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MATHS [4 marks sum]

Mensuration · ICSE STD 7 (ICSE - English Medium). 28 questions.

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[4 marks sum]

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Question 14 Marks
A floor is 40 m long and 15 m broad. It is covered with tiles, each measuring 60 cm by 50 cm. Find the number of tiles required to cover the floor.
Answer

$\begin{aligned} & \text { Length of floor }(\mathrm{l})=40 \mathrm{~m} \\ & \text { breadth of floor }(\mathrm{b})=15 \mathrm{~m} \\ & \therefore \text { Area of floor }=\mathrm{I} \times \mathrm{b}=40 \times 15=600 \mathrm{~m}^2 \\ & \text { Length of one tile }=60 \mathrm{~cm}=\frac{6}{10} \mathrm{~m} \\ & \text { and breadth }=50 \mathrm{~cm}=\frac{5}{10} \mathrm{~m} \\ & \therefore \text { Area of one tile }=\frac{6}{10} \times \frac{5}{10} \\ & =\frac{30}{100}=\frac{3}{10} \mathrm{~m}^2 \\ & \therefore \text { Number of tiles }=\frac{\text { Total area of floor }}{\text { Area of one tile }} \\ & =\frac{600}{\frac{3}{10}}=\frac{600 \times 10}{3}=2000\end{aligned}$
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Question 24 Marks
From each corner of a rectangular paper (30 cm x 20 cm) a quadrant of a circle of radius 7 cm is cut. Find the area of the remaining paper i.e., shaded portion.
Answer

$ \begin{aligned} & \text { Length of paper }(I)=30 \mathrm{~cm} \\ & \text { and breadth }(b)=20 \mathrm{~cm} \\ & \therefore \text { Area of rectangular paper }=I \times b \\ & =30 \times 20=600 \mathrm{~cm} \end{aligned} $
Radius of each quadrant at the corner $=7 \mathrm{~cm}$
Area of 4 quadrants $=4 \times \frac{1}{4} \pi r^2$
$ =\pi r^2=\frac{22}{7} \times 7 \times 7=154 \mathrm{~cm}^2 $
$\therefore$ Area of remaining paper
$ =600-154=446 \mathrm{~cm}^2 $
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Question 34 Marks
The inner circumference of a circular track is 264 m and the width of the track is 7 m. Find:
(i) the radius of the inner track.
(ii) the radius of the outer circumference.
(iii) the length of the outer circumference.
(iv) the cost of fencing the outer circumference at the rate of ₹50 per m.
Answer
Inner circumference of the circular track $=264 \mathrm{~m}$
(i) $\therefore$ Inner radius $(r)=\frac{C}{2 \pi}$
$ =\frac{264 \times 7}{2 \times 22}=\frac{1848}{44}=42 \mathrm{~cm} $
(ii) Width of the track $=7 \mathrm{~m}$
$\therefore$ Outer radius $(\mathrm{R})=42+7=49 \mathrm{~m}$
(iii) Outer circumference $=2 \pi R$
$ =2 \times \frac{22}{7} \times 49=308 \mathrm{~m} $
(iv) Rate of fencing = ₹  50  per metre
$\therefore$ Total cost of fencing outer circumference  = ₹$50 \times 308=$ ₹15,400
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Question 44 Marks
The sides of a triangle are 21 cm, 17 cm, and 10 cm. Find its area.
Answer
Let $a=21 \mathrm{~cm}, b=17 \mathrm{~cm}$ and $c=10 \mathrm{~cm}$
$ \begin{aligned} & \therefore a+b+c \\ & =21 \mathrm{~cm}+17 \mathrm{~cm}+10 \mathrm{~cm}=48 \mathrm{~cm} \\ & s=\frac{a+b+c}{2}=\frac{48}{2}=24 \mathrm{~cm} \end{aligned} $
Area of the triangle
$ \begin{aligned} & =\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{24(24-20)(24-17)(24-10)} \\ & =\sqrt{24 \times 3 \times 7 \times 14} \\ & =\sqrt{2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 2 \times 7} \\ & =2 \times 2 \times 3 \times 7 \\ & =84 \mathrm{~cm}^2 \end{aligned} $
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Question 54 Marks
The diagonal of a rectangular board is 1 m and its length is 96 cm. Find the area of the board.
Answer
Length of diagonal $(A B)=96 \mathrm{~cm}$
Diagonal $(A C)=1 \mathrm{~m}=100 \mathrm{~cm}$

In right-angled triangle $A B C$,
By Applying Pythagoras Theorem,
$ \begin{aligned} & (A C)^2=(A B)^2+(B C)^2 \\ & =(100)^2=(96)^2+B C^2 \\ & 10000=9216=B C^2 \\ & 10000-9216=B C^2 \\ & \sqrt{784}=B C \\ & \therefore B C=28 \mathrm{~cm} \end{aligned} $
Area of the rectangular board
$ \begin{aligned} & =1 \times b \text { or } A B \times B C \\ & =96 \times 28 \\ & =2688 \mathrm{~cm}^2 \end{aligned} $
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Question 64 Marks
If the area of a rhombus is $112 cm^2$ and one of its diagonals is 14 cm , find its other diagonal.
Answer
Area of rhombus $=112 \mathrm{~cm}^2$
One diagonal $=14 \mathrm{~cm}$
Let second diagonal $=\mathrm{x}$
Then, area $=\frac{\text { Product of diagonal }}{2}$
$ \begin{aligned} & \Rightarrow 112=\frac{14 \times \mathrm{x}}{2} \\ & \Rightarrow x=\frac{112 \times 2}{14}=\frac{224}{14} \\ & \Rightarrow x=16 \end{aligned} $
$\therefore$ Second diagonal $=16 \mathrm{~cm}$
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Question 74 Marks
The base of a parallelogram is thrice it height. If its area is $768 cm^2$, find the base and the height of the parallelogram.
Answer
Area of the parallelogram $=768 \mathrm{~cm}^2$
Let the height of the parallelogram $=x$
Then base $=3 x$
$\therefore$ Area $=$ Base $\times$ Height
$\Rightarrow 768=3 \mathrm{x} \times \mathrm{x}$
$\Rightarrow 768=3 x^2$
$\Rightarrow \mathrm{x}^2=\frac{768}{3}=256 \mathrm{~cm}$
$\therefore x=\sqrt{16 \times 16}=16 \mathrm{~cm}$
$\therefore$ Base $=3 \times 16=48 \mathrm{~cm}$
and height $=x=16 \mathrm{~cm}$
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Question 84 Marks
The length and breadth of the rectangular piece of land area in the ratio of 5 : 3. If the total cost of fencing it at the rate of ₹48 per metre is ₹19,200, find its length and breadth.
Answer
Ratio in length and breadth of a rectangular piece of land $=5: 3$
Cost of fencing = ₹ $19,200$
and rate = ₹ $48$ per $\mathrm{m}$
$\therefore$ Perimeter $=\frac{19200}{48}=400 \mathrm{~m}$
Let length $=5 \mathrm{x}$.
Then breadth $=3 \mathrm{x}$
$\therefore$ Perimeter $=2(\mathrm{l}+\mathrm{b})$
$400=2(5 \mathrm{x}+3 \mathrm{x})$
$400=2 \times 8 x=16 x$
$\therefore 16 \mathrm{x}=400$
$\Rightarrow x=\frac{400}{16}=25$
$\therefore$ Length of the land $=5 \mathrm{x}=5 \times 25=125 \mathrm{~m}$ and breadth $=3 \mathrm{x}=3 \times 25=75 \mathrm{~m}$.
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Question 94 Marks
Each side of a square is 44 cm. Find its perimeter. If this perimeter is equal to the circumference of a circle, find the radius of the circle.
Answer
The side of a square $=44 \mathrm{~cm}$
$\therefore$ Its perimeter $=4 \times$ Side
$ =4 \times 44=176 \mathrm{~cm} $
Since, Its is given that, Circumference of a circle $=$ Perimeter of a square
$\therefore$ Circumference of a circle $=176 \mathrm{~cm}$
Let, the radius of the circle $=r$
$ \begin{aligned} & \Rightarrow 2 \pi r=176 \mathrm{~cm} \\ & r=\frac{176 \times 7}{2 \times 22}=28 \mathrm{~cm} \end{aligned} $
$\therefore$ The radius of the circle $=28 \mathrm{~cm}$
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Question 104 Marks
The perimeters of two squares are in the ratio 8:15, find the ratio between the lengths of their sides.
Answer
Let the perimeter of the first square $=8 x$
$\therefore$ Side of the first square $=\frac{\text { Perimeter }}{4}=\frac{8 \mathrm{x}}{4}$
and the perimeter of the second square $=7 x$
$\therefore$ Side of the second square $=\frac{\text { Perimeter }}{4}$
$ =\frac{15 x}{4} $
Now, the ratio between the sides of the square $=\frac{8 x}{4}: \frac{15 x}{4}$
$ =8: 15 $
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Question 114 Marks
The circumferences of two circles are in the ratio 5: 7, find the ratio between their radius.
Answer
The ratio of the circumference of the circle $=5: 7$
Let circumference of the first ratio $=5 x$
$ \begin{aligned} & \therefore 2 \pi r=5 x \\ & \Rightarrow r=\frac{5 x}{2 \pi} \end{aligned} $
and the circumference of the second ratio $=7 x$
$ \begin{aligned} & \therefore 2 \pi r=7 x \\ & \Rightarrow r=\frac{7 x}{2 \pi} \end{aligned} $
The ratio between their radius $=\frac{5 \mathrm{x}}{2 \pi}: \frac{7 \mathrm{x}}{2 \pi}$
$ =5: 7 $
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Question 124 Marks
The radius of two circles are in the ratio 3: 5, find the ratio between their circumferences.
Answer
The ratio of the radius of the circles = 3: 5
Let the radius of the first circle = 3x
and radius of the second circle = 5x
∴ Circumference of the first circle = 2πr
= 2π × 3x = 6πx
and circumference of the second circle
= 2πr = 2π × 5x = 10x
∴ The ratio between their circumference
= 6πx : 10πx
= 16 : 10
= 3 : 5
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Question 134 Marks
The length and breadth of a rectangular field are in the ratio 7 : 4. If its perimeter is 440 m, find its length and breadth. Also, find the cost of fencing it @ ₹150 per m.
Answer
Given : Perimeter $=440 \mathrm{~m}$
Let the length of rectangular field $=7 x$ and breadth $=4 x$
$ \begin{aligned} & 2(l+b)=\text { Perimeter } \\ & 2(7 x+4 x)=440 m \\ & 2(11 x)=440 m \\ & 22 x=440 m \end{aligned} $
$ \begin{aligned} & x=\frac{440}{22} \\ & x=20 m \end{aligned} $
$\therefore$ Length $=7 \mathrm{x}=7 \times 20=140 \mathrm{~m}$
Breadth $=4 x=4 \times 20=80 m$
Cost of fencing per $m=$ ₹ 150
Cost of fencing $440 \mathrm{~m}=$ ₹ 150 x   440 = ₹ 66,000 
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Question 144 Marks
The radius of two circles are $20 \mathrm{~cm}$ and $13 \mathrm{~cm}$. Find the difference between their circumferences. (Take $\left.\pi=\frac{22}{7}\right)$
Answer
Radius of $1^{\text {st }}$ circle $=20 \mathrm{~cm}$
Circumference of the circle $=2 \pi r$
$ \begin{aligned} & =2 \times \frac{22}{7} \times 20 \\ & =40 \times \frac{22}{7} \\ & =125.7 \mathrm{~cm} \end{aligned} $
The radius of $2^{\text {nd }}$ circle $=13 \mathrm{~cm}$
Circumference of the circle $=2 \pi \mathrm{r}$
$ \begin{aligned} & =2 \times \frac{22}{7} \times 13 \\ & =26 \times \frac{22}{7} \\ & =81.7 \end{aligned} $
$\therefore$ The difference of circumference of two circles
$ =125.7-81.7 \mathrm{~cm}=44 \mathrm{~cm} $
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Question 154 Marks
A square lawn is surrounded by a path $2.5 \ m$ wide. If the area of the path is $165\ m^2$ find the area of the lawn.
Answer

Area of path $=165 \ m^2$
Width of path $=2.5\ m$
Let the side of square lawn $=xm$
$\therefore$ Outer side $=x+2 \times 2.5$
$ =(x+5) m $
$\therefore$ Area of path $=(x+5)^2-x^2$
$ \Rightarrow x^2+10 x+25-x^2=165 $
$ \Rightarrow 10 x=165-25=140 $
$ \Rightarrow x=\frac{140}{10}=14\ m$
$\therefore$ Side of lawn $=14\ m$
and area of lawn $=(14)^2m^2=196\ m^2$
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Question 164 Marks
The length and breadth of a rectangular piece of land are in the ratio $5 : 3$. If the total cost of fencing it at the rate of $₹\ 24$ per meter is $₹\ 9600$, find its $:$
$(i)$ length and breadth
$(ii)$ area
$(iii)$ cost of levelling at the rate of $₹\ 60$ per $m^2$.
Answer
Ratio in length and breadth of a rectangular piece of land $=5: 3$
Cost of fencing $= ₹ \ 9600$
And rate $= ₹ \ 24$ per $m$
Perimeter $=\frac{Total \ cost\  of\ fencing}{Rate\ per\ m}$
$=\frac{9600}{24}$
$=400\ m$
Let length $=5x$
Then breadth $=3x$
$\therefore$ Perimeter $=2(l+b)$
$400=2 \times(5 x+3 x)$
$\therefore 16 x=400$
$x=\frac{400}{16}=25$
$(i)$ Length of land $=5x=5 \times 25=125\ m$
and breadth $=3 x=3 \times 25=75\ m$
$(ii)$ Area $=1 \times b$
$=125 \times 75=9375\ m^2$
$(iii)$ Cost of levelling at rate $₹\ 60$ per $m^2$
$=₹\  60 \times 9375\ m^2= ₹\  5,62,500$
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Question 174 Marks
The ratio between the areas of two circles is 16 : 9. Find the ratio between their :
(i) radius
(ii) diameters
(iii) circumference
Answer
(i) Let radius of first circle $=r_1$
and radius of second circle $=r_2$
Given that ratio of the areas of circles $=16: 9$
$\Rightarrow \frac{\pi r_1^2}{\pi r_2^2}=\frac{16}{9}$
$\Rightarrow \frac{\pi r_1^2}{\pi r_2^2}=\frac{4^2}{3^2}$
$\Rightarrow \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{4}{3}$
(ii) Let the diameter of first circle $=d_1$
and diameter of second circle $=\mathrm{d}_2$
Since, we know that diameter $=2 \times$ radius
$\therefore d_1=2 \times r_1=2 \times 4 x=8 x$
and $d_2=2 \times r_2=2 \times 3 x=6 x$
Now, the ratio between the diameter of two circles $=d_1: d_2$
$ =8 x: 6 x=4: 3 $
(iii) Now, consider the ratio of circumference of the circles
$ =\frac{2 \pi r_1}{2 \pi r_2}=\frac{r_1}{r_2}=\frac{4}{3} $
$\therefore$ The ratio between the circumference of two circles $=4: 3$
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Question 184 Marks
The sides of a rectangular park are in the ratio $4 : 3$. If its area is $1728\ m^2,$ find
$(i)$ its perimeter
$(ii)$ cost of fencing it at the rate of $₹\ 40$ per meter.
Answer
Ratio in the sides of a rectangle $=4: 3$
Area $=1728\ m^2$
Let length $=4x,$ and breadth $=3x$
$\therefore$ Area $=1 \times b$
$1728=4x \times 3x$
$\Rightarrow 12 x^2=1728$
$\Rightarrow x^2=\frac{1728}{12}$
$\Rightarrow x^2=144=(12)^2$
$\therefore x=12$
$\therefore$ Length $=4x=4 \times 12=48\ m$
Breadth $=3 \mathrm{x}=3 \times 12=36 \mathrm{~m}$
$(i)$ Now perimeter $=2(l+b)$
$=2(48+36) \ m$
$=2 \times 84=168\ m$
$(ii)$ Rate of fencing $= ₹\  40$ per metre
Total cost $=168 \times 40= ₹ \ 6720$
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Question 194 Marks
The ratio between the radius of two circles is $5 : 7$ Find the ratio between their:
$(i)$ circumference
$(ii)$ areas
Answer
$(i)$ The ratio of the radius of the circles $=5: 7$
Let radius of first circle $=5 \mathrm{x}$
and radius of second circle $=7 \mathrm{x}$
$\therefore$ Circumference of first circle $=2 \pi r$
$ =2 \pi \times 5 x=10 \pi x $
and circumference of second circle
$ =2 \pi \times 7 x=14 \pi x $
$\therefore$ Ratio between their circumference
$ =10 \pi x: 14 \pi x$
$ =10: 14=5: 7 $
$(ii)$ Area of first circle $=\pi r^2$
$ =\frac{22}{7} \times 5 \mathrm{x} \times 5 \mathrm{x}=\frac{550}{7} \mathrm{x}^2 $
and area of second circle $=\pi r^2$
$ =\frac{22}{7} \times 7 \mathrm{x} \times 7 \mathrm{x}=\frac{1078}{7} \mathrm{x}^2 $
Ratio between their areas
$=\frac{550}{7} x^2: \frac{1078}{7} x^2$
$=550: 1078 \ldots \ldots \ldots . . .\ ($Dividing by  $22)$
$=25: 49$
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Question 204 Marks
The sides of a triangle are in the ratio 15 : 13 : 14 and its perimeter is 168 cm. Find the area of the triangle.
Answer
Perimeter of the triangle $=168 \mathrm{~cm}$
Sum of ratio of sides $=15+13+14=42$
Let the first side $=\frac{168 \times 15}{42}=60 \mathrm{~cm}$
Second side $=\frac{168 \times 13}{42}=52 \mathrm{~cm}$
Third side $=\frac{168 \times 14}{42}=56 \mathrm{~cm}$
Now, $s=\frac{a+b+c}{2}$
$ =\frac{60+52+56}{2}=\frac{168}{2}=84 $
$\therefore$ Area $=\sqrt{s(s-a)(s-b)(s-c)}$
$ =\sqrt{84(84-60)(84-52)(84-56)} $
$ =\sqrt{84 \times 24 \times 32 \times 28} $
$ =\sqrt{2 \times 2 \times 3 \times 7 \times 2 \times 2 \times 2 \times 3 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7} $
$ =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7 $
$ =1344 \mathrm{~cm}^2 $
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Question 214 Marks
The legs of a right$-$angled triangle are in the ratio $4 : 3$ and its area is $4056\ cm^2$. Find the length of its legs.
Answer
Area of right-angled triangle $=4056\ cm^2$
Legs of a right-angled triangled are in the ratio i.e. $4: 3$
Let one leg $($Base$) \ =3 x$
Then second leg $($altitude$)\  =4 x$

Area $=\frac{1}{2} \times$ Base $\times$ Altitude
$ =\frac{1}{2} \times 3x \times 4x=6x^2$
$ \therefore 6x^2=4056 $
$ x^2=\frac{4056}{6}=676 $
$ x=\sqrt{676}=\sqrt{26 \times 26} $
$\therefore x=26\ cm$
$\therefore$ One leg $($ base $)=3 x=3 \times 26=78\ cm$
and second leg $($altitude$) \ 4 x=4 \times 26=104\ cm$
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Question 224 Marks
Find the area of an isosceles triangle whose base is 16 cm and the length of each of the equal sides is 10 cm.
Answer
In isosceles $\triangle A B C$
Base $B C=16 \mathrm{~cm}$
and $A B=A C=10 \mathrm{~cm}$

$\begin{aligned} & \text { Let } A D \perp B C \text { and } B D=\frac{1}{2} B C=\frac{16}{2} \\ & \therefore B D=8 \mathrm{~cm} \\ & \text { In right } \triangle A B D \\ & A B^2=A D^2+B D^2 \ldots \ldots \ldots \ldots .(\text { Pythagoras Theorem) } \\ & (10)^2=A D^2+(8)^2 \\ & 100=A D^2+64 \\ & 100-64=A D^2 \\ & 36=A D^2 \\ & A D=\sqrt{36}=\sqrt{6 \times 6} \\ & \therefore A D=6 \mathrm{~cm} \\ & N o w \text { the area of triangle }=\frac{\text { Base } \times \text { Altitude }}{2} \\ & =\frac{16 \times 6}{2}=48 \mathrm{~cm} 2\end{aligned}$
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Question 234 Marks
Find the area of a right-angled triangle whose hypotenuse is 13 cm long and one of its legs is 12 cm long.
Answer
In right-angled $\triangle A B C$,
Base $B C=12 \mathrm{~cm}$
and hypotenuse $\mathrm{AC}=13 \mathrm{~cm}$

Applying Pythagoras Theorem,
$ \begin{aligned} & (A C)^2=(A B)^2+(B C)^2 \\ & (13)^2=(A B)^2+(12)^2 \\ & 169=(A B)^2+144 \\ & (A B)^2=169-144 \\ & (A B)^2=25 \\ & \therefore A B=\sqrt{25} \\ & =\sqrt{5 \times 5}=5 \mathrm{~cm} \end{aligned} $
Now, area of $\triangle A B C=\frac{1}{2}$ base $\times$ altitude
$ =\frac{1}{2} \times 12 \times 5=30 \mathrm{~cm}^2 $
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Question 244 Marks
The adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 6 cm, find the distance between the shorter sides.
Answer
In parallelogram $A B C D$
$ A B=D C=15 \mathrm{~cm} $
$ B C=A D=10 \mathrm{~cm} $

Distance between longer sides $A B$ and $D C$ is $6 \mathrm{~cm}$
i.e., perpendicular $\mathrm{DL}=6 \mathrm{~cm}$
$\mathrm{DM} \perp \mathrm{BC}$
Area of parallelogram $=$ Base $\times$ Altitude
$ =A B \times D L=15 \times 6=90 \mathrm{~cm}^2 $
Again let $\mathrm{DM}=\mathrm{xcm}$
$\therefore$ Area of parallelogram $A B C D=B C \times D M$
$ =10 \times x=10 x \mathrm{~cm}^2 $
$\therefore 10 \times \mathrm{cm}^2=90 \mathrm{~cm}^2$
$\Rightarrow x=\frac{90}{10}=9 \mathrm{~cm}$
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Question 254 Marks
Find the area of the rhombus, if its diagonals are 30 cm and 24 cm.
Answer
Given, diagonal $\left(d_1\right)=30 \mathrm{~cm}$
Other diagonal $\left(\mathrm{d}_2\right)=24 \mathrm{~cm}$

If $A C$ and $B D$ are the diagonals of a rhombus its
Area $=\frac{1}{2} \times$ Product of it diagonals
$ \begin{aligned} & =\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD} \\ & =\frac{1}{2} \times \mathrm{d}_1 \times \mathrm{d}_2 \\ & =\frac{1}{2} \times 30 \times 24 \mathrm{~cm}^2 \\ & =15 \times 24=360 \mathrm{~cm}^2 \end{aligned} $
$\therefore$ Area of rhombus $=360 \mathrm{~cm}^2$
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Question 264 Marks
Find the perimeter of a rectangle with length = 24 cm and diagonal = 25 cm
Answer
Length of a rectangle $(\mathrm{I})=24 \mathrm{~cm}$, Diagonal $=25 \mathrm{~cm}$

Let breadth of the rectangle $=\mathrm{b} \mathrm{m}$
Applying Pythagoras Theorem in triangle $A B C$,
We get, $(A C)^2=(A B)^2+(B C)^2$
$ \begin{aligned} & (25)^2=(24)^2+(b)^2 \\ & 625=576+(b)^2 \\ & 625-576=b^2 \\ & 49=b^2 \\ & \sqrt{7 \times 7}=b \\ & \therefore b=7 \mathrm{~cm} \end{aligned} $
Now, perimeter of the rectangle
$ \begin{aligned} & =2(1+b) \\ & =2(24+7) \\ & =2(31) \\ & =62 \mathrm{~cm} \end{aligned} $
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Question 274 Marks
The length of a rectangular field is 30 m and its diagonal is 34 m. Find the breadth of the field and its perimeter.
Answer
Length $=30 \mathrm{~m}$
Diagonals $=34 \mathrm{~m}$

Let the breadth of the rectangle $=b \mathrm{~m}$
Applying Pythagoras Theorem in triangle $A B C$,
We get,
$ \begin{aligned} & A C^2=A B^2+B C^2 \\ & (34)^2=(30)^2+b^2 \\ & 1156=900+b^2 \\ & 1156-900=b^2 \\ & 256=b^2 \\ & \Rightarrow b \sqrt{256}=16 \mathrm{~m} \\ & \text { Perimeter }=2(1+b) \\ & =2(30+16) \\ & =2 \times 46 \\ & =92 \mathrm{~m} \end{aligned} $
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Question 284 Marks
Find the radius of the circle whose circumference is equal to the sum of the circumferences of the circles having radius 15 cm and 8 cm.
Answer
For circle with radius $=15 \mathrm{~cm}$
Circumference of circle $=2 \pi r$
$=2 \times \frac{22}{7} \times 15 \mathrm{~cm}$
$ =\frac{660}{7} \mathrm{~cm} $
For circle with radius $=8 \mathrm{~cm}$
Circumference of circle $=2 \pi \mathrm{r}$
$=2 \times \frac{22}{7} \times 8 \mathrm{~cm}$
$=\frac{352}{7} \mathrm{~cm}$
Sum of the circumferences of these two circles
$ =\frac{660}{7} \mathrm{~cm}+\frac{352}{7} \mathrm{~cm}=\frac{1012}{7} \mathrm{~cm} $
If the required radius $=R \mathrm{~cm}$
$ \begin{aligned} & \text { Its circumference }=2 \pi R \\ & =2 \times \frac{22}{7} \times \mathrm{R} \mathrm{cm} \\ & =\frac{44}{7} \mathrm{R} \mathrm{cm} \end{aligned} $
Given, $\frac{44}{7} \mathrm{R}=\frac{1012}{7}$
$ \Rightarrow \mathrm{R}=\frac{7}{44} \times \frac{1012}{7} \mathrm{~cm} $
$ =23 \mathrm{~cm} $
$\therefore$ Required radius $=23 \mathrm{~cm}$
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[4 marks sum] - MATHS STD 7 Questions - Vidyadip