Perimeter and Area Test

Result

Perimeter and Area Test - 17

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A circular park is surrounded by a road $$21\ m$$ wide. If the radius of the park is $$105\ m$$, find the area of the road.

A
$$13860\ m^{2}$$

B
$$30492\ m^{2}$$

C
$$7623\ m^{2}$$

D
$$15246\ m^{2}$$

Solution

$$O$$ is the centre of a circular park of radius $$OA=105\ m$$ with a concentric surrounding circular road of width $$AB=21\ m$$
To find the area of the road, first we need to find area of the park and then subtract the area of park from total area including road.
Let's find area of a park.
Park forms a circle with radius$$(r)=105\ m$$
$$\therefore$$ Area of the park $$=\pi r^2=\pi (105)^2$$
The outer circle represents total area which includes park and road with radius$$(R)=105+21=126$$
$$\therefore$$ Total area $$=\pi R^2=\pi(126)^2$$
Thus, area of the road $$=$$ Total area $$-$$ Area of park
$$=\pi(126)^2-\pi(105)^2=\pi(126^2-105^2)=4851\pi=15246\ m^2$$
Hence, area of road $$=15246\ m^2$$
Question 2
1 / -0

The perimeter of a quadrant of a circle of radius $$r$$ is:

A
$$\dfrac{\pi r}{2}$$

B
$$2\pi r$$

C
$$\dfrac{r}{2}\left [ \pi +4\right ]$$

D
$$2\pi r+\dfrac{r}{2}$$

Solution

Perimeter of quadrant of a circle $$=\cfrac { 1 }{ 4 } \times 2\pi \times r+2r$$
$$=\cfrac { \pi r }{ 2 } +2r\\ =\cfrac { r }{ 2 } \left[ \pi +4 \right] $$
Question 3
1 / -0

The base and the corresponding altitude of a parallelogram are $$10\: cm$$ and $$3.5\: cm$$, respectively. The area of the parallelogram is

A
$$30\: cm^2$$

B
$$35\: cm^2$$

C
$$70\: cm^2$$

D
$$ 17.5\:cm^2$$

Solution

The area of the parallelogram is base $$\times$$ height $$cm^2$$

Area of the parallelogram$$=(10)(3.5)=35\:cm^2$$.
Question 4
1 / -0

If the base of a parallelogram is $$8\ cm$$ and its altitude is $$5\ cm$$, then its area is equal to

A
$$15\ cm^{2}$$

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

C
$$40\ cm^{2}$$

D
$$10\ cm^{2}$$

Solution

Area of parallelogram $$=Base \times height $$
$$=8\times 5\\ =40\ { cm }^{ 2 }$$
Question 5
1 / -0

A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running around it on the outside. Find the cost graveling the path at Rs 4 per square meter.

A
$$Rs 2002$$

B
$$Rs 2003$$

C
$$Rs 2004$$

D
$$Rs 2000$$

Solution

Given : Radius ($$r$$) $$=21m$$
Radius ($$R$$) $$=24.5m$$

Required area $$=\pi (R^{2}-r^{2})$$

Required area $$=\pi (24.5^{2}-21^{2})=500.5$$

Cost $$=500.5\times 4$$

$$=2002Rs$$
Question 6
1 / -0

In fig. 3, the area of the shaded region is :

A
$$\pi \left ( r_1+r_2 \right )$$

B
$$\pi \left ( r_1^{2}+r_2^{2} \right )$$

C
$$\pi \left ( r_1-r_2 \right )$$

D
$$\pi \left ( r_2^{2}-r_1^{2} \right )$$

Solution

The area of the given shaded region will be :
The area of the smaller circle subtracted from the larger circle
Area of larger circle is $$\pi r_{2}^{2}$$ and
Area of smaller circle is $$\pi r_{1}^{2}$$
Then, the ares of shaded region is $$\pi r_{ 2 }^{ 2 } -\pi r_{ 1 }^{ 2 }=\pi(r_{2}^{2}-r_{1}^{2})$$
Question 7
1 / -0

If a circular grass lawn of $$35\ m$$ in radius has a path $$7\ m$$ wide running around it on the outside, then the area of the path is

A
$$1450\ m^2$$

B
$$1576\ m^2$$

C
$$1694\ m^2$$

D
$$3368\ m^2$$

Solution

Radius of bigger circle(with the path) = $$35 + 7 = 42\ m.$$
So, $$R = 42$$ $$cm$$ & $$r = 35 $$ $$cm$$
Thus area of the path $$=$$ Area of bigger circle $$-$$ Area of smaller circle
$$\therefore$$ Required area $$= \pi (42)^2 - \pi (35)^2 = \dfrac{22}{7} \times (42 + 35)(42 - 35) = 22 \times 77 = 1694\ m^2$$
Question 8
1 / -0

A pond that is $$100$$ m in diameter is surrounded by a circular grass walk that is $$2$$ m wide. How many square meters of grass is there for the walk?

A
$$98 \pi $$

B
$$100 \pi $$

C
$$102 \pi $$

D
$$204 \pi $$

Solution

Area of grass on walk $$ = $$ Area of outer circle $$-$$ Area of the pond
$$ = \pi \times 52^2 - \pi\times 50^2$$
$$ = \pi (2704 - 2500)$$
$$ = \pi\times 204$$
So, Area of grass for the walk $$ = 204\pi m^2$$
Question 9
1 / -0

Find the area of a circular ring formed by two concentric circles whose radii are $$5.7\ cm$$ and $$4.3\ cm$$ respectively (Take $$\pi =3.1416$$)

A
$$43.98$$ sq. cm

B
$$53.67$$ sq. cm

C
$$47.24$$ sq. cm

D
$$38.54$$ sq. cm

Solution

Let the radii of the outer and inner circles be $$r_2$$ and $$r_1$$ respectively; we have
$$Area =\pi r_2^2 -\pi r_1^2 =\pi (r_2^2 - r_1^2)$$
$$=\pi (r_2 -r_1)(r_2+r_1)$$
$$=\pi (5.7 -4.3)(5.7 + 4.3)=\pi \times 1.4 \times 10$$ sq. cm
$$=3.1416 \times 14 sq. cm. =43.98$$ sq. cm
Question 10
1 / -0

A track is in the form of a ring whose inner circumference is $$352\ m$$ and the outer circumference is $$396\ m$$. The width of the track is:

A
$$44\ m$$

B
$$14\ m$$

C
$$22\ m$$

D
$$7\ m$$

Solution

Let the radii of inner and outer circles be $$r$$ and $$R$$, respectively.
Circumference of inner circle $$ = 352\ m$$
$$\therefore 2\pi r = 352$$
$$2\times \dfrac{22}{7}\times r=352$$
$$\Rightarrow r = 56\ m$$
Similarly, Circumference of outer circle $$ = 396 m$$
$$\therefore 2\pi R = 396$$
$$2\times \dfrac{22}{7}\times R=396$$
$$\Rightarrow R = 63\ m$$
Now, Width of the track $$ = R - r$$
$$ = 63 - 56$$
$$ = 7\ m$$
So, width of the track $$ = 7\ m$$