm}$

So 

Curved surface area = π rl

$=22 \times 25$

$=550 \mathrm{~cm}^{2}$

Question 4

Ans: Given,

Radius = 5cm

Volume = 50π $\mathrm{Cm}^{3}$

Volume= $\frac{1}{3}π r^{2} h$

$50 \pi=\frac{1}{3} \pi \times 5 \times 5 \times \mathrm{h}$

$50 \pi=\frac{\pi}{3} \times 25 \mathrm{~h}$

$\frac{50 \times \pi \times 3}{\pi \times 25}=h$

$\frac{150}{25}=h$

$6=h$

$h=6 \mathrm{~cm}$

Hence, the height of the cone is 6cm

Question 5

Ans: Given,

Radius = 11.3cm

Curved surface area = $710(cm)^{2}$

Curved surface area $=\pi rl$

$710=\frac{355}{113} \times \frac{11.3}{10} \times l$

$710=\frac{355}{10} l$

$\frac{710 \times 10}{355}=l$

$\frac{7100}{355}=l$

$20=l$

$l=20 \mathrm{~cm}$

Hence , slant height is 20cm

Question 6

Ans:  Let the radius be r and height be h

Volume =$\frac{1}{3} \pi r^{2} h$

According to question 

Radius =$\frac{r}{2}$

and height= h

Volume = $\frac{1}{3} \pi\left(\frac{8}{2}\right)^{2} h$

$=\frac{1}{3} \pi \frac{r^{2}}{4} h=$ $\frac{\pi^{2} h}{12}$

∴ Ratio = $\frac{\pi r^{2} h}{12}=\frac{1}{3} \pi r^{2} h$

$=\frac{\frac{\pi r^{2} h}{12}}{\frac{1 \pi x^{2} h}{3}}$

$=\frac{1}{4}=1: 4 .$

Question 7

Ans: Given 

Curved surface area = $264 m^{2}$

Slant height = 12m

Curved surface area = πrl

$264=\frac{22}{7} \times r \times 12$

$\frac{264 \times 7}{22 \times 12}=r$

$7=r$

$r=7 \mathrm{~m}$

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\ l^{2}-r^{2} &=h^{2} \end{aligned}$

Question 8

Ans: According to question, 

Height (h) = $2 \times$ diameter 

Diameter = $\frac{\text { height }}{2}$

$\therefore \quad \operatorname{Radius}=\frac{h}{2} \times \frac{1}{2}=\frac{h}{4}$.

Volume $=36 \pi \mathrm{cm}^{3}$

$\frac{1}{3} \pi r^{2} h=36 \pi$

$\frac{1}{3} \times \pi \times \frac{h}{4} \times \frac{h}{4} \times \pi=36 \pi .$

$\frac{h^{3}}{48} \pi=36 \pi$

$h^{3}=\sqrt{728}$

$h=\sqrt[3]{1728}=\sqrt{12 \times 12 \times 12}=12 . \mathrm{cm}$

Question 9

Ans: Given,

Radius and height of cone are in ratio = 3:4

Let radius be 3x 

and height be 4x 

Volume = 301.44 $(cm)^{3}$

$\because \quad$ Volume =\frac{1}{3} \pi r^{2} h$

$301.44=\frac{1}{3} \times 3.14$  $\times(3x)^{2}$  $\times$(4x)$

$301.44=3.14 \times 12 x^{3}$

$\frac{301.44}{3.14 \times 12}=x^{3}$

$\frac{30144}{3768}=x^{3}$

$8 =x^{3}$

$x^{3}=8$

$x=\sqrt[3]{8}$

$x=\sqrt{2 \times 2 \times 2}$

$x=2$

$\begin{aligned} \therefore \text { Radius } &=3 \times \\ &=3 \times 2 \\ &=6 \mathrm{~cm} . \end{aligned}$

and Hight $=4 x$

$=4 \times 2$

=8

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\ l &=\sqrt{(8)^{2}+(6)^{2}} \\ l &=\sqrt{64+36} \\ l &=\sqrt{100} \\ l &=10 \mathrm{~cm} \end{aligned}$

Question 10

Ans: Given ,

The ratio of radius and slant height of cone = 4:7

Curved surface area $=792 \mathrm{~cm}^{2}$

Let radius be 4x and 

slant height be 7x

Curved surface area = πrl

$792=\frac{22}{7} \times 4 x \times 7 x$

$792=22 \times 4 x^{2}$

$792=88 x^{2}$

$\frac{792}{88}=x^{2}$

$9=x^{2}$

$x^{2}=9$

$x=\sqrt{9}$

$x=3$

Radius =4x

$=4 \times 3$

$=12 \mathrm{~cm}$

Question 11

Ans: Given, 

The ratio of radii of two cones = 3:5

Let r1 be 3x and r2 be 5x

Volume of first cone = $\frac{1}{3} \pi r_{1}^{2} h$

$=\frac{1}{3}π  (3x)^{2}\times h$

= $3 \pi x^{2} h .$

Volume of second cone =  $\frac{1}{3} \pi r_{2}^{2} h$

$\begin{aligned} &=\frac{1}{3} \pi \times (5 x)^{2} x h \\=& \frac{1}{3} \pi 25 x^{2} h \\=& \frac{25 \pi x^{2} h}{3} \end{aligned}$

ration of their volumes 

$=3 \pi x^{2}h \frac{25 \pi x^{2} 1}{3}$

$=\frac{3 \pi x^{2} h}{\frac{25 \pi x^{2} h}{3}}$

$=\frac{9}{25}$

$=9: 25$

Question 12

Ans: Given, 

Circumference = 44m

Height = 10m

Circumference = 2πr

$44=2 \times \frac{22}{7} \times r$

$\frac{44 \times 7}{44}=r$

$7=r$,

$\therefore l^{2}=h^{2}+r^{2} .$

$l=\sqrt{(10)^{2}+(7)^{2}}$

$l=\sqrt{100+49}$

$l=\sqrt{149 \mathrm{~m} .}$

∴ Curved surface area = πrl

$=\frac{22}{7} \times 7 \times \sqrt{149}$

$\begin{aligned} &=22 \sqrt{149}\\=& 22 \times 12.2 . \\=& 268.4 \mathrm{~m}^{2} . \end{aligned}$

Width of canvas(d)=  2cm= $\frac{2}{100} \mathrm{~m}$

$\therefore$ Area $=l \times b$

$268.4=l \times \frac{2}{100}$

$\frac{26840}{2}=l$

=13429=l

Hence, the length of canvas is 13420m

Question 13

Ans: Given, 

Radius =7m

Height = 24m

$\because$ Slant height $(l)=\sqrt{h^{2}+r^{2}}$

$l=\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=\sqrt{2.5 \times 25}$

$l=25 \mathrm{~m}$

Curved surface area =  πrl

$=\frac{22}{7} \times 7 \times 25$

$=22 \times 25$

$=550 \mathrm{~m}^{2}$

Width of canvas (b) = 5m

Area = L $\times$ B

$550=L \times 3$

$\frac{550}{5}=L$

$110=L$

$L=110 \mathrm{~m}$

$\therefore$ Length $=110 \mathrm{~m}$.

Question 14

Ans: Given,

Volume =  $1232 \mathrm{~m}^{3} .$

Area of base = $154 \mathrm{~m}^{2}$

Area of base = $\pi r^{2}$

$154=$ $\frac{22}{7} \times r^{2}$

$\frac{154 \times 7}{22}=r^{2}$

$49=r^{2}$

$\sqrt{49}=r$

$7=r$

$r=7 \mathrm{~m}$

Volume $=\frac{1}{3} \pi r^{2} h$

$1232=$ $\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h$

$\frac{1232 \times 3}{22 \times 7}=h $

$8 \times 3=h$

$24=h$

$h=24 m$

$l^{2}=h^{2}+x^{2}$

$l=\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=\sqrt{25 \times 25}$

$l=25 \mathrm{~m}$

ஃ Curved surface area = πrl

$=\frac{22}{7} \times 7 \times 25$

$=22 \times 25$

$=550 \mathrm{~m}^{2}$

∴ The area of canvas $=550 \mathrm{~m}^{2}$

Question 15

Ans: Given, 

In the tent, accommodate is available for 11 persons and each person must have $4r^{2}$ of space on the ground 

So , aera of base of the tent = $11 \times{4}$

$44 m^{2}$

Air is required for each person to breadth = $20m^{3}$

Volume of air = $20 \times 11$

$=220 \mathrm{~m}^{3}$

Area of base = π$r^{2}$

$44=\frac{22}{7} \times r^{2}$

$\frac{44 \times 7}{22}=r^{2}$

$r^{2}=14 \mathrm{~m} .$

Volume = $=\frac{1}{3} \pi{r}^{2} h$

$220=\frac{1}{3}\times $\frac{22}{7}\times 14 \times h$

$220=\frac{44h}{3}$

$\frac{220 \times 3}{44}=h$

$15=h$

h=15 m

Hence , the height of the cone = 15m

Question 16

Ans: False , because the volume of a cone is one third $\left(\frac{1}{3}\right)$ of the volume of cylinder of the same radius and height 

Question 17

Ans: Given,

Height of cylinder = 9cm

and radius= $\frac{40}{2} cm=20 c m$

Height of cone = 108cm

∴ Volume of cylinder = $\pi r^{2} h$

$=\frac{22}{7} \times 20 \times 20 \times 9$

$=\frac{440 \times 180}{7}$

$=\frac{79200}{7} \mathrm{~cm}^{2} .$

ஃ Volume of cone $\frac{79200}{7}$ (Given volume of cylinder is equal to volume of cone)

Volume of cone = $=\frac{1}{3} \pi r^{2}$h

$\frac{79200}{7}=\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 108$

$\frac{79200 \times 7 \times 3}{7 \times 22 \times 108}=r^{2}$

$\frac{79200 \times 21}{7 \times 22 \times 108}=r^{2}$

$\frac{79200}{792}=r^{2}$

$100=r^{2}$

$r^{2}=100$

$r=\sqrt{100}$

$r=10$

Hence, the radius of cone is 10cm

Question 18

Ans: Given, 

Radius, of cone and cylinder (r) = 7m,

Height of the cylinder (h1)=8m

And height of the conical of (h2)= 4m

(IMAGE TO BE ADDED)

∴ l =$\sqrt{h2^{2}+r^{2}}$

$=\sqrt{(4)^{2}+(7)^{2}}$ $=\sqrt{16+49}$ $=8.06M$

∴ Area of canvas = Curved surface area of cylinder + Curved surface area of cone 

= 2πrh_{1}$ + πrl

$=2 \times \frac{22}{7} \times 7 \times 8+\frac{22}{7} \times 7 \times 8.06 .$

$=352+177.32 .$

$=529.32 . \mathrm{m}^{2}$

Question 19

Ans: Given, 

Height of cylinder = 3m 

Radius of cylinder = $\frac{105}{2} \mathrm{~m}$

and slant height = 53m

Total area of the canvas = Curved surface area of cylinder +curved surface area of cone 

= 2πrh+πrl
$=22 \times \frac{22}{7} \times $3+\frac{2 x}{7}\times\frac{105}{2} \times 53$

$=22 \times 45+11 \times 795 .$

$=990+8745$

$=9735 \mathrm{~m}^{2}$

Question 20

Ans: Given, 

Edge of a cube = 9cm

Diameter of cone = 9cm (because equal to edge of cube)

Then radius = $\frac{9}{2} cm$

and height of cone = 9cm (because equal to edge of cube)

Volume of largest cone = $\frac{1}{3}πr^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 9$

$=\frac{99 \times 27}{14}$

$=\frac{2673}{14}$

$=190.93 . \mathrm{cm}^{3}$

Question 21

Ans:  Given,

Height of cylinder(h1) = 3m, 

Total height of tent = 13.4m

Height of conical part (h2) = 13.5-3

=10.5m

(IMAGE TO BE ADDED)

Radius = 14 m

$l=\sqrt{h_{2}^{2}+r^{2}}$

$l=\sqrt{(10.5)^{2}+(14)^{2}}$

$l=\sqrt{110.25+196}$

$l=\sqrt{306.25}$

$l=17.5 \mathrm{~m}$

Question 22

Ans:  (IMAGE TO BE ADDED)

Given, 

Height of the cylinder (h1)= 32cm

and radius (r1) = 18cm

Height of the conical= 24 cm

Volume of sand in it = π$r_{1}^{2}$ h

$=\frac{22}{7} \times 18 \times 18 \times 32$

$=\frac{396 \times 576}{7}$

$=\frac{228096 \mathrm{cm}^{3}}{7}$

(IMAGE TO BE ADDED)

So volume of conical heap of the sand = $\frac{228096 \mathrm{~cm}^{3}}{7}$

Volume of conical heap = $\frac{1}{3} \pi r^{2} h$

$\frac{228096}{7}=$ $\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24$

$\frac{22.8096}{7}=$  $\frac{176}{7} r^{2}$

$\frac{228096 \times 7}{7 \times 176}$ $=r^{2}$

$\frac{228096}{176}=r^{2}$

$1296=r^{2} .$

$r^{2}=1296$

$r=\sqrt{1296}$

$r=36 \mathrm{~cm}$

(i) Radius of cone = 36cm

(ii) $l=\sqrt{h^{2}+r^{2}}$

$l=\sqrt{(24)^{2}+(36)^{2}}$

$l=\sqrt{576+1296}$

$l=\sqrt{1872}$

l=43.3cm

Height the slant height of heap is 43.3cm

Question 23

Ans:  (IMAGE TO BE ADDED)

Given, 

Height cylinder = 8cm

Radius of cylinder = 6cm

Volume of cylinder = $\pi r^{2} h$

$=31416 \times 6 \times 6 \times 8$

$=18.8496 \times 48$

$=904.7808 \mathrm{~cm}^{3}$

Radius of cone = 6cm 

Height of cone = 8cm

Volume of cone = $\frac{1}{3} \pi r^{2}h$

$=62832 \times 48$

$=301.5936 \mathrm{~cm}^{3}$

 

∴ Volume of remaining solid = Volume of cylinder - Volume of cone 

$=904.7808-301.5936$

$=603.1872 \mathrm{~cm}^{3}$

Question 24

Ans: (IMAGE TO BE ADDED)

Given,

Radius , of cylinder = 3cm 

and height of cylinder = 5cm 

Volume of cylinder = $\pi r^{2} h$

$=\frac{21}{7} \times 3 \times 3 \times 5$

$=\frac{990}{7} \mathrm{~cm}^{2}$

Radius of conical portion = $\frac{3}{2} c m$

and Height of conical portion = $\frac{8}{9} c h$

Volume of conical portion = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{8}{9}$

$=\frac{44}{21} \mathrm{~cm}^{3}$

Metal in the remaining part = Volume of cylinder - Volume of conical portion 

$=\frac{990}{7}-\frac{44}{21}$

$=\frac{2970-44}{21}$

$=\frac{2926}{21}$

According to the question 

$\frac{2926}{21}: \frac{44}{21}$

$\frac{\frac{2926}{21}}{\frac{4 4}{21}}$

$=\frac{2926 \times 21}{21 \times 44}$

$\frac{1463}{22}=$

$=133: 2$

Question 25

Ans:   (IMAGE TO BE ADDED)

Given, 

Radius of cylinder = $\frac{7}{2} \mathrm{~cm}$

Its height = 8cm

Volume of cylinder = $\pi r^{2} h$

$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 8$

$=22 \times 14$

$=308 \mathrm{~cm}^{3} .$

Radius of cone $=\frac{7}{4} \mathrm{~cm}$

Its height = 8cm

Volume of cone = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7}\times \frac{7}{4} \times \frac{7}{4} \times 8$

$=\frac{77}{3} \mathrm{~cm}^{3}$

Volume of water required to fill the vessel= Volume of cylinder -Volume of cone 

$=308-\frac{77}{3}$

$=\frac{924-77}{3 .}$

$=\frac{847}{3}$

$=28 \frac{1}{3} \mathrm{~cm}^{3}$

According to question,

Height of cone = $1 \frac{3}{4}=\frac{7}{7} \mathrm{Cm} x$

its radius = 2cm

Volume of cone =  $\frac{1}{3} \pi r^{2} h$

=$\frac{22}{3} \mathrm{~cm}^{3}$

Change in volume of cones 

$=\frac{77}{3}-\frac{22}{3}$

$=\frac{77-22}{3}$

$=\frac{55}{3} \mathrm{~cm}^{3}$

Let the drop in water level be h cm

Volume = π$r^{2} h$

$\frac{55}{3}=$ $\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h $

$\frac{55 \times 2}{3 \times 11 \times 7}=h$

$\frac{10}{21}=h $

$h=\frac{10}{21} \mathrm{~cm}$

S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 B

  Exercise 15 B

Question 1 

Ans: (i) Given,

r= 7cm, 'h=8cm

∴ Volume = $\pi r^{2} h$

$=\frac{22}{7} \times 7 \times 7+8$

$=22 \times 56$

$=1232 \mathrm{~cm}^{3}$

(ii) Given,

$\begin{aligned}&r=7 \mathrm{~cm} \\&h=12 \mathrm{~cm} .\end{aligned}$

$\therefore$ Volume $=\pi r^{2} h$

$\begin{aligned} &=\frac{22}{7} \times 7 \times 7 \times 12 \\ &=22 \times 84 \\ &=1848 \mathrm{~cm}^{3} \end{aligned}$

(iii) Given,

$r=14 \mathrm{~cm}$

$\mathrm{h}=16 \mathrm{~cm} .$

$\therefore \quad$ Volume $=\pi r^{2} h$

$-\frac{22}{7} \times 14 \times 14 \times 6$

$=44 \times 224$

$=9856 \mathrm{~cm}^{3}$

(iv) Given, 

r= 21cm,

h=40cm

Volume = $\pi r^{2} h$,

$=\frac{22}{7} \times 21 \times 21 \times 40$

$=66 \times 840$

$=\quad 55440 \mathrm{~cm}^{3}$

Question 2

Ans: (a) Given, 

Volume, = $44 \mathrm{Cm}^{3}$

Height $=3.5 \mathrm{~cm}$

ஃ Volume = $\pi r^{2}h$

$44=\frac{22}{7}\times r^{2}\times \frac{3.5}{10}$

$ 44=11 r^{2}$

$ \frac{44}{11}=r^{2}$

$4=r^{2}$

$r^{2}=4 .$

$r=\sqrt{4} .$

$r=2 \mathrm{cm} $

∴ Diameter = 2r 

$=2 \times 2$

$=4 c \mathrm{~m}$

(b) Given, 

Volume , = $385 \mathrm{~cm}^{3}$

Height $=1 \mathrm{dm}=10 \mathrm{~cm}$

$\therefore \quad$ Volume $=\pi r^{2} h$

$385=\frac{22}{7} \times r^{2} \times 10$

$\frac{49}{4}=r^{2}$

$r^{2}=\frac{49}{4}$

$r=\sqrt{\frac{49}{4}}$

$r=\frac{7}{2}$

∴ Diameter = 2r = $2 \times \frac{7}{2}=7 \mathrm{~cm}$

 

Question 3

Ans: (a) Given,

Volume = $66 \mathrm{~cm}^{3}$

Radius = 2cm

∴ Volume = π$(r^{2}$h

$\frac{66 \times 7}{22 \times 4}=h$

$\frac{21}{4}=h$

$h=\frac{21}{4}$

$h=5.25 \mathrm{~cm}$

(b) Given,

Volume = 4litres= 4000 $\mathrm{Cm}^{3}$

Radius= 5cm

$\therefore \quad$ Volume $=\pi r^{2} h$

$4000=\frac{22}{7} \times 5 \times 5 \times h$

$\frac{4000 \times 7}{22 \times 25}=h$

$\frac{80 \times 7}{11}$

$\frac{560}{11}=h$

$h=\frac{510}{11} \mathrm{~cm}$

Question 4

Ans: Given,

Height(h) = 7m

Radius(r) = $\frac{20}{2}=100 \mathrm{~cm}$ $=\frac{10}{100}$ $=\frac{1}{10}m$

$\therefore$ Volume $=\pi r^{2} h$

$=\frac{11}{50}$

According to question,

Total weight $=\frac{11}{50} \times 225$

$=\frac{99}{2}$

$=49.5 \mathrm{~kg}$

Question 5

Ans: Giver,

Internal radius. $(r)=3 \mathrm{~cm}$

Thickness of pipe= $1 \mathrm{~cm}$

$\therefore$ Outer radius $(R)=3+1$=4cm

Length = 6cm

$\begin{aligned} \therefore \quad \text { Volume } &=\pi R^{2} h-\pi r^{2} h . \\ &=\pi h\left(R^{2}-r^{2}\right) . \\ &=\frac{22}{7} \times 6\left((4)^{2}-(3)^{2}\right) \end{aligned}$

$=\frac{132}{7}(16-9)$

$=\frac{132}{7} \times 7 $

$=132 \mathrm{cm}^{3}$

Question 6

Ans: Given,

Sum of radius, of the base and the height of a cylinder (h+r) = 37cm,

Total surface area = $1628 \mathrm{~cm}^{2}$

∴ Total surface area = $2πr h+2πr^{2}$

$1628=2 \pi r(h+r)$

$1628=$ $2 \times \frac{22}{7} \times r \times 37 .$

r=7

$\therefore r=7 \mathrm{~cm} .$

So, $h+r=37$

$h+7=37$.

$h=377$

$h=30 \mathrm{~cm}$.

$\begin{aligned} \text { Volume } &=\pi r^{2} h \\ &=\frac{22}{7} \times 7 \times 7 \times 30 \\ &=22 \times 210 \\ &=4620 \mathrm{~cm}^{3} \end{aligned}$

Question 7

Ans: Given

Capacity of a cylindrical tank = $6160 \mathrm{m}^{3}$

Radius $(r)  \frac{28}{2}=14 \mathrm{~m}$

$\therefore$ Volume $=\pi r^{2} h$

$6160=\frac{22}{7} \times 14 \times 14 \times h$

$\frac{6160}{22 \times 281}=h $

h= 10m

Area of curved surface of taken inner sides = $2 \pi r h$

$=2 \times \frac{22}{7} \times 14 \times 10$

$=44 \times 20$

$=880 \mathrm{m}^{2}$

Question 8

Ans: Given,

Curved surface area of cylinder = $4400 \mathrm{~cm}^{2}$

Circumference $=110 \mathrm{~cm} $

Circumference $=2 \pi r$

$110=2 \times \frac{22}{7} \times r$

$\frac{110 \times 7}{2 \times 22}=r$

$\frac{35}{2}$=r

$r=\frac{35}{2} \mathrm{~cm}$

(i) Curved Surface area $=2$ πrh

4400= $=\frac{2}\times \frac{22}{7} \times \frac{35}{2} \times h$

$\frac{4400}{22 \times 5}=h$

$40=h$

$h=40$ cm

hence the height of cylinder is 40cm

(ii) Volume = $\pi r^{2} h$

$=110 \times 350$

$=38500 \mathrm{~cm}^{3}$

Question 9

Ans: (i) Given ,

Height of the wall = 20 meter 

Radius = $\frac{2}{2}=1$

$\because$ Volume =  πr^{2}$

$\begin{aligned} &=\frac{22}{7} \times 1 \times 1 \times 20 \\=& \frac{440}{7} \\=& 62 \frac{6}{7} \mathrm{~m}^{3} . \end{aligned}$

(ii) Curved surface area = $2 π r h$

$\begin{aligned} &=2 \times \frac{22}{7} \times 1 \times 20 \\=& \frac{44 \times 20}{7} \\=& \frac{880}{7} \end{aligned}$

Rate of plastering the inner surface = Rs 5  per m².

Total cost = 880\7 x 5

$=\frac{4400}{7}$

$=7628.57 $

Question 10

Ans: Giver

Radius of cylinder $=\frac{20}{2}=10 \mathrm{~cm}

Curved surface area = $1000 \mathrm{~cm}^{2}$

(i) Curved surface area $=2 \pi r$

$1000=2 \times \frac{22}{7} \times 10 \times h$

$\frac{175}{11}=h$

$h=\frac{175}{11}$

$h=15.9 \mathrm{~cm} .$

(ii) $\begin{aligned} \text { Volume } &=\pi r^{2}h \\ &=3.14 \times 10 \times 10 \times 15-9 . \\ &=31.4 \times 159 . \\ &=4992.6 \mathrm{~cm}^{3}\end{aligned}$

Question 11

Ans: Given, 

Radius =  $\frac{35}{2} \mathrm{~cm}$

height $=1.2 \mathrm{~m}=1.2 \times 100 \mathrm{~cm}=120 \mathrm{~cm}$.

(i) Outer lateral surface area = $2 \pi r h$

$=110 \times 120$

$=13200 \mathrm{~cm}^{2} $

(ii) Capacity = $\pi r^{2} h$

 $=55 \times 2100$

$=115500 \mathrm{~cm} .$

=115.5 liters 

Question 12

Ans: Given, 

Radius of cylindrical glass = $\frac{8}{2}=4 \mathrm{~cm}$

and its height = 15cm

Radius, of cylindrical vessel = $\frac{30}{2}=15 \mathrm{~cm}$

and its height = 80 cm

Volume of cylindrical glass= $=\pi r^{2} \mathrm{~h}$

$\pi \times 4 \times 4 \times 15$

$=240 \pi$

Volume of cylindrical vessel $=\pi r^{2} h$

$=\pi \times 15 \times 15 \times 80$

$=\pi \times 225 \times 80$

$=18000 \pi$.

$\therefore$ Number of glasses $=\frac{\text { Volume of vessel }}{\text { Volume of glass}}

$=\frac{18000 \pi}{240 \pi}$ 

$=75$ glasses

Question 13

Ans: Given, 

R= 22m

r= 20 

h=$\frac{7}{100} m$

$\pi R^{2} h-\pi{r}^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times \frac{7}{100}(R+r)(R-r)$

$=\frac{22}{7}\times{7}{100}(22) \times 18=18.48 \mathrm{~mm}$

Question 14

Ans: Given

Radius of iron cylindrical block = $=\frac{0.5m}{20}$ =$\frac{1}{4} \mathrm{~m}=$0.25m= $0.25\times $100cm  =25cm

Length = 3.5m= 3.5 $\times 100$cm = 350 cm

Volume of block =  $\pi r^{2} b$

$=\frac{22}{7} \times 25 \times 25 \times 350$

$=550 \times 1250$

$=687500 \mathrm{~cm}^{3}$

So , volume of base = $687500 \mathrm{~cm}^{3}$

Area of square base $=25 \times 25$

$=625 \mathrm{~cm}^{2}$

Height of bar= $\frac {Volume of bar}{Area of square base}$

$=\frac{687500}{625}$

$=1100 \mathrm{~cm} $

$=\frac{1100}{100}$

$=11 \mathrm{~m}$

Question 15

Ans: Given, 

Length of swimming pole (l) = 70m

Breadth (b)= 44m

and depth (h) =  3m

Volume = lbh

$=70 \times 40 \times 3$

$=924012^{3}$

Radius of pipe= $\frac{14}{2} c m=7 cm=\frac{7}{160} \mathrm{~m}$

Volume = $πr^{2}h

9240= $\frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times h$ (Volume = 9240)

$\frac{9240 \times 100 \times 100}{22 \times 7}=h$

$\frac{92400000}{154}=h .$

$600000=h .$

$\therefore h=600000$

Let be the distance 

$\therefore \quad$ Distance $=$ speed \times $ Time.

$600000=2 \times$ Time. (speed = 2m)

$\frac{600000}{2}=$ Time.

$83 \frac{1}{3}$ hours

Question 16

Ans: Given, 

External radius of a hollow cylinder = $\frac{12}{2}=6 \mathrm{~cm}$

Internal radius of a hollow cylinder = 6_ 0.25

= 5.75cm

Length (h)= 15cm

Volume of hollow cylinder = $\pi R^{2} h-\pi r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 15\left((6)^{2}-\left(5.70^{2}\right)\right.$

$=\frac{330}{7}(36-33.0625) .$

$=\frac{330}{7} \times 2.9375$

$=\frac{969.375}{7 }$

Radius of solid cylinder $\frac{2}{2}=1 \mathrm{~cm}$ (given).

$\begin{aligned} \therefore \text { Volume } &=\pi r^{2} h \\ \frac{969-375}{7} &=\frac{22}{7} \times 1 \times 1 \times h \end{aligned}$

$\frac{969.375}{22}=h$ 

$\therefore h=\frac{969.375}{22}$

$h=44.0625 \mathrm{~cm} $

Question 17

Ans: Given, 

Internal radius of tube= $\frac{11.2}{20} \mathrm{~cm}=5.6 \mathrm{~cm}$

$\operatorname{Length}(h)=21 \mathrm{~cm}$.

Thickness $=0.4 \mathrm{~cm}$.

∴ Outer radius 5.6+0.4

=6cm

Volume of Metal = $\pi R^{2} h - r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$\left.=\frac{22}{7}\times 21 \times(6)^{2}-(5-6)^{2}\right)$

$=66 \times(36-31.36)$

$=66 \times 464$

$=306.24 \mathrm{cm}^{3}$

$=306.2 \mathrm{~cm}^{3}$

Question 18

Ans: Given, 

Volume of water = 1000 lit = $1000 \times 1000 \mathrm{~cm}^{3}=1000000 \mathrm{~cm}^{3}$

Radius of pip = 0.6cm

∴ Volume = $\pi r^{2} h$

$1000000=\frac{22}{7} \times 0-6 \times 0.6 \times h .$

$\frac{1000000 \times 7}{22 \times 0-6 \times 0.6}=h .$

$\frac{700000000}{7.92}=h .$

$883838.38=h .$

$h=883838-38 \mathrm{~cm}$

Let height (h) be the distance, 

Distance = speed $\times$ times 

$883838.38=8 \times$ time.

$\frac{88383838}{800}=$ time

$110479.7975=$ Time

Time = 110479.7975sec

Time = $\frac{110479.7975}{60 \times 60}$ hours 

$=\frac{11047967975}{36000060}$ hours

$=30.69$ hours 

Question 19

Ans: Given,

Radius =$\frac{28}{2} \mathrm{~cm}=14 \mathrm{~cm}$

Height = 72cm

$\therefore$ Volume $\pi r 2 h$

$=\frac{22}{7} \times 14 \times 14 \times 72$

$=44 \times 1008$

$=44352 \mathrm{~cm}^{3}$

Length f tank = 66cm

Breadth of tank = 28cm

Volume = $=l \times b \times h$

$\begin{aligned} 44352 &=66 \times 28 \times h . \\ 44352 &=1848 h . \\ \frac{44352}{1848} &=h . \\ 24 &=h \\ \therefore h &=24 \mathrm{~cm} . \end{aligned}$

Hence, the height of the water level in the tank is 24 cm

Question 20

Ans: Given,

Radius of cylindrical vessel = $\frac{14}{2} c m=7 cm$ 

Height of water = $8 \frac{9}{14}(cm )=\frac{121}{14}$

Volume of water= $\pi r^{2} h$

$=1331 \mathrm{~cm}^{3}$

Volume of cube = $=1331 \mathrm{~cm}^{3}$

Volume of cube= $a^{3}$

$1331=a^{3}$

$\sqrt[3]{1331}=a$

$\sqrt{11 \times 11 \times 11}=a$

$a=11$

Hence , the length of the edge is 11cm

Question 21

Ans: Let the radius of the cylinder = r

And height = h

Volume = $\pi r^{2} h$

If the radius is halved 

$\therefore \quad r=\frac{r}{2}$

Height = h

$\therefore Volume=\pi\left(\frac{r}{2}\right)^{2}h $

$=\pi \frac{r^{2}}{4} h$

According to question 

$=\frac{\pi r^{2}}{4} h=\pi r^{2} h$

$=\frac{1}{4}$= 1:4

Question 22

Ans: Given,

Length of sheet = 22cm

and breadth = 12cm

If it is folded breadth wire, 

∴ Circumference = 12cm

and height = 22cm

Circumference = 2πr

$12=2 \times \frac{22}{7} \times r$

$\frac{21}{11}=r$

$y=\frac{21}{11} \mathrm{~cm}$

∴ Volume =πz $=r^{2} h$

$=\frac{22}{7} \times \frac{21}{11} \times \frac{21}{11} \times 22$

$=\frac{462 \times 42}{77}$

$=\frac{19404}{72}$

$=269.5 \mathrm{~cm}^{3}$

IF it is folded length wire 

$\therefore$ circumference  $=22 \mathrm{~cm}$. 

and Height: $12 \mathrm{~cm}$.

So circumference =$2 \pi r$

$22=2 \times \frac{22}{7} \times r$

$\frac{22 \times 7}{2 \times 22}=r$

$\frac{7}{2}=r$

$r=\frac{7}{2} cm$

Volume =$\left.\pi r^{2}\right)h$

$\frac{27}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12$

$=22 \times 21$

$=482cm^{3}$

According to question, 

$=462-269-5$

$=192.5 \mathrm{~cm}^{3}$

Question 23

Ans: Given, 

Depth of a wall = 20m

Radius = $\frac{7}{2} \mathrm{~m}$

Volume of earth = $\pi r^{2} h$

$=22 \times 35$

$=770 \mathrm{m}^{3}$

Length of platform= 22m,

Breadth = 14m

Volume of platform = $770 \mathrm{~m} 3$

Volume = lbh 

770= $22 \times 14 \times h$

$\frac{770}{22 \times 14}=h$

$\frac{770}{308}=h$

$2-5=h$

$h=2.5 \mathrm{~m}$

Hence , the height of the platform is 2.5m

Question 24

Ans: Given,

Height of cylindrical barrel of pen = 7cm

Radius = $\frac{5}{2} \cdot mm =$ $\frac{5}{2} \times \frac{1}{10} \mathrm{~cm}$=1\4cm

Volume of ink in it =  $\pi r^{2} h$

$-\frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times 7$

$=\frac{11}{8} \mathrm{~cm}^{3}$

Volume of link in bottle =  $\frac{1}{5}l=\frac{1}{5} \times 1000 \mathrm{~cm}=200 \mathrm{cm}^{3}$

$\therefore$ Total number of barrels $=200 \div \frac{11}{8}$

$=200 \times \frac{8}{11}$

$=\frac{1600}{11}$

Word written in one barrel = 310 words 

Total number of words = $\frac{1600}{11} \times 320$

$=\frac{496000}{11 .}$

$=45080.90 .$

$=45090$ words 

Question 25

 

Ans: Given , 

Length of a rectangular box= 40cm,

Breadth = 30cm and 

Height = 25cm

Volume = lbh 

$=40 \times 30 \times 25$

$=1200 \times 25$

$=30000 \mathrm{~cm}^{3}$

 Radius of cylindrical tin= 17.5cm

Volume = $30000 \mathrm{~cm}^{3}$

Volume $=\pi r^{2} h$

$30000=3-14 \times 17.5 \times 17.5 \times h $

$30000=3-14 \times 306.25 \dot{x h}$

$30000=961.625h $

$\frac{30000000}{961625}=h$

$31 \cdot 2$

$h=31-2 \mathrm{~cm}$

Hence the height of the cylindrical tin is 31.2cm

Question 26

 

Ans: Given, 

Diameter of a circular tank = 17.5m

So, the radius = $\frac{175}{20}=\frac{35}{4}=$ 8.75m

Outer radius = 8.75+4

= 12.75m

Height= 2m 

Volume of the embankment = $\pi R^{2} h-\pi r^{2} h$

$\begin{aligned} & \pi h\left(R^{2}-r^{2}\right) \\=& \frac{22}{7} \times 2\left((12-75)^{2}-(8.75)^{2}\right) \\=& \frac{44}{7}(162.5625-76.5625) \\=& \frac{44}{7} \times 86 . \\=& \frac{3784}{7} \\ 540.57 \mathrm{~m}^{3} . \end{aligned}$

 

∴ Volume of earth of the tank = $540.57 \mathrm{~m}^{3}$

$\begin{aligned} \therefore \text { Volyme } &=\pi r^{2} h \\ 540.57 &=\frac{22}{7} \times 8.75 \times 8.75 \times h . \end{aligned}$

$540.57=\frac{1684.375 \mathrm{~h}}{7}$

$2.25=h$

$h=2.25 \mathrm{~m} .$

Hence the depth of the circular tank is 2.25m

Question 27

 

Ans: Given, 

Speed of water = $7 m=700 \mathrm{~cm}$

Internal radius= $\frac{2}{2} c m=1 c m$

Radius of tank = 40cn

Time =$\frac{1}{2}$ hour $=$ $\frac{1}{2} \times 60 \times 60=\frac{3600}{2} \mathrm{sec}=$

=1800sec 

∴ Distance  (h) = Times into speed 

$=1800 \times 700$

$=1260000 \mathrm{~cm}$

Volume $=\pi r^{2} \mathrm{~h}$. 

=$\frac{22}{7}1 \times1 \times \times 1260000$

$=3960000 \mathrm{~cm}^{3}$

∴ Volume of water in the tank =$3960000 \mathrm{cm}^{3} $

Volume $=\pi r^{2} \mathrm{h}$

$3960000=\frac{22}{7} \times 40 \times 40 \times h$.

$\frac{17325}{22}=h$

$787.5=h$

$h=787.5 \mathrm{cm} $

Question 28

 

Ans: Given, 

Radius of pipe = $\frac{7}{2} cm=\frac{7}{200}$ m

Speed = 36 km\hr = $36 \times \frac{1000}{60}$ = 600m

Radius of tank = $35 \mathrm{~cm}=\frac{35}{100} \mathrm{~m}$

Height = 1m

∴ Volume of tank =π$r^{2}h$

$\frac{22}{7} \times \frac{35}{100} \times \frac{35}{100} \times 1$

$=\frac{77}{200} \mathrm{~m}^{3}$

$\because$ Volume $=\pi r^{2} h .$

$\frac{77}{200}=\frac{22}{7} \times \frac{7}{200} \times \frac{7}{200} \times h .$

$\frac{2200}{221}=h$

$100=h .$

$h=100$

Let height (h) be the distance 

Distance = speed into Time

100= $600 \times Time$

Time $=\frac{100}{600}$

Time $=0.167 \mathrm{~min} $

Question 29

 

Ans: Given, 

Radius of coin =$\frac{1.5}{20} \mathrm{~cm}=\frac{3}{4} \mathrm{~cm}$

and its thickness = 0.2 cm

Volume of one coin = $\pi r^{2} h$

$=\frac{22}{7} \times \frac{3}{4} \times \frac{3}{4} \times 0.2 .$

$=\frac{39.6}{112}$

$=0.35 \mathrm{~cm}^{3}$

Radius of cylinder = $\frac{4.5}{20}=\frac{9}{4} \mathrm{~cm}$

Height = 10cm

Volume of cylinder = $\pi r^{2} h$

$\begin{aligned} &=\frac{22}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \\=& \frac{17820}{112} \\=& 159.11 \mathrm{~cm}^{3} . \end{aligned}$

Volume of cylinder = volume of one coin X Number of coin

159.11 = $0.35 \times$ Number of coin.

$\frac{159 \cdot 11}{0.35}$ =Number of coin.

$454.6=$ Number of coin

Number of coin $=454.6$

Question 30

 

Ans: Given

Weight of  $1cm^{3}=21 g$

Length of pipe = $1 \mathrm{~m}=100 \mathrm{~cm}$

Internal radius = $\frac{3}{2} \mathrm{~cm}=1.5 \mathrm{~cm}$

External radius = 1.5+1= 2.5cm

∴ Volume = Volume of outer - Volume of inner surface 

$π R^{2} h$-$π r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 100\left((2.5)^{2}-(1.5)^{2}\right)$

$=\frac{2200}{7} \times(6.25-2.25) .$

$=\frac{2200}{7} \times 4 $

$=\frac{8800}{7} \mathrm{~cm}^{3} $

∴ Total Weight of metal = $\frac{8800}{7} \times {21}$

$=26400 \mathrm{~g}$