Perimeter and Area Test

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Perimeter and Area Test - 25

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Question 1
1 / -0

A bicycle wheel has diameter 1m. If the bicycle travels one kilometer, then the number of revolutions the wheel make is.

A
$$\dfrac {1}{\pi }$$

B
$$\dfrac {100}{\pi }$$

C
$$\dfrac {500}{\pi }$$

D
$$\dfrac {1000}{\pi }$$

Solution

Let the number of revolution of the wheel is $$n$$.
Since, $$\text{Radius} = \dfrac{1}{2} $$
Then,
$$n\ \times$$ Circumference of wheel = Distance travelled by bicycle
$$n \times 2\pi \times \frac {1}{2}=1\ \text{km}$$
$$n \times \pi =1000\ \text{m}$$
$$n=\frac {1000}{\pi }\ \text{m}$$
Question 2
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In the figure shown here, $$BDEC$$ is a rectangle. $$ABC$$ and $$DEF$$ are straight lines. The area, in cm$$^2$$ , of the whole figure is ______ .

A
$$120$$

B
$$128$$

C
$$136$$

D
$$142$$

Solution

Area of the whole figure= area of rectangle BDEC+ area of triangle ABD+ area of triangle ECF
$$=DE\times EC+\dfrac { 1 }{ 2 } \times AB\times BD+\dfrac { 1 }{ 2 } \times CE\times EF$$
$$=8\times 10+\dfrac { 1 }{ 2 } \times 8\times 3+\dfrac { 1 }{ 2 } \times 8\times 11$$
$$=80+12+44$$
$$=136{ cm }^{ 2 }$$
Question 3
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The areas of two concentric circles forming a ring are 154 sq. cm and 616 sq. cm. The breadth of the ring is

A
$$21 \ cm$$

B
$$56 \ cm$$

C
$$14 \ cm$$

D
$$7 \ cm$$

Solution

Breadth of the ring is equal to the difference between the radius of the outer circle and the radius of the inner circle
Let the radius of the outer circle be $$r_2$$ and that of the inner circle be $$r_1$$

We know that, the area of a circle is $$\pi r^2$$

Given that the area of outer circle is $$616 \ \displaystyle cm^{2}$$
$$\displaystyle \therefore \ \pi r_{2}^{2}=616 \ cm^{2}$$

$$\displaystyle \Rightarrow r_{2}^{2}=\frac{616\times 7}{22}\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$$

$$=196 \ cm^2$$

$$\displaystyle \therefore \ r_{2}=14 \ cm $$

Also, the area of the inner circle is $$\displaystyle 154 \ cm^{2}$$

$$\displaystyle \therefore \ \pi r_{1}^{2}= 154 \ cm^2$$

$$\displaystyle \Rightarrow r_{1}^{2}=\frac{154\times 7}{22}\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$$

$$=49 \ cm^2$$

$$\displaystyle \therefore r_{1}=7 \ cm.$$

$$\displaystyle \therefore \ $$ The required breadth $$\displaystyle = r_{2}-r_{1}$$

$$=14-7$$

$$=7 \ cm$$

Hence, the breadth of the ring is $$7 \ cm$$.
Question 4
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The perimeter of circle is $$\displaystyle \pi $$ cm, then the area

A
$$\displaystyle \frac{\pi^{2}}{4}$$

B
$$\displaystyle \frac{\pi^{2}}{2}$$

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

D
$$\displaystyle \frac{\pi}{4}$$

Solution

Given, Circumference

of a circle of radius r $$ = 2\pi r = \pi $$

$$ r = \dfrac {1}{2} $$

Area of a circle of radius r $$ =

\pi {r}^{2} $$

Area of circle of radius $$ \dfrac {1}{2} = \pi { ( \dfrac {1}{2} )}^{ 2 } = \dfrac { \pi }{ 4 } $$
Question 5
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The cost of fencing a circular field at the rate of Rs. 12 per meter is Rs. 1320. The field is to be ploughed at Rs. 2 per $$\mathrm {m^{2}}$$, then cost of ploughing is: $$\displaystyle \left ( \text{use }\pi =\frac{22}{7} \right )$$

A
$$Rs. 3850$$

B
$$Rs.1925$$

C
$$Rs. 660$$

D
$$Rs. 2585$$

Solution

Given that at a rate of $$Rs.\ 12$$ per meter, the cost of fencing the circular field is $$Rs. 1320$$
Hence, circumference of the field $$ = \dfrac {1320}{12} = 110 \ m$$
We know that, circumference of a circle is $$2\pi r$$
$$\therefore \ 2\pi r=110$$

$$\Rightarrow 2 \times \dfrac {22}{7} \times r = 110 $$

$$ \Rightarrow r = \dfrac {35}{2} \ m $$

We also know that the area of a circle is $$\pi r^2$$
So, area of the field $$ = \dfrac {22}{7} \times \dfrac {35}{2} \times \dfrac {35}{2} = \dfrac {1925}{2} \ {m}^{2} $$

Given that the cost of ploughing one $$m^2$$ area is $$Rs. \ 2$$.
Hence, cost of ploughing the field at $$ Rs. \ 2 $$ per $$ {m}^{2} = \dfrac {1925}{2} \times 2 = Rs. \ 1925 $$.

Hence, option B is correct.
Question 6
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Four circular cardboard pieces, each of radius 7 cm are placed in such a way that each piece touches the two other pieces. The area of the space enclosed by the four pieces is:

A
$$\displaystyle 21{ cm }^{ 2 }$$

B
$$\displaystyle 42{ cm }^{ 2 }$$

C
$$\displaystyle 84{ cm }^{ 2 }$$

D
$$\displaystyle 168{ cm }^{ 2 }$$

Solution

Area of enclosed space = Area of square - 4 x Area of quadrant
$$ = \quad side \times \quad side - 4 \times \dfrac {πr^2}{4}$$
$$\displaystyle ={ \left( 14 \right) }^{ 2 }-4$$$$\displaystyle \left( \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 } \right) $$
$$\displaystyle =196-154$$
$$\displaystyle =42{ cm }^{ 2 }$$
Question 7
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One side of a parallelogram is $$8 \ cm$$. If the corresponding altitude is $$6\ cm$$, then its area is given by

A
$$24 \ cm^2$$

B
$$36 \ cm^2$$

C
$$40 \ cm^2$$

D
$$48 \ cm^2$$

Solution

As we know,
Area of Parallelogram $$=Base \times Height$$
Here, $$Base=8\ cm, Height=6\ cm$$
So, area of the Parallelogram $$=8\times 6\ cm^2$$$$=48\ cm^2$$.
Question 8
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The perimeter of sheet of paper in the shape of a quadrant of a circle is 25 cm , then area of the paper is

A
$$\displaystyle \frac{77}{2}cm^{2}$$

B
$$\displaystyle \frac{77}{2}\left ( \frac{25}{18} \right )^{2}cm^{2}$$

C
$$\displaystyle \frac{231}{2}cm^{2}$$

D
None

Solution

Let the radius of the quadrant be $$r$$

Given,
Perimeter of the square paper $$ = \dfrac {\pi r}{2} + 2r =25 cm $$
$$\Rightarrow \dfrac {22}{7} \times \dfrac {r}{2} + 2r = 25 $$
$$ \Rightarrow \dfrac {22r + 28r}{14} = 25 $$
$$ \Rightarrow \dfrac {50r}{14} = 25 $$
$$ \Rightarrow r = 7 cm $$

Area of the square paper quadrant of radius $$r$$
$$ = \dfrac {\pi {r}^{2}}{4} \\ = \dfrac {\dfrac {22}{7} \times 7 \times 7}{4} = \dfrac {77}{2} {cm}^{2} $$
Question 9
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A race track is in the form of a ring whose inner circumference is $$440 \ \mathrm{m} $$ & outer circumference is $$506 \ \mathrm{m}$$ The width of the track is:

A
$$ 7 \ \mathrm{m}$$

B
$$ 21 \ \mathrm{m}$$

C
$$ 10.5 \ \mathrm{m}$$

D
$$ 14 \ \mathrm{m}$$

Solution

Let the radius of the outer circle be R and inner circle be r.

Outer circumference $$ = 506 \ \mathrm{m} $$

$$\implies 2 \times \frac {22}{7}

\times R = 506 \ \mathrm{m} $$
$$ R = 80.5\ \mathrm{ m} $$

Inner circumference $$ = 440 m $$

$$\implies 2 \times \frac {22}{7}

\times r = 440 \ \mathrm{m} $$
$$ r = 70 \ \mathrm{ m} $$

Hence, width of the track $$ = (80.5 - 70)\ \mathrm{m} = 10.5 \ \mathrm{ m} $$
Question 10
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If the radius of the circle is increased by $$100%$$, then the area is increased by

A
$$100$$%

B
$$200$$%

C
$$300$$%

D
$$400$$%