Perimeter and Area Test

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Perimeter and Area Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

Perimeter of a triangle is the sum of the lengths of all the ............ sides.

A
$$4$$

B
$$2$$

C
$$3$$

D
$$6$$

Solution

Perimeter of a triangle is the sum of the lengths of all the three sides.
Question 2
1 / -0

What is the area of the given figure? $$ABCD$$ is a rectangle and $$BDE$$ is an isosceles right triangle.

A
$$ab$$

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

C
$$cab$$

D
$$b\left(a + \dfrac b2\right)$$
Question 3
1 / -0

If the radii of two concentric circles are $$15\ \text{cm}$$ and $$13\ \text{cm}$$, respectively, then the area of the circulating ring in sq. cm will be:

A
$$176$$

B
$$178$$

C
$$180$$

D
$$200$$

Solution

$$R = 15\ \text{cm}, r = 13\ \text{cm}. $$
Area of the circulating ring $$= \pi (R^2-r^2)$$
$$= \cfrac {22}{7} ({15}^2-{13}^2)\ \text{cm}^2$$
$$= \cfrac {22}{7} \times 28 \times 2\ \text{cm}^2$$
$$=176\ \text{cm}^2$$
Question 4
1 / -0

The diameter of each circle shown in the given figure is 'd' then the area of the square is given by

A
$$9\displaystyle d^{2}$$

B
$$9 d$$

C
$$3\displaystyle d^{2}$$

D
$$\displaystyle \frac{3}{4} d^{2}$$

E
$$\displaystyle \frac{4}{3} d^{2}$$

Solution

There are 3 circles of diameter 'd
So side of the square is 3d and
Hence area is $$side^2=\displaystyle 9d^{2}$$
Question 5
1 / -0

The area of a circle is 301.84 $$\displaystyle cm^{2} $$ Then its radius is

A
$$9.2$$ cm

B
$$9.3$$ cm

C
$$9.8$$ cm

D
9.6 cm

Solution

Area of the circle = $$\displaystyle \pi r^{2}$$
According to the question,
$$\displaystyle \pi r^{2}=301.84$$
$$\displaystyle \Rightarrow \frac{22}{7}r^{2}=301.84$$
$$\displaystyle r^{2}=\frac{7\times 301.84}{22}=96.04$$
$$\displaystyle \therefore r=\sqrt{96.04}=9.8\ cm$$
Question 6
1 / -0

A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular park is $$132$$ m. Its width is

A
$$20$$ m

B
$$21$$ m

C
$$22$$ m

D
$$24$$ m

Solution

Let $$r_1$$ and $$C_1$$ be the radius and the circumference of the inner circle.
Let $$r_2$$ and $$C_2$$ be the radius and the circumference of the outer circle.
The width of the circular path is $$(r_2-r_1)$$ m.
Given, $$C_2-C_1=132$$ m
$$\Rightarrow 2\pi r_2 - 2\pi r_1=132$$
$$\Rightarrow 2\pi (r_2-r_1)=132$$
$$\Rightarrow r_2-r_1=\cfrac{66}{\dfrac{22}{7}}$$
$$\Rightarrow r_2-r_1=21$$
Thus, width of the circular path is $$21$$ m.
Question 7
1 / -0

The radius of a circle whose area is equal to the sum of the areas of two circles of radii 5 cm and 12 cm is

A
13 cm

B
14 cm

C
15 cm

D
17 cm

Solution

We know that area of a circle of radius $$r = πr ^2 $$
By given condition,
$${ A }_{ 3 }={ A }_{ 1 }+{ A }_{ 2 }$$ .....(1)
Since, $$r_1 = 5$$ & $$r_2= 12$$
So, by (1), we have
$$ { \pi r }^{ 2 }={ \pi 5 }^{ 2 }+{ \pi 12 }^{ 2 }\\\\ { r }^{ 2 }={ 5 }^{ 2 }+{ 12 }^{ 2 }\\\\ r=\sqrt { 25+144 } =\sqrt { 169 } =13 \text{ cm}$$
Question 8
1 / -0

The radius of a circle is increased by 1 cm. Then the ratio of new circumference to the new diameter is

A
$$\pi :3$$

B
$$\pi :2$$

C
$$\pi :1$$

D
$$\pi :\frac { 1 }{ 2 } $$

Solution

New radius = $$(r+1)\ \mathrm{cm} $$
Ratio = $$2\pi (r+1):2(r+1)=\pi :1$$
Question 9
1 / -0

From a square metal sheet of side $$28\;cm$$, a circular sheet is cut off. Find the radius of the largest possible circular sheet that can be cut. Also find the area of the remaining sheet.

A
$$14\;cm,\;148\;cm^2$$.

B
$$14\;cm,\;168\;cm^2$$.

C
$$12\;cm,\;168\;cm^2$$.

D
$$14\;cm,\;164\;cm^2$$.

Solution

Area of square sheet $$=$$ $${ (28) }^{ 2 }=784{ cm }^{ 2 }$$

The largest circle of diameter equals to the side of square can be cut off from the square sheet.

$$\therefore $$ Radius of circular sheet $$=$$ $$\dfrac { 28 }{ 2 } =14cm$$

Area of remaining sheet $$=$$ $$784-\pi { r }^{ 2 }=784-\dfrac { 22 }{ 7 } \times 14\times 14$$

$$=$$ $$784 - 616$$ = $$168$$$${ cm }^{ 2 }$$
Question 10
1 / -0

If the perimeter and area of a circle are numerically equal then the radius of the circle is

A
$$6$$ units

B
$$\displaystyle \pi $$ units

C
$$4$$ units

D
$$2$$ units

Solution

$$\textbf{Step 1: Substitute area of circle and perimeter of circle formula in the given condition}$$
$$\text{Area of circle}=\pi r^2$$
$$\text{Perimeter of circle}=2\pi r$$
$$\text{Given that area and perimeter are numerically equal}$$
$$\implies \pi r^2=2\pi r$$
$$\implies r=2$$

$$\textbf{Thus, the required radius of the circle is 2 units}$$

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Re-Attempt Weekly Quiz Competition

Perimeter and Area Test - 11

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Perimeter and Area Test - 11

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SHARING IS CARING

Question 1

$$4$$

$$2$$

$$3$$

$$6$$

Solution

Question 2

$$ab$$

$$\displaystyle ab^{2}$$

$$cab$$

$$b\left(a + \dfrac b2\right)$$

Question 3

$$176$$

$$178$$

$$180$$

$$200$$

Solution

Question 4

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

$$9 d$$

$$3\displaystyle d^{2}$$

$$\displaystyle \frac{3}{4} d^{2}$$

$$\displaystyle \frac{4}{3} d^{2}$$

Solution

Question 5

$$9.2$$ cm

$$9.3$$ cm

$$9.8$$ cm

9.69.6 cm

Solution

Question 6

$$20$$ m

$$21$$ m

$$22$$ m

$$24$$ m

Solution

Question 7

13 cm

14 cm

15 cm

17 cm

Solution

Question 8

$$\pi :3$$

$$\pi :2$$

$$\pi :1$$

$$\pi :\frac { 1 }{ 2 } $$

Solution

Question 9

$$14\;cm,\;148\;cm^2$$.

$$14\;cm,\;168\;cm^2$$.

$$12\;cm,\;168\;cm^2$$.

$$14\;cm,\;164\;cm^2$$.

Solution

Question 10

$$6$$ units

$$\displaystyle \pi $$ units

$$4$$ units

$$2$$ units

Solution

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9.6 cm