Perimeter and Area Test

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Perimeter and Area Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

The radius of a circle is increased by $$5$$ units. What is ratio of the new circumference and the new diameter?

A
$$x-5$$

B
$$\pi +5$$

C
$$\pi$$

D
$$\pi -1$$

Solution

Let the radius of the circle be $$r$$.
$$\therefore$$ New circumference $$=2\pi(r+5)$$
and new diameter $$=2(r+5)$$
$$\therefore$$ Ratio$$=\cfrac{2\pi(r+5)}{2(r+5)}$$
$$=\pi$$
Question 2
1 / -0

If circumference of a circle is $$\displaystyle 3\pi $$, then its area is

A
$$\displaystyle \frac{7\pi }{2}$$

B
$$\displaystyle 9\pi ^{2}$$

C
$$\displaystyle 4\pi ^{2}$$

D
$$\displaystyle \frac{9\pi }{4}$$

Solution

$$\Rightarrow$$ Circumference of a circle $$=3\pi$$
$$\Rightarrow$$ $$2\pi r=3\pi$$
$$\Rightarrow$$ $$2r=3$$
$$\Rightarrow$$ $$r=\dfrac{3}{2}$$
$$\Rightarrow$$ Area of a circle $$=\pi r^2$$
$$=\pi\times \left(\dfrac{3}{2}\right)^2$$

$$=\dfrac{9}{4}\pi$$
Question 3
1 / -0

$$ \displaystyle \frac{1}{5}\, of\,10 km= $$ _____m

A
2

B
200

C
20

D
2000

Solution

$$ \displaystyle \frac{1}{5} $$ of 10 km
$$ \displaystyle \frac{1}{5}\times10 \ km $$= 2 km=2000 m
Question 4
1 / -0

A wall is made up of square bricks of unit length. Then its area is _____

A
1 sq unit

B
3 sq units

C
20 sq units

D
24 sq units

Solution

As 24 square bricks are required for the wall
So Area = 24 $$ \times $$ area of one brick =$$ 24 \times 1$$= 24 sq units
Question 5
1 / -0

Area of a rectangle is $$120 $$ $$m^2$$ and the breadth is $$5 m$$ . Then its length is

A
$$204 m$$

B
$$24 m$$

C
$$28 m$$

D
$$26 m$$

Solution

We know, area of rectangle = $$ l \times b $$
Since, area $$= 120$$
$$\therefore $$ $$l \times b = 120$$
$$\Rightarrow $$ length =$$l =\displaystyle \frac{120}{5}=24 m$$
Question 6
1 / -0

A figure is formed by putting two squares one on the other as shown below. If the length of each side of the two squares is $$8$$ cm, then the perimeter of the formed figure is

A
$$56$$ cm

B
$$64$$ cm

C
$$32$$ cm

D
$$48$$ cm

Solution

New dimensions are $$b = 16$$ & $$ l=8 $$
$$\displaystyle \therefore $$Perimeter of the figure formed
$$\displaystyle = 16+8+16+8= \displaystyle 48$$ cm
Question 7
1 / -0

The total boundary length of circle is called

A
area

B
volume

C
circumference

D
diameter

Solution

Boundary length of circle is called circumference which is the same as perimeter of circle.
Question 8
1 / -0

Circumference of a circle is equal to

A
$$\displaystyle \pi r$$

B
$$2\displaystyle \pi r$$

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

D
$$\displaystyle 2+\frac{\pi r}{2}r$$

Solution

The circumference of a circle whose radius is equal to $$r$$ is given by $$2\pi r.$$
Question 9
1 / -0

The circumference of a circle is called

A
chord

B
segment

C
boundary

D
None

Solution

Perimeter of circle = Circumference ---eq.1
Perimeter = Boundary ---eq.2
from eq.1 & eq.2
Circumference = Boundary
$$ \therefore $$ The circumference of a circle is called boundary.
Question 10
1 / -0

The ______ of a circle is called the circumference.

A
area

B
volume

C
perimeter

D
radius

Solution

The perimeter of a circle is called the circumference.
Since, the circumference of a circle is the distance across the boundary of the circle.

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Re-Attempt Weekly Quiz Competition

Perimeter and Area Test - 13

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Perimeter and Area Test - 13

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SHARING IS CARING

Question 1

$$x-5$$

$$\pi +5$$

$$\pi$$

$$\pi -1$$

Solution

Question 2

$$\displaystyle \frac{7\pi }{2}$$

$$\displaystyle 9\pi ^{2}$$

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

$$\displaystyle \frac{9\pi }{4}$$

Solution

Question 3

2

200

20

2000

Solution

Question 4

1 sq unit

3 sq units

20 sq units

24 sq units

Solution

Question 5

$$204 m$$

$$24 m$$

$$28 m$$

$$26 m$$

Solution

Question 6

$$56$$ cm

$$64$$ cm

$$32$$ cm

$$48$$ cm

Solution

Question 7

area

volume

circumference

diameter

Solution

Question 8

$$\displaystyle \pi r$$

$$2\displaystyle \pi r$$

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

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

Solution

Question 9

chord

segment

boundary

None

Solution

Question 10

area

volume

perimeter

radius

Solution

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