Perimeter and Area Test

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Perimeter and Area Test - 18

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Question 1
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If area of circular field is $$6.16 \text{ m}^2$$, then its diameter will be

A
$$6.8 \ \text{m}$$

B
$$2.8 \ \text{m}$$

C
$$12 \ \text{m}$$

D
$$6.84 \ \text{m}$$

Solution

Area of circular field $$=6.16\,\text{m}^2$$

Area of a circle $$=\pi r^2$$

$$\Rightarrow$$ $$6.16=\dfrac{22}{7}\times r^2$$

$$\Rightarrow$$ $$6.16\times \dfrac{7}{22}=r^2$$

$$\Rightarrow$$ $$r^2=\dfrac{43.12}{22}$$

$$\Rightarrow$$ $$r^2=1.96$$

$$\Rightarrow$$ $$r=1.4\,\text{m}$$

$$\Rightarrow$$ $$d=2r=1.4\times 2=2.8\text{ m}$$

$$\therefore$$ The diameter of a circular field is $$2.8\,\text{m}.$$
Question 2
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A circular grassy plot of land, $$42\ m$$ in diameter, has a path $$3.5\ m$$ wide running round it on the outside. Find the cost of gravel ling the path at Rs. $$4$$ per square metre.

A
Rs. $$2002$$

B
Rs. $$2003$$

C
Rs. $$2004$$

D
Rs. $$2000$$

Solution

We have, Radius of the plot $$ = 21\ m$$
Radius of plot including the path $$ = 21 + 3.5 = 24.5\ m$$
Area of the path $$ = \pi (24.5^2) - \pi (21^2)$$
$$ = \pi(24.5^2 - 21^2)$$
$$ = \pi (24.5 - 21) (24.5 + 21)$$
$$ = \cfrac {22}{7} \times 3.5\times 45.5$$
$$ = 500.5\ m^2$$
Total cost of gravelling the path $$ = 500.5\times 4$$
$$ =$$ Rs. $$2002$$
Question 3
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The circumference of a circular ground is $$88$$ metres. A strip of land, $$3$$ metres wide, inside and along circumference of the ground is to be levelled. What is the budgeted expenditure if the levelling costs Rs. $$7$$ per square metre?

A
$$1050$$

B
$$1125$$

C
$$1325$$

D
$$1650$$

Solution

Circumference of ground $$ = 88\ m$$
$$\Rightarrow 2\pi r = 88$$
$$\Rightarrow r = 14\ m$$ ....take $$\pi = \dfrac{22}{7}$$
Area of strip to be levelled $$ = $$ Area of ground $$-$$ Area of inner circle
$$ = \pi 14^2 - \pi 11^2$$
$$ = 235.72\ m^2$$ ....take $$\pi = \dfrac{22}{7}$$
Total cost of levelling the strip (if the cost of levelling $$1\ m^2$$ is Rs. $$7$$) $$ = 235.72\times 7$$
$$ =$$ Rs. $$1650$$
Question 4
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A flooring tile has the shape of a parallelogram whose base is $$24\: cm$$ and the corresponding height is $$10\: cm$$. How many such tiles are required to cover a floor of area $$1080$$ $$m^2$$?

A
$$20000$$

B
$$35000$$

C
$$45000$$

D
$$65000$$

Solution

Area of the parallelogram $$=$$ Base$$\times $$Height

So area of each tile $$=24\times 10=240 \:cm^2$$

Area of the floor $$=1080 \: m^2=(1080\times 100\times 100) \: cm^2$$

$$\therefore$$ Required number of tiles $$=\dfrac{\text{Area of the floor}}{\text{Area of each tile}}=\dfrac{10800000}{240}=45000$$
Question 5
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The adjacent sides of a parallelogram are 10 cm and 12 cm. The diagonal joining the ends of these sides is 14 cm. Its area is

A
$$48 \sqrt{6}$$ $$cm^2$$

B
$$48 \sqrt{5}$$ $$cm^2$$

C
$$48 \sqrt{3}$$ $$cm^2$$

D
none of these

Solution

Area of the parallelogram $$ABCD=2\times \text{ area } \triangle BCD$$

Now, in $$\triangle BCD$$
$$S=\dfrac { 12+10+14 }{ 2 } =18\text{ cm}$$

Area of $$\triangle BCD=\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) }$$
$$=\sqrt { 18\left( 18-12 \right) \left( 18-10 \right) \left( 18-14 \right) } $$
$$=\sqrt { 18\times 6\times 8\times 4 } $$
$$=\sqrt { 3\times 6\times 6\times 2\times 4\times 4 } $$
$$=4\times 6\sqrt { 6 } $$
$$=24\sqrt { 6 } \text{ cm}^{ 2 }$$

$$\therefore $$ Area of the parallelogram $$ABCD=2\times 24\sqrt { 6 } =48\sqrt { 6 } \text{ cm}^{ 2 }$$
Question 6
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The cost of fencing a circular field at the rate of $$Rs\:.240\: per\: metre$$ is $$Rs. \: 52,800$$ . The field is to be ploughed at the rate of $$Rs. 12.50 \: per \: m^{2}$$. Find the cost of ploughing the field.

A
$$Rs. \:48,125$$

B
$$Rs. \:38,125$$

C
$$Rs. \:10,125$$

D
$$Rs. \:11,125$$

Solution

Cost of fencing at the rate of Rs.$$240$$ per meter=$$Rs.52800$$
Therefore,
Circumference=$$\dfrac{52800}{240}$$ = $$220$$

Let r be the radius of the field
Therefore,
$$2\pi r=220$$

$$\Rightarrow r=\dfrac{220\times 7}{2\times 22}$$

$$\Rightarrow r=35$$

Therefore,
Area of the field=$$\pi r^2$$

=$$\dfrac{22}{7}\times 35\times 35$$

=$$3850 m^2$$

Cost of fencing the field = $$12.5\times 3850$$

=$$Rs.48125$$
Question 7
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The circumference of a circular field is $$308\:m$$ . Find its area.

A
$$7546 \: m^{2}$$

B
$$6546 \: m^{2}$$

C
$$7046 \: m^{2}$$

D
$$7846 \: m^{2}$$

Solution

Let the radius of the circle be $$r$$ cm
Circumference of the circle=$$308$$
Therefore,
$$2\pi r=308$$
$$=>r=\dfrac{308\times 7}{2\times 22}$$
$$=>r=49m$$
Therefore,
Area of the circle=$$\pi \times (49)^2$$
=$$\frac{22}{7}\times 2401$$
=$$7546m^2$$
Question 8
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The diameter of a circle is $$28 \:cm$$. Find its area.

A
$$616 \: cm^{2}$$

B
$$516 \: cm^{2}$$

C
$$116 \: cm^{2}$$

D
$$216 \: cm^{2}$$

Solution

The diameter of a circle is $$28cm$$

Therefore,
Radius of the circle=$$\dfrac{28}{2}=14cm$$

Therefore,
Area of the circle=$$\pi (14)^2$$

=$$\dfrac{22}{7}\times196$$

=$$616cm^2$$
Question 9
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Directions For Questions

The shaded region of the given diagram represents the lawn in the form of a house.
On the three sides of the lawn there are flower-beds having a uniform width of 2 m.

...view full instructions

Find the length and the breadth of the lawn.

A
26 m and 10 m

B
35 m and 12 m

C
41 m and 16 m

D
49 m and 18 m

Solution

Since, in the given figure
Outer length$$=30m$$ and Outer breadth$$=12m$$
As per given, On the three sides of the lawn there are flower-beds having a uniform width of 2 m.
$$\therefore$$ length of lawn$$=30-2-2=26m$$
and
breadth of lawn$$=12-2=10m$$
Question 10
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The diagram, given below, shows two paths drawn inside a rectangular field 80 m long and 45 m wide. The widths of the two oaths are 8 m and 15 m as shown. Find the area of the shaded portion.

A
$$1195\, m^{2}$$

B
$$8895\, m^{2}$$

C
$$1155\, m^{2}$$

D
$$1895\, m^{2}$$

Solution

Given, rectangular field is 80 m long and 45 m wide.
Two paths are drawn inside the rectangular field. The widths of the two oaths are 8 m and 15 m.
The two paths intersect. The length and breath of the intersection is $$8$$ m and $$15$$ m.

Area of the shaded region $$=(8\times 80) + (15\times 45) - (15\times 8)$$
$$=640+675-120$$ sq. m
$$=1195$$ sq. m