Perimeter and Area Test

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Perimeter and Area Test - 29

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Weekly Quiz Competition

Question 1
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Calculate the area of a circular ring whose outer and inner radii are $$12$$ and $$10\ cm.$$

A
$$118.16$$ $$cm^2$$

B
$$128.16$$ $$cm^2$$

C
$$138.16$$ $$cm^2$$

D
$$148.16$$ $$cm^2$$

Solution

Area of a circular ring $$=\pi(R^2-r^2)$$
$$=3.14(12^2-10^2)$$
$$= 3.14 \times 44$$
$$= 138.16$$ $$cm^2$$
Question 2
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Find the width of the ring.

A
$$2\ cm$$

B
$$3\ cm$$

C
$$4\ cm$$

D
$$5\ cm$$

Solution

Width of the ring, $$d =$$ outer radius $$-$$ inner radius
$$d = R - r$$
$$=10 - 5 = 5\ cm$$
Question 3
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The diameter of a circle is $$1$$. Calculate the area of the circle.

A
$$\dfrac {\pi}{8}$$

B
$$\dfrac {\pi}{4}$$

C
$$\dfrac {\pi}{2}$$

D
$$\pi$$

E
$$2\pi$$

Solution

formula to calculate area of a circle is $$\pi { r }^{ 2 }$$.
Given, diameter $$d=1$$

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

$$\Rightarrow$$ Area of circle $$=$$ $$\pi { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }=\dfrac { \pi }{ 4 } $$ sq. units.
Question 4
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Find the area of the shaded portion

A
$$4 cm^{2}$$

B
$$5 cm^{2}$$

C
$$6 cm^{2}$$

D
$$7 cm^{2}$$

Solution

We know that the area of a triangle is given by the formula $$A=\dfrac{1}{2} bh$$, where b is the length of the base and h is the triangles height.
$$ar (ADBE)=ar(ABC)-ar(DEC)$$
$$=\dfrac{1}{2}\times 2\times 6-\dfrac{1}{2}\times 1\times 2$$
$$=6-1=\boxed{5cm^2}$$
Question 5
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The radius of a circle whose circumference is $$\pi$$ is

A
$$\dfrac{1}{2}$$

B
$$1$$

C
$$2$$

D
$$\pi$$

E
$$2 \pi $$

Solution

Given that the circumference of circle is $$\pi $$.
We know that the circumference of circle is $$2\pi r$$, where $$r$$ is radius of circle.
$$\therefore$$ Circumference $$=\pi= 2 \pi r$$
$$\therefore r=\dfrac {1}{2}$$
Question 6
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The area of the parallelogram $$ABCD$$ is

A
$$20$$

B
$$10$$

C
$$5\sqrt{3}$$

D
$$10\sqrt{3}$$

E
$$18$$

Solution

From the given diagram, we conclude that the height of the parallelogram $$ABCD$$ is $$4$$ and its base is given by $$5$$.
We know that, Area of parallelogram $$=\text{base} \times \text{height}$$
$$\therefore$$ Area of parallelogram $$ABCD$$ is given by $$\text{base}\times \text{height}=4\times 5=20$$
Hence, the answer is $$20$$.
Question 7
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What is the area of a circle with a diameter of $$16$$?

A
$$8$$ $$\pi$$

B
$$16$$ $$\pi$$

C
$$64$$ $$\pi$$

D
$$128$$ $$\pi$$

E
$$256$$ $$\pi$$

Solution

Given, diameter $$=d=16$$
We need to find area of circle.
Area of circle $$=\pi r^2$$
Therefore, radius $$=r= \dfrac {16}{2} = 8 $$
Area of the circle $$ = \pi r^2 = \pi (8)^2 = 64\pi $$
Question 8
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In the figure given above, $$\overline {AC}$$ is a diameter of the large circle and B lies on $$\overline {AC}$$ so that $$\overline {AB}$$ is a diameter of the small circle. If $$AB = 1$$ and $$BC = 2$$, Calculate the area of the shaded region.

A
$$\dfrac {\pi}{4}$$

B
$$\pi$$

C
$$2\pi$$

D
$$\dfrac {9\pi}{4}$$

E
$$9\pi$$

Solution

To find the area of the shaded region, we have to find the area of the bigger circle and the area of the smaller circle.
For large circle, $$AC$$ is diameter measuring $$3$$.
$$AC=AB+BC=1+2=3$$
Area of larger circle is $$=$$ $$\pi { R }^{ 2 }$$
$$\pi { \left( \dfrac { 3 }{ 2 } \right) }^{ 2 }=\pi \dfrac { 9 }{ 4 } $$
Area of smaller circle is $$=\pi { r }^{ 2 }$$
$$\pi { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }=\pi \dfrac { 1 }{ 4 } $$
Area of shaded region is $$=$$ Area of larger circle $$-$$ Area of smaller circle
$$=\dfrac { 9\pi }{ 4 } -\dfrac { \pi }{ 4 } $$

$$=\dfrac { 9\pi -\pi }{ 4 } =\dfrac { 8\pi }{ 4 } =2\pi $$
Question 9
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A circle of radius $$x$$ has an area twice that of a square of side $$ a.$$ The equation used to find the radius of the circle is

A
$$\pi a^2 = 2x^2$$

B
$$\pi x^2 = 2a^2$$

C
$$\pi x^2 = 4a^2$$

D
$$\pi a^2 = 4x^2$$

E
$$4\pi x^2 = a^2$$

Solution

The area of square whose side length is $$a$$ is $${a}^{2}$$
So the area of circle is $$2{a}^{2}$$
Since, the radius of circle be $$x$$ , then the area is $$\pi {x}^{2} = 2{a}^{2}$$
Question 10
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The figure shows a rectangle $$ABCD$$.
Calculate the unshaded area.

A
$$4x+3y$$

B
$$8x+5y$$

C
$$12xy$$

D
$$22xy$$

Solution

For the bigger rectangle,
Length of the rectangle $$AD=BC =6x$$ cm
and breadth of the rectangle $$AB=CD=4y$$ cm
So, area of rectangle $$=6x\times 4y\ sq.cm$$
$$=24xy\ sq.cm$$

For the shaded rectangle
Length of shaded rectangle $$=2x$$ cm
and breadth of shaded rectangle $$=y$$ cm
So, area of shaded rectangle $$=2x\times y \ sq.cm$$
$$=2xy\ sq.cm$$

Hence, the area of unshaded part $$=(24 xy-2 xy)\ sq.cm=22 xy\ sq.cm$$