Perimeter and Area Test

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Perimeter and Area Test - 24

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Question 1
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$$ABCD$$ is a square field of side $$80\;m$$. If $$EF$$ is perpendicular to $$DC$$, find the area of $$\Delta EDC$$ and the area of the shaded portion respectively.

A
$$3200\;m^2,\;3200\;m^2$$

B
$$3200\;m^2,\;3500\;m^2$$

C
$$3600\;m^2,\;3200\;m^2$$

D
$$3200\;m^2,\;3600\;m^2$$

Solution

ar $$\Delta EDC=\dfrac { 1 }{ 2 } DC\times EF$$

$$=$$ $$\dfrac { 1 }{ 2 } \times 80\times 80$$

$$=$$ $$3200$$ $${ m }^{ 2 }$$

area of shaded portion $$=$$ Area of square - $$3200$$ $${ m }^{ 2 }$$

$$=$$ $${ \left( 80\times 80 \right) }-3200$$

$$=$$ $$3200$$ $${ m }^{ 2 }$$
Question 2
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Mrs. Kaushik has a square plot with the measurement as shown in the figure. she wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of $$Rs.\;55\;per\;m^2$$.

A
$$Rs.\;17,875$$

B
$$Rs.\;17,675$$

C
$$Rs.\;12,875$$

D
$$Rs.\;17,825$$

Solution

Area of the garden = Area of the outer square - Area of the inner rectangle
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=25\times25\;m^2-20\times15\;m^2$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(625-300)\;m^2=325\;m^2$$
Cost of developing a garden @ $$Rs.\;55\;per\;m^2=Rs.\;(55\times325)=Rs.\;17,875$$.
Question 3
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In figure, $$AB$$ and $$CD$$ are two perpendicular diameters of a circle with centre $$O$$. If $$OA$$$$=$$$$7cm$$, find the area of the shaded region. [Use $$\pi=\displaystyle\frac{22}{7}$$]

A
$$31.4cm^2$$

B
$$42.5cm^2$$

C
$$54.2cm^2$$

D
$$66.5cm^2$$

Solution

Area of shaded region $$=$$ area of larger circle $$-$$ area of smaller circle $$-$$ area of the triangle

$$=\pi { R }^{ 2 }-\pi { r }^{ 2 }-\dfrac { 1 }{ 2 } \times DC\times BO$$

$$=\pi { R }^{ 2 }-\pi { r }^{ 2 }-\dfrac { 1 }{ 2 } \times DC\times BO$$

$$=\dfrac { 22 }{ 7 } \left( { 7 }^{ 2 }-{ \left( 3.5 \right) }^{ 2 } \right) -\dfrac { 1 }{ 2 } \times 14\times 7$$

$$=\dfrac { 22 }{ 7 } \left( 49-12.25 \right) -49$$

$$=(\dfrac { 22 }{ 7 } \times 36.75)-49$$

$$= 115.5 - 49$$

$$= 66.5$$ $${ cm }^{ 2 }$$
Question 4
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If the diameter of a semicircular protractor is $$14cm$$ then find its perimeter.

A
$$36cm$$

B
$$76cm$$

C
$$112cm$$

D
$$162cm$$

Solution

Diameter $$=14cm$$
Radius=$$\displaystyle\frac{Diameter}{2}=\frac{14cm}{2}=7cm$$
Length of the semicircular part $$=\pi r=\displaystyle\frac{22}{7}\times (7)=22cm$$
Total perimeter=length of semicircular part +Diameter
$$=22cm+14cm=36cm$$
Thus, the perimeter of the protractor is $$36cm$$
Question 5
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The area of the shaded portion in the given figure is

A
$$7.5\pi$$ sq.units

B
$$6.5\pi$$ sq.units

C
$$5.5\pi$$ sq.units

D
$$4.5\pi$$ sq.units

Solution

$$\textbf{Step 1 : Find the required area}$$
$$\text{Area of the outer semi-circle =} \dfrac{\pi}{2}\times 5^2$$

$$\text{Area of the inner semi-circle =} \dfrac{\pi}{2}\times (5-1)^2$$

$$\therefore \text{ Area of shaded region = Area of outer semicircle - Area of inner semicircle}$$
$$=\dfrac{\pi}{2}\times 5^2-\dfrac{\pi}{2}\times 4^2$$

$$= \dfrac{\pi}{2}(25-16)$$
$$= \text{4.5 $\pi$ unit}^2$$

$$\boldsymbol{\textbf{Hence, Area of shaded portion is 4.5 $\pi$} \textbf{ sq. unit}}$$
Question 6
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A circular ground whose diameter is $$140$$ meters is to be fenced by wire three times around its circumference. Find the length of wire needed.

$$[$$use $$\displaystyle \pi = \frac {22}{7}$$$$]$$

A
$$440m$$

B
$$1320m$$

C
$$660m$$

D
None of these

Solution

Diameter of circular ground $$=140m$$.
Radius of circular ground $$=70m$$.
Circumference of circular ground $$=2\pi r$$

$$\displaystyle =2\times \frac {22}{7}\times 70$$
$$=440m$$
Length of wire needed to fenced three times
$$=440\times 3$$
$$=1320m$$
Question 7
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The inner circumference of a circular tracks is $$220\ m$$. The track is $$7$$ $$m$$ wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs. $$2$$ per metre. Use $$\pi=\displaystyle\frac{22}{7}$$

A
Rs. $$947$$

B
Rs. $$726$$

C
Rs. $$612$$

D
Rs. $$528$$

Solution

Circumference of inner side = $$220 m$$

$$\Rightarrow 2\pi r=220\Rightarrow r=\dfrac { 220\times 7 }{ 44 } =35m$$

Now, width of track $$=7m$$

$$\therefore $$ Outer radius $$=35+7 = 42 m$$

Therefore outer circumference $$=2\pi R=2\times \dfrac { 22 }{ 7 } \times 42=264m$$

$$\therefore $$ Cost of fencing $$=$$ Rs. $$\left( 264\times 2 \right) $$ $$=$$ Rs. $$528$$
Question 8
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The radii of two circles are $$8$$cm and $$6$$cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.

A
$$32cm$$

B
$$23cm$$

C
$$15cm$$

D
$$10cm$$

Solution

Let the radius of required circle is $$r$$ cm.
So, $$\pi r^2=\pi r^2_1+\pi r^2_2$$
$$r^2=r^2_1+r^2_2$$
$$r^2=(8)^2+(6)^2$$
$$r^2=100$$
$$r=10cm$$
Question 9
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If the radius of circle is $$\displaystyle \pi $$, then its area will be:

A
$$\displaystyle \pi $$

B
$$\displaystyle { \pi }^{ 2 }$$

C
$$\displaystyle { \pi }^{ 3 }$$

D
$$\displaystyle 3\pi $$

Solution

Area of circle $$=$$ $$\pi { r }^{ 2 }=\pi \times { \left( \pi \right) }^{ 2 }={ \pi }^{ 3 }$$
Question 10
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Find the area of a circle whose circumference is $$22cm$$.

A
$$35.2cm^2$$

B
$$38.5cm^2$$

C
$$41.7cm^2$$

D
$$47.6cm^2$$

Solution

Let $$r$$ be the radius of the circle.

Then, Circumference$$=22cm$$

$$\displaystyle \Rightarrow 2\pi r=22$$

$$\displaystyle\Rightarrow 2\times \frac{22}{7}\times r=22$$

$$\Rightarrow r=\displaystyle\frac{7}{2}cm$$

$$\therefore$$ Area of the circle$$=\displaystyle \pi r^2$$

$$=\displaystyle\frac{22}{7}\times\frac{7}{2}\times \frac{7}{2}cm^2$$

$$=38.5 \, cm^2$$