Perimeter and Area Test

Result

Perimeter and Area Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

In figure, what is the approximate area of parallelogram DAWN?

A
11.57

B
13.64

C
14.63

D
17.25

E
20.00

Solution

area of parallelogram $$=ab\sin \theta$$
$$AD\times DN \sin 47^o$$
$$4\times 5\times \sin 47^o$$
$$20\times 0.731$$
$$14.63$$
Question 2
1 / -0

A lawn $$30$$ m long and $$16$$ m wide is surrounded by a path $$2$$ m wide. What is the area of the path?

A
$$200$$ sq. m

B
$$280$$ sq. m

C
$$300$$ sq. m

D
$$320$$ sq. m

Solution

Given, length of lawn $$=16$$ m, width $$=2$$ m
We need to find the area of the path.
Figure is shown according to question having sides $$=16+2+2 , 30+2+2 = 20m, 34m$$
Thus area of path $$= (34\times 20 - 30\times 16)m^{2}$$
$$= (680 - 480) $$
$$= 200$$ sq. m
Question 3
1 / -0

Two similar parallelograms have corresponding sides in the ratio $$1:k$$. What is the ratio of their areas?

A
$$1:3{k}^{2}$$

B
$$1:4{k}^{2}$$

C
$$1:{k}^{2}$$

D
$$1:2{k}^{2}$$

Solution
Question 4
1 / -0

A wire of length 36 cm is bent in the form of a semicircle. What is the radius of the semicircle?

A
$$9\ cm$$

B
$$8\ cm$$

C
$$7\ cm$$

D
$$6\ cm$$

Solution

Length of wire is in the form of semicircle.
Given, length of wire $$=36\ cm$$
The length of the wire will cover the diameter as well as the circumference of the semicircle.
$$\therefore \ $$Length of the wire $$= \dfrac {1}{2} (2\pi r) + 2r\\$$
$$\Rightarrow 36 = (\pi + 2)r\\$$
$$\Rightarrow 36 = \left (\dfrac {22}{7} + 2\right ) r = \dfrac {36}{7}\times r\\$$
$$\Rightarrow r = 7\ cm$$.

Hence, the radius of the semicircle is $$7\ cm$$.
Question 5
1 / -0

If the area of a semicircle is 84 cm$$^2$$, then the area of the circle is .......................

A
144 cm$$^2$$

B
42 cm$$^2$$

C
168 cm$$^2$$

D
288 cm$$^2$$

Solution

We know that the area of semicircle is one half of a circle .
Given area of semicircle is $$84 cm^{2}$$
Then area of circle $$=2\times 84$$
$$=168 cm^{2}$$
Question 6
1 / -0

In the following figure, what is the area of the shaded circle inside of the square?

A
512

B
256

C
16

D
50.24

E
12.57

Solution
Question 7
1 / -0

Take a circle with radius $$5$$ m, then determine the area of given circle with suitable units.

A
$$78.53\ m$$

B
$$78.53\ m^{2}$$

C
$$78.53\ cm$$

D
$$78.53\ cm^{2}$$

Solution

Area of circle$$=\pi { r }^{ 2 }=\cfrac { 22 }{ 7 } \times 5\times 5=78.57m^2$$
Question 8
1 / -0

Find the circumference of a circle whose radius is-
$$35\ cm$$

A
$$220\ cm$$

B
$$20\ cm$$

C
$$22\ cm$$

D
None of these

Solution

Given: Radius of circle $$(r)=35\space\mathrm{cm}$$
$$\therefore$$ Circumference of circle $$=2\pi r=2\times\dfrac{22}{7}\times35=220\space\mathrm{cm}$$
So, $$\text{A}$$ is the correct option.
Question 9
1 / -0

Calculate perimeter of a right angled triangle whose non-hypotenuse sides are $$6$$ cm and $$8$$ cm.

A
$$18\ cm$$

B
$$14\ cm$$

C
$$24\ cm$$

D
$$11\ cm$$

Solution

In right $$\triangle PQR$$, using Pythagoras theorem,
$$P{ R }^{ 2 }=P{ Q }^{ 2 }+Q{ R }^{ 2 }={ 6 }^{ 2 }+{ 8 }^{ 2 }=36+64=100$$
$$PR=\sqrt { 100 } =10㎝$$
$$\therefore $$Perimeter of $$PQR=PQ+QR+RP=6+8+10=24 cm$$
Question 10
1 / -0

In the given figure, each square contains four square grid of equal dimensions. Then calculate the area of the given figure.

A
$$16\ cm^{2}$$

B
$$20\ cm^{2}$$

C
$$24\ cm^{2}$$

D
$$28\ cm^{2}$$

Solution

Area of figure$$=4\times $$ area of $$1$$ square
$$=4{ a }^{ 2 }$$
Put $$a=2$$, then area $$=4\times { 2 }^{ 2 }=16 cm^2$$