[3 marks sum]

Question 13 Marks

One side of a parallelogram is $20 cm$ and its distance from the opposite side is $16 cm $. Find the area of the parallelogram.

Answer

Area of parallelogram = Base × Height
$= AB \times DE$
$= 20 cm \times 16 cm = 320 cm^2$
$\therefore $ Area of parallelogram $= 320 cm^2$

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Question 23 Marks

In the figure given below, find the area of shaded region: (All measurements are in cm)

Answer

Outer length $=30 cm$
and Outer breadth $=20 cm$
$\therefore$ Outer area $=1 \times b$
$=30 \times 20=600 cm^2$
Inner length $=12 cm$ and inner breadth $=12 cm$
Inner area $= I \times b =12 \times 12=144 cm^2$
Area of shaded portion
$=$ Area of outer figure - Area of inner figure $=600-144=456 cm^2$

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Question 33 Marks

In the figure given below, find the area of shaded region: (All measurements are in cm)

Answer

Outer length $=20 cm$
and breadth $=16 cm$
$\therefore$ Outer area $=1 \times b$
$=20 \times 16 cm^2=320 cm^2$
Inner length = 15 cm
and $\operatorname{Inner}$ breadth $=10 cm$
$\therefore$ Inner area $=15 \times 10=150 cm^2$
$\therefore$ Area of shaded region $=$ Area of whole region - Area of unshaded region
$=320-150 cm^2=170 cm^2$

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Question 43 Marks

The length and breadth of a rectangular paper are 35 cm and 22 cm. Find the area of the largest circle which can be cut out of this paper.

Answer

Length $=35 \mathrm{~cm}$
Breath $=22 \mathrm{~cm}$

The largest circle which can be cut from the rectangle will have,
radius $=\frac{22}{2} \mathrm{~cm}=\frac{\mathrm{D}}{2}=\mathrm{R}=11 \mathrm{~cm}$
Area of circle $=\pi r^2$
$ \begin{aligned} & =\frac{22}{7} \times 11 \times 11 \\ & =380.28 \mathrm{~cm}^2 \end{aligned} $

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Question 53 Marks

A wire is along the boundary of a circle with a radius of 28 cm. If the same wire is bent in the form of a square, find the area of the square formed.

Answer

Radius of circular wire $=28 \mathrm{~cm}$
$\therefore$ Circumference $=2 \pi r$
$ =2 \times \frac{22}{7} \times 28 \mathrm{~cm}=176 \mathrm{~cm} $
$\therefore$ Perimeter of the square formed by this wire $=176 \mathrm{~cm}$
$\therefore$ Side $(\mathrm{a})=\frac{176}{4}=44 \mathrm{~cm}$
Area of square so formed $=a^2=(44)^2 \mathrm{~cm}^2$
$ =1936 \mathrm{~cm}^2 $

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Question 63 Marks

A metal wire, when bent in the form of a square of largest area, encloses an area of $484 cm^2$. Find the length of the wire. If the same wire is bent to the largest circle, find:
(i) radius of the circle formed.
(ii) area of the circle.

Answer

Area of the square made wire $=484 \mathrm{~cm}^2$
$ \therefore \text { Length (side) }=\sqrt{\text { Area }}=\sqrt{484}=22 \mathrm{~cm} $
(i) Perimeter of wire $=4 \times$ Side
$ =4 \times 22=88 \mathrm{~cm} $
$\therefore$ Circumference of circular wire $=88 \mathrm{~cm}$
$\therefore$ Radius $(r)=\frac{C}{2 \pi}=\frac{88 \times 7}{2 \times 22} \mathrm{~cm}=14 \mathrm{~cm}$
(ii) $\therefore$ Area of the circle $=\pi r^2$
$ =\frac{22}{7} \times 14 \times 14=616 \mathrm{~cm}^2 $

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Question 73 Marks

The diameter of the wheel of a car is 70 cm. How many revolutions will it make to travel one kilometer?

Answer

$ \begin{aligned} & \text { Diameter of a car wheel }(\mathrm{d})=70 \mathrm{~cm} \\ & \therefore \text { Circumference }=\pi \mathrm{d}=\frac{22}{7} \times 70 \\ & =220 \mathrm{~cm} \end{aligned} $
No. of revolutions in $1 \mathrm{~km}=\frac{220}{100} \mathrm{~m}$
$ \begin{aligned} & =1 \mathrm{~km} \div\left(\frac{220}{100}\right) \mathrm{m} \\ & =1000 \times \frac{100}{220} \mathrm{~m} \\ & =\frac{5000}{11}=454 \frac{6}{11} \mathrm{~km} \end{aligned} $

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Question 83 Marks

The circumference of a circle exceeds its diameter by 18 cm. find the radius of the circle.

Answer

Let $c$ be the circumference and $d$ be the diameter of the circle.
$ \begin{aligned} & \therefore c=d+18 \\ & \Rightarrow d \pi=d+18 \Rightarrow d \pi-d=18 \\ & d(\pi-1)=18 \\ & \Rightarrow d\left(\frac{22}{7}-1\right)=18 \\ & \Rightarrow d\left(\frac{15}{7}\right)=18 \\ & \Rightarrow d=\frac{18 \times 7}{15}=\frac{126}{15}=8.4 \mathrm{~cm} \\ & \therefore \text { Radius }=\frac{d}{2}=\frac{8.4}{2}=4.2 \mathrm{~cm} \end{aligned} $

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Question 93 Marks

The area of an equilateral triangle is $(64 \times \sqrt{3}) \mathrm{cm}^2$. Find the length of each side of the triangle.

Answer

Area of equilateral triangle $=64 \sqrt{3} \mathrm{~cm}^2$
Let each side $=a$
Then, $\frac{\sqrt{3}}{4} \mathrm{a}^2=64 \sqrt{3}$
$ \begin{aligned} & a^2=\frac{64 \sqrt{3} \times 4}{\sqrt{3}}=256 \\ & a=(16)^2 \end{aligned} $
$ \therefore a=16 \mathrm{~cm} $
$\therefore$ Each side $=16 \mathrm{~cm}$

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Question 103 Marks

Find the area of an equilateral triangle whose each side is $16 \mathrm{~cm}$. (Take $\sqrt{3}=1.73$ )

Answer

Side of the equilateral triangle $=16 \mathrm{~cm}$

$\begin{aligned} & \therefore \text { Area }=\frac{\sqrt{3}}{4}(\mathrm{a})^2 \\ & =\frac{\sqrt{3}}{4} \times 16 \times 16 \\ & =1.73 \times 4 \times 16=110.72 \mathrm{~cm}^2\end{aligned}$

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Question 113 Marks

The area of a right-angled triangle is $160 cm^2$. If its one leg is $16$ cm long, find the length of the other leg.

Answer

Area of the right-angled triangle $=160 \mathrm{~cm}^2$
Let base $($ one side $)=16 \mathrm{~cm}$

$\therefore$ Altitude (second side)
$ =\frac{\text { Area } \times 2}{\text { Base }}=\frac{160 \times 2}{16}=\frac{320}{16}=20 \mathrm{~cm} $

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Question 123 Marks

One side of a parallelogram is $18\ cm$ and its area is $153 cm^2$. Find the distance of the given side from its opposite side.

Answer

Area of parallelogram $A B C D=153 \mathrm{~cm}^2$
Side (Base) $A B=18 \mathrm{~cm}$

$\therefore$ Distance $\mathrm{DL}$ between $\mathrm{AB}$ and $\mathrm{DC}$ ..... (altitude)
$ =\frac{\text { Area }}{\text { Base }}=\frac{153}{18}=\frac{17}{2} \mathrm{~cm}=8.5 \mathrm{~cm} $

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Question 133 Marks

Find the area of a rectangle whose length and breadth are $25 \ m$ and $16 \ cm.$

Answer

Length of rectangle $= 25 \ cm$
Breadth of rectangle $= 16 \ cm$

Area of rectangle $=1 \times$ b or
$AB \times BC$ $=25 \times 16 cm^2=400 cm^2$
$\therefore$ Area of rectangle $=400 cm^2$

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Question 143 Marks

If P = perimeter of a rectangle, l = its length and b = its breadth; find l, if P = 96 m and b = 28 m.

Answer

Perimeter of a rectangle $=96 \mathrm{~m}$
Breadth $(b)=28 \mathrm{~m}$
Let length $=$ I
$ P=2(I+b) $
$ I=\frac{P-2 b}{2} $
$ =\frac{96-2 \times 28}{2} $
$ =\frac{96-56}{2} $
$ =\frac{40}{2} $
$ =20 \mathrm{~m} $
$ \therefore I=20 \mathrm{~m} $

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Question 153 Marks

If P = perimeter of a rectangle, l= its length and b = its breadth find b, if P = 88 cm and l = 24 cm

Answer

Perimeter of a rectangle $=88 \mathrm{~cm}$
Length $(I)=24 \mathrm{~cm}$
Let breadth $=\mathrm{b}$
$P=2(l+b)$
$b=\frac{P}{2}-1$
$b=\frac{88}{2}-24$
$=44-24$
$=20 \mathrm{~cm}$
$\therefore$ Breadth of a rectangle $=20 \mathrm{~cm}$

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Question 163 Marks

A wire is in the shape of square of side 20 cm. If the wire is bent into a rectangle of length 24 cm, find its breadth.

Answer

Side of square $=20 \mathrm{~cm}$
Perimeter of square $=4 \times 20=80 \mathrm{~cm}$
Or perimeter of rectangle $=80 \mathrm{~cm}$
Length of a rectangle $=24 \mathrm{~cm}$
$\therefore$ Perimeter of a rectangle $=2(l+b)$
$b=\frac{80}{2}-24$
$b=40-24=16 \mathrm{~cm}$

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Question 173 Marks

The lengths of the sides of two squares are in the ratio 8:15, find the ratio between their perimeters.

Answer

Let the side of first square = 8x
∴ Perimeter of first square = 4 × Side = 4 × 8x = 32x
and the side of second squares = 15x
∴ Perimeter of second square = 4 × Side = 4 × 15x = 60x
Now, the ratio between their perimeter = 32x : 60x= 8 : 15

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Question 183 Marks

The circumference of a circle is eigth time the circumference of the circle with a radius of 12 cm. Find its diameter.

Answer

Let the radius of the bigger circle $=R$, and Let the radius of the smaller circle $=r$.
$ \mathrm{r}=12 \mathrm{~cm} $
According to the problem,
$ \begin{aligned} & 2 \pi R=8 \times 2 \pi r \\ & 2 \pi R=8 \times 2 \pi \times 12 \\ & R=\frac{8 \times 2 \pi \times 12}{2 \pi} \\ & R=8 \times 12 \\ & R=96 \end{aligned} $
Diameter $=2 \times 96$
$ =192 \mathrm{~cm} $

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Question 193 Marks

Find the diameter of a circle whose circumference is equal to the sum of circumference of circles with radius 10 cm, 12 cm, and 18 cm.

Answer

Let the radius of the circle = R cm
∴ 2πR = 2π × 10 + 2π × 12 + 2π × 18
On dividing each term by 2π, we get:
R = 10 + 12 + 18 = 40 cm
∴ Radius of the circle obtained = 40 cm
And, its diameter = 2 × Radius
= 2 × 40 cm = 80 cm

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Question 203 Marks

The perimeter of a square and the circumference of a circle are equal. If the length of each side of the square is 22 cm, find:
(i) perimeter of the square.
(ii) circumference of the circle.
(iii) radius of the circle.

Answer

(i) Side of square $=22 \mathrm{~cm}$
Perimeter of square $=4 \times$ side
$ =4 \times 22=88 \mathrm{~cm} $
(ii) Circumference of circle
Given, Perimeter of square $=$ Circumference of circle
$ =88 \mathrm{~cm} $
(iii) Circumference of circle $=88 \mathrm{~cm}$
$\therefore$ Radius $=\frac{\mathrm{C}}{2 \pi}=\frac{88 \times 7}{2 \times 22}=\frac{616}{44}=14 \mathrm{~cm}$

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Question 213 Marks

The diameter of a circle is $42 \mathrm{~cm}$, find its perimeter. If the perimeter of the circle is doubled, what will be the radius of the new circle? (Take $\pi=\frac{22}{7}$ )

Answer

Given, the Diameter of a circle $=42 \mathrm{~cm}$
$\therefore$ Radius of circle $=\frac{42}{2}=21 \mathrm{~cm}$
The perimeter of the circle $=2 \pi r$
$ \begin{aligned} & =2 \times \frac{22}{7} \times 21 \\ & =132 \mathrm{~cm} \end{aligned} $
If the perimeter of the circle doubled
$ =2 \times 132=264 \mathrm{~cm} $
Radius $=\frac{\mathrm{C}}{2 \pi}=\frac{264}{2 \times \frac{22}{7}}$
$ =\frac{264 \times 7}{2 \times 22} $
$ =42 \mathrm{~cm} $

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Question 223 Marks

The diameter of a circular field is $56 \mathrm{~m}$. Find its circumference and cost of fencing it at the rate of ₹80 per m. $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$

Answer

Given, Diameter of a circular field $=56 \mathrm{~m}$
$ \therefore \text { Radius }=\frac{56}{2}=28 \mathrm{~m} $
Circumference of the circle $=2 \pi r$
$ \begin{aligned} & =2 \times \frac{22}{7} \times 28 \mathrm{~m} \\ & =2 \times 22 \times 4 \mathrm{~m} \\ & =176 \mathrm{~m} \end{aligned} $
The cost of fencing $176 \mathrm{~m}$ is,
$=176 m \times$ ₹ 80 per m
= ₹ 14,080

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Question 233 Marks

The radius of a circle is $21 \mathrm{~cm}$. Find the circumference (Take $\pi=3 \frac{1}{7}$ ).

Answer

Given, radius $(r)=21 \mathrm{~cm}$ and $\pi=\frac{22}{7}$
Circumference of the circle $=2 \pi r$
$ \begin{aligned} & =2 \times \frac{22}{7} \times 21 \mathrm{~cm} \\ & =2 \times 22 \times 3 \mathrm{~cm} \\ & =132 \mathrm{~cm} \end{aligned} $

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Question 243 Marks

The length and the breadth of a rectangle are 36 cm and 28 cm. If its perimeter is equal to the perimeter of a square, find the side of the square.

Answer

$ \begin{aligned} & \text { Length of rectangle }=36 \mathrm{~cm} \\ & \text { Breadth of rectangle }=28 \mathrm{~cm} \\ & \text { Perimeter of the rectangle }=2(l+b) \\ & =2(36+28) \\ & =2(64) \\ & =128 \mathrm{~cm} \end{aligned} $
Given, the perimeter of the square $=$ perimeter of rectangle $=128 \mathrm{~cm}$
$\therefore$ Side of the square $=\frac{\text { Perimeter }}{4}$
$ =\frac{128}{4} $
$ =32 \mathrm{~cm} $

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Question 253 Marks

The cost of fencing a square field at the rate of the cost of fencing 440 m = ₹150 × 440 = ₹75 per meter is the cost of fencing 440 m = ₹150 × 440 = ₹67,500. Find the perimeter and the side of the square field.

Answer

Length of the fence $\times$ its rate $=$ = ₹ 67,500
$\Rightarrow$ Length of the fence $=$ ₹ $\frac{67500}{75}=900 m$
$\therefore$ Perimeter of a square field $=$ length of its fence $=900 m$
Since, perimeter of a square $=4 \times$ Length of its side
$\Rightarrow$ Length of the side of the square
$
=\frac{\text { Perimeter }}{4}=\frac{900}{4}=225 m
$

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Question 263 Marks

The length and the breadth of a rectangular plot are $135\ m$ and $65\ m.$ Find, its perimeter and the cost of fencing it at the rate of ₹ $60$ per m.

Answer

Given:
Length $(I)=135 m$
Breadth $(b)=65 m$

$ \text { Perimeter }=2(I+b)$
$=2(135+65)$
$=2(200)$
$=400 m $
$\therefore$ Perimeter of rectangular plot is $=400 m$
Cost of fencing per $m$= ₹ 60
$\therefore$ Cost of fencing $400\ m = ₹ 60 \times 400\ m = ₹ 24000$

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MATHS [3 marks sum]

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