Perimeter and Area Test

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Perimeter and Area Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The circumference of a circle is $$44$$ m then the area of the circle is

A
$$\displaystyle 6084.5\:m^{2}$$

B
$$\displaystyle 276.5\:m^{2}$$

C
$$\displaystyle 154\:m^{2}$$

D
$$\displaystyle 44\:m^{2}$$

Solution

Circumference of circle: $$44$$ m
Let the radius of circle be $$r$$.
Therefore, $$2\pi r = 44$$
$$\Rightarrow r = 7$$ m
Area of circle:
$$\pi r^2 = \dfrac{22}7\times7^2 = 154\ m^2$$
Question 2
1 / -0

A circular park has a path of uniform width around it. The difference between outer and inner circumference of the circular path is $$132\ m$$. Its width is _____ $$\displaystyle \left(\pi=\frac{22}{7}\right)$$

A
$$22\ m$$

B
$$20\ m$$

C
$$21\ m$$

D
$$24\ m$$

Solution

Given $$\displaystyle 2\pi r_{2}-2\pi r_{1}=132$$
$$\displaystyle \Rightarrow r_{2}-r_{1}=\frac{132}{2\pi }=\frac{132\times 7}{44}=21\ m$$
Question 3
1 / -0

Find the area of the parallelogram whose base is $$17\ cm$$ and height $$0.8\ m$$?

A
$$\displaystyle 13.6\:cm^{2}$$

B
$$\displaystyle 1360\:cm^{2}$$

C
$$\displaystyle 13.6\:m^{2}$$

D
$$\displaystyle 1360\:m^{2}$$

Solution

Base $$= 17\ cm$$
Height $$= 0.8\ m =0.8 \times 100 = 80\ cm$$
Area of parallelogram
$$= b \times h$$
$$= 17\times 80$$
$$= 1360\ cm^2$$
Question 4
1 / -0

If the diameter of a circle is $$14$$ cm, then its circumference is

A
$$66\;cm$$

B
$$44\;cm$$

C
$$33\;cm$$

D
$$55\;cm$$

Solution

Since, diameter $$= d=14\;cm$$
$$\displaystyle C=\pi d=\frac{22}{7}\times 14=44\;cm$$
Question 5
1 / -0

The base of a parallelogram is three times its height. If the area of the parallelogram is $$75$$ sq cm, then its height is

A
$$5 cm$$

B
$$\displaystyle 5\sqrt{2}cm$$

C
$$\displaystyle 3\sqrt{2}cm$$

D
$$15 cm$$

Solution

let height of ||g be 'b', then base of ||g be 3b.
area of ||g$$=$$base$$\times $$height
$$75c{ m }^{ 2 }=3b\times b\\ { b }^{ 2 }=25$$
$$b=5$$cm, therefore height $$=5$$cm
Question 6
1 / -0

The length and breadth of a rectangle are in the ratio $$3:2$$. If the sides of the rectangle are extended on each side by $$1 m$$, the ratio of length to breadth becomes $$10:7$$. Find the area of the original rectangle in square meters.

A
$$2350\ m^{2}$$

B
$$1150\ m^{2}$$

C
$$54\ m^{2}$$

D
$$1000\ m^{2}$$

Solution

Let length of rectangle$$=3x$$
breadth of rectangle $$=2x$$
Now,
$$ \cfrac { 3x+1 }{ 2x+1 } =\cfrac { 10 }{ 7 } \\ 21x+7=20x+10\\ x=3$$
$$\therefore$$ length$$=3\times 3=9m,$$ breadth$$=2\times 3=6m$$
Area of original rectangle
$$=(9\times 6){ m }^{ 2 }\\ =54\ { m }^{ 2 }$$
Question 7
1 / -0

In the given figure, $$P$$ is the center of the circle. The area and perimeter respectively, of the figure are

A
$$3256 cm^{2}$$, $$234.2 cm$$

B
$$2492 cm^{2}$$, $$229.2 cm$$

C
$$2942 cm^{2}$$, $$234.2 cm$$

D
$$3256 cm^{2}$$, $$229.2 cm$$

Solution

Area of figure$$=$$area of circle + area of rectangular - area of quarter circle.
$$ =\pi { r }^{ 2 }+l\times b-\cfrac { 1 }{ 4 } \times \pi { r }^{ 2 }\\ =\cfrac { 3 }{ 4 } \pi { r }^{ 2 }+l\times b=\cfrac { 3 }{ 4 } \times 3.14\times 20\times 20+50\times 40\\ =942+2000\\ =2942c{ m }^{ 2 }$$
&
Perimeter of figure
$$=2\pi r-\cfrac { 2\pi r }{ 4 } +2(l+b)-2r\\ =\cfrac { 3\pi r }{ 2 } +2(l+b)-2r\\ =\cfrac { 3\times 3.14\times 20 }{ 2 } +2(50+40)-2\times 20\\ =94.2 +180-40\\ =234.2cm$$
Question 8
1 / -0

A field is in the form of a parallelogram whose base is $$420\ m$$ and altitude is $$36 m$$. Find the cost of watering at $$10$$ paise per sq. m.

A
Rs. $$15.120$$

B
Rs. $$1512$$

C
Rs. $$151.20$$

D
None

Solution

Base $$= 420\ m$$
Height $$= 36\ m$$
Area $$= b \times h = 420 \times 36$$
$$= 15,120 \displaystyle \ m^{2}$$
The cost of watering per sq m
$$= 10$$ paise
Cost of watering the field $$= 15120 \times 0.1$$
Rs. $$1512$$
Question 9
1 / -0

The ratio between the length and breadth of a rectangular garden is $$5:3$$. If the perimeter of the garden is $$160$$ meters, what will be the area of $$5\text{ m}$$ wide road around its outside?

A
$$600\displaystyle \text{ m}^{2}$$

B
$$1200\displaystyle \text{ m}^{2}$$

C
$$900\displaystyle \text{ m}^{2}$$

D
$$1000\displaystyle \text{ m}^{2}$$

Solution

Let the length and breadth of the rectangular garden be $$5x$$ meter and $$3x$$ meters.
Given $$2(5x + 3x) = 160 $$
$$ \Rightarrow 16x = 160 $$
$$ \Rightarrow x = 10$$
$$\therefore $$ Length of garden $$= 50 \text{ m}$$ and Breadth of garden $$= 30 \text{ m}$$
Area of $$5 \text{ m}$$ wide road around its outside $$=$$ Area of outer rectangle $$-$$ Area of inner rectangle
$$[\because $$ length of outer rectangle $$= 50+2\times 5$$ and breadth of outer rectangle $$=30+2\times 5 ]$$
Area $$= (50 + 10) \times (30 + 10) - 50 \times 30$$
$$= 60 \times 40 - 50 \times 30$$
$$= (2400 - 1500)$$
$$ = 900 \text{ m}^{2} $$
$$\therefore$$ Area of road is $$900\text{ m}^2$$
Question 10
1 / -0

Circumference of a circle is equal to

A
$$\pi\;r$$

B
$$2\pi\;r$$

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

D
$$\displaystyle2+ \frac{\pi r}{2}r$$

Solution

$$c=\pi d=2\pi r$$

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Perimeter and Area Test - 21

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Perimeter and Area Test - 21

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SHARING IS CARING

Question 1

$$\displaystyle 6084.5\:m^{2}$$

$$\displaystyle 276.5\:m^{2}$$

$$\displaystyle 154\:m^{2}$$

$$\displaystyle 44\:m^{2}$$

Solution

Question 2

$$22\ m$$

$$20\ m$$

$$21\ m$$

$$24\ m$$

Solution

Question 3

$$\displaystyle 13.6\:cm^{2}$$

$$\displaystyle 1360\:cm^{2}$$

$$\displaystyle 13.6\:m^{2}$$

$$\displaystyle 1360\:m^{2}$$

Solution

Question 4

$$66\;cm$$

$$44\;cm$$

$$33\;cm$$

$$55\;cm$$

Solution

Question 5

$$5 cm$$

$$\displaystyle 5\sqrt{2}cm$$

$$\displaystyle 3\sqrt{2}cm$$

$$15 cm$$

Solution

Question 6

$$2350\ m^{2}$$

$$1150\ m^{2}$$

$$54\ m^{2}$$

$$1000\ m^{2}$$

Solution

Question 7

$$3256 cm^{2}$$, $$234.2 cm$$

$$2492 cm^{2}$$, $$229.2 cm$$

$$2942 cm^{2}$$, $$234.2 cm$$

$$3256 cm^{2}$$, $$229.2 cm$$

Solution

Question 8

Rs. $$15.120$$

Rs. $$1512$$

Rs. $$151.20$$

None

Solution

Question 9

$$600\displaystyle \text{ m}^{2}$$

$$1200\displaystyle \text{ m}^{2}$$

$$900\displaystyle \text{ m}^{2}$$

$$1000\displaystyle \text{ m}^{2}$$

Solution

Question 10

$$\pi\;r$$

$$2\pi\;r$$

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

$$\displaystyle2+ \frac{\pi r}{2}r$$

Solution

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