Perimeter and Area Test

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Perimeter and Area Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The perimeter of the following shaded portion of the figure is

A
40 m

B
40.07 m

C
35.72m

D
35 m

Solution

Then length and breadth of rectangle is 17 m and 4 m respectively.
Then perimeter of rectangle=$$2( L+B)=2(17+4)=42$$
The radius of one fourth circle given in figure is 1 m
Then perimeter of one fourth circle=$$\dfrac{1}{4}\times (2\pi r)=\dfrac{1}{4}\left ( 2\pi (1) \right )=\dfrac{1}{2}\pi $$
Then perimeter of four one fourth circle=$$4\times \dfrac{1}{2}\pi =2\times 3.14=6.28 m$$
Then perimeter of shaded figure $$= 42-6.28=35.72 m$$
Question 2
1 / -0

If the radius of a circle is 35 m, then the circumference of a circle is:

A
$$110\ m$$

B
$$220\ m$$

C
$$70\ m$$

D
None of these

Solution

Given that, the radius of a circle is, $$r = 35\ m$$
To find out: The circumference of the circle.

We know that, the circumference of a circle is $$\displaystyle 2 \pi r$$

$$\therefore \ $$ Circumference of the given circle $$=2 \times \dfrac{22}{7} \times 35\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$$
$$\therefore \ $$ Circumference $$= 220\ m$$

Hence, if the radius of a circle is $$35\ m$$, the circumference of the circle is $$220\ m$$.
Question 3
1 / -0

Find the area of the parallelogram whose base is 16 cm and height 0.4 m ?

A
440 $$cm^2$$

B
740 $$cm^2$$

C
640 $$cm^2$$

D
None of these

Solution

Base = 16 cm
Height = 0.4 m
= 0.4 $$\times$$ 100 = 40 cm
Area of parallelogram = b $$\times$$ h
= 16 $$\times$$ 40
= 640 $$cm^2$$
Question 4
1 / -0

What will be area of a parallelogram with base 6 cm and altitude 3.5 cm?

A
28 square cm

B
20 square cm

C
21 square cm

D
None of these

Solution

Area of parallelogram = Base $$\times$$ Altitude
= 6 cm $$\times$$ 3.5 cm
= 21 square cm
Question 5
1 / -0

Find the area of a circle whose radius is 7 cm (in cm)

A
$$\displaystyle 42\pi $$

B
$$\displaystyle 49\pi $$

C
$$\displaystyle 54\pi $$

D
$$\displaystyle 60\pi $$

Solution

Area of a circle $$ = \pi {r}^{2} $$, where r is the radius of the circle
Since $$r = 7$$
So, area of the given circle $$ = \pi \times 7 \times 7 = 49 \pi $$
Question 6
1 / -0

A well of diameter $$150\text{ cm}$$ has a $$30\text{ cm}$$ wide parapet running around it. Find the area of the parapet.

A
$$16971.428$$ $$\text{cm}^2$$

B
$$1697.142$$ $$\text{cm}^2$$

C
$$169.714$$ $$\text{cm}^2$$

D
$$16971.42$$ $$\text{cm}^2$$

Solution

Diameter of well $$= 150\text{ cm}$$
Radius of well = $$\displaystyle \frac{150}{2} = 75 \text{ cm}$$
Width of parapet $$= 30 \text{ cm}$$
Now, parapet of the well forms two circles.
One is the inner circle and the second is the outer circle.
Radius of inner circle $$(r) = 75 \text{ cm}$$
Radius of outer circle $$R =$$ inner radius $$+$$ width of parapet $$= 75 + 30 = 105 \text{ cm}$$
Area of parapet $$=$$ Area of outer circle $$-$$ Area of inner circle
$$=\pi R^2 - \pi r^2 = \pi (R^2 - r^2)$$
$$=\displaystyle \frac{22}{7} [(105)^2 - (75)^2]$$
$$=\displaystyle \frac{22}{7} \times 5400 $$
$$=\displaystyle \frac{118800}{7} $$
$$ = 16971.428 \text{ cm}^2$$
Question 7
1 / -0

If a square of area $$3$$ square units is inscribed in a circle, then the area of the circle is

A
$$\displaystyle \frac{3}{2}\pi\ sq.cm$$

B
$$\displaystyle 9\pi ^{2}\ sq.cm$$

C
$$\displaystyle 6\pi\ sq.cm$$

D
$$\displaystyle \sqrt{3}\pi \ sq.cm$$

Solution

Given, area of a square $$=3$$ sq.cm
Let $$a$$ be side of square.
$$\Rightarrow$$ $$a^2=3$$
$$\Rightarrow$$ $$a=\sqrt{3}$$ cm

Diagonal $$=a\sqrt{2}$$
$$=\sqrt{3}\times \sqrt{2}$$ cm
$$=\sqrt{6}$$ cm

Diameter of circle $$=$$ diagonal of square $$=\sqrt{6}$$ cm

Radius $$=\dfrac{\sqrt{6}}{2}\ cm$$
$$\Rightarrow r=\dfrac{\sqrt{3}}{\sqrt2}\ cm$$

$$\therefore$$ Area of the circle $$=\pi r^2$$
$$=\pi\times \left(\dfrac{\sqrt{3}}{\sqrt2}\right)^2\ cm^2$$
$$=\dfrac{3}{2}\pi\ cm^2$$

Hence, the area of the circle is $$\dfrac{3}{2}\pi\ cm^2$$
Question 8
1 / -0

If a circle is inscribed in a square of area $$4$$ sq units, then the area of the circle is

A
$$\displaystyle \pi $$

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

C
`$$\displaystyle \frac{\pi }{4}$$

D
$$\displaystyle \frac{3\pi }{4}$$

Solution

$$\Rightarrow$$ Area of a square $$=4$$ square units.
Let $$a$$ be side of square.
$$\Rightarrow$$ $$a^2=4$$
$$\Rightarrow$$ $$a=2$$ units.
Diameter of circle and side of square are equal.
$$\therefore$$ $$r=\dfrac{2}{2}=1$$ unit.
$$\Rightarrow$$ Area of a circle inscribed in a square $$=\pi r^2$$
$$=\pi\times (1)^2$$
$$=\pi$$
Question 9
1 / -0

The diameter of a wheel is $$98\ cm$$ The number of revolutions it will have to make to cover a distance of $$1540\ m$$ is

A
500

B
600

C
700

D
800

Solution

Given the diameter of wheel is $$98\ cm$$
Then radius of wheel =$$\dfrac{98}{2}=49\ cm$$
Then circumference of wheel $$=2\pi r$$
$$=2\times \dfrac{22}{7}\times 49=308\ cm$$
the number of revolutions to be made to cover a distance of $$1540\ m=\dfrac{1540\times 100\ cm}{308\ cm}$$
$$=\dfrac{154000}{308}=500$$
Question 10
1 / -0

If P represents the area and W represents the circumference of the circle, then P in terms of W is

A
$$\displaystyle \frac{2\pi }{W}$$

B
$$\displaystyle \frac{4\pi^{2} }{W}$$

C
$$\displaystyle \frac{2\pi^{2} }{W}$$

D
$$\displaystyle \frac{W^{2} }{4\pi }$$

Solution

Since P is the area so P = $$\displaystyle \pi r^{2}$$ ....(1)
W is the perimeter thus W = $$\displaystyle 2\pi r\Rightarrow r=\frac{W}{2\pi }$$
Put the value of $$r$$ in (1)
$$\displaystyle \Rightarrow P=\pi \left (\frac{W}{2\pi } \right )^{2}$$
$$\displaystyle P=\pi \frac{W^{2}}{4\pi ^{2}}$$
$$\displaystyle \Rightarrow P= \frac{W^{2}}{4\pi }$$

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Perimeter and Area Test - 27

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Perimeter and Area Test - 27

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Question 1

40 m

40.07 m

35.72m

35 m

Solution

Question 2

$$110\ m$$

$$220\ m$$

$$70\ m$$

None of these

Solution

Question 3

440 $$cm^2$$

740 $$cm^2$$

640 $$cm^2$$

None of these

Solution

Question 4

28 square cm

20 square cm

21 square cm

None of these

Solution

Question 5

$$\displaystyle 42\pi $$

$$\displaystyle 49\pi $$

$$\displaystyle 54\pi $$

$$\displaystyle 60\pi $$

Solution

Question 6

$$16971.428$$ $$\text{cm}^2$$

$$1697.142$$ $$\text{cm}^2$$

$$169.714$$ $$\text{cm}^2$$

$$16971.42$$ $$\text{cm}^2$$

Solution

Question 7

$$\displaystyle \frac{3}{2}\pi\ sq.cm$$

$$\displaystyle 9\pi ^{2}\ sq.cm$$

$$\displaystyle 6\pi\ sq.cm$$

$$\displaystyle \sqrt{3}\pi \ sq.cm$$

Solution

Question 8

$$\displaystyle \pi $$

$$\displaystyle \frac{\pi }{2}$$

`$$\displaystyle \frac{\pi }{4}$$

$$\displaystyle \frac{3\pi }{4}$$

Solution

Question 9

500

600

700

800

Solution

Question 10

$$\displaystyle \frac{2\pi }{W}$$

$$\displaystyle \frac{4\pi^{2} }{W}$$

$$\displaystyle \frac{2\pi^{2} }{W}$$

$$\displaystyle \frac{W^{2} }{4\pi }$$

Solution

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