Mensuration Test

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Mensuration Test - 29

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Weekly Quiz Competition

Question 1
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The area of a trapezium is $$240$$ $$ ft^{2}$$ and its $$ parallel side_{1} = 35$$ ft, $$parallelside_{ 2} = 13$$ ft. Find the height.

A
$$10$$ ft

B
$$20$$ ft

C
$$30$$ ft

D
$$40$$ ft

Solution

Area of a trapezoid = $$\frac{1}{2}(b_{1}+ b_{2})\times h$$
$$b_{1} $$= $$ 35$$ ft
$$b_{2} $$= $$ 13$$ ft
h = ?
$$240$$ = $$\frac{1}{2}(35 + 13)\times h$$
$$240$$ = $$\frac{1}{2} \times 48\times h$$
$$240 = 24 \times h$$
Height, $$h = 10$$ ft.
Question 2
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Calculate the area of a trapezium whose parallel sides are $$ 14$$ cm and $$5$$ cm, and the height is $$20$$ cm.

A
$$180$$ $$cm^{2}$$

B
$$190$$ $$cm^{2}$$

C
$$290$$ $$cm^{2}$$

D
$$170$$ $$cm^{2}$$

Solution

Area of a trapezium = $$\cfrac{1}{2}(b_{1}+ b_{2})\times h$$
Given: $$b_{1} = $$ $$14$$ cm
$$b_{2} =$$ $$5$$ cm
$$h=20$$ cm
$$A=\cfrac{1}{2}(14 + 5)\times 20$$
$$A=\cfrac{1}{2} \times 19\times 20$$
$$A=190$$ $$cm^{2}$$
Question 3
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Find the area of a trapezium whose parallel sides are $$21$$mm and $$17$$ mm and a height of $$10$$ mm.

A
$$190$$ $$m^{2}$$

B
$$190$$ $$cm^{2}$$

C
$$190$$ $$mm^{3}$$

D
$$190$$ $$mm^{2}$$

Solution

Area of a trapezoid = $$\frac{1}{2}(b_{1}+ b_{2})\times h$$
$$b_{1} = $$ $$21$$ mm
$$b_{2} = $$ $$17$$ mm
h = $$10$$ mm
= $$\frac{1}{2}(21 + 17)\times 10$$
= $$\frac{1}{2} \times 38\times 10$$
= $$190$$ $$mm^{2}$$
Question 4
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A swimming pool is $$200$$ft long and $$14$$ ft wide and its depth being $$20$$ ft respectively. Find the volume of the pool.

A
$$55,000 ft^{3}$$

B
$$54,000 ft^{3}$$

C
$$56,000 ft^{3}$$

D
$$57,000 ft^{3}$$

Solution

Volume of the pool $$= l \times b \times h$$
$$= 200 \times 14 \times 20$$
$$= 56,000 ft^{3}$$
Question 5
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The area of a trapezium is $$210$$ $$ in^{2}$$ and its parallel sides are $$30$$ in and $$12$$ in. Find the height.

A
$$8 \ in$$

B
$$9 \ in$$

C
$$10\ in$$

D
$$11 \ in$$

Solution

Area of a trapezoid = $$\frac{1}{2}(b_{1}+ b_{2})\times h$$
$$b_{1} = $$ 30 in
$$b_{2} = $$ 12 in
h = ?
$$210$$ = $$\frac{1}{2}(30 + 12)\times h$$
$$210$$ = $$\frac{1}{2} \times 42\times h$$
$$210 = 21 \times h$$
Height, $$h = 10 \ in$$.
Question 6
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Find the area of the figure shown above.

A
$$20$$ $$cm^2$$

B
$$21$$ $$cm^2$$

C
$$24$$ $$cm^2$$

D
$$42$$ $$cm^2$$

Solution

$$ABCD$$ is a trapezium.
Given: $$AB=4$$ cm, $$CD=6$$ cm, Height $$=4$$cm
Area of trapezium $$=\dfrac{1}{2}\times $$ Sum of parallel side $$\times$$ height
$$=\dfrac{1}{2}\times (6+4) \times 4=20 \space cm^2$$
Question 7
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The curved surface area of the cylinder is $$2\pi r h$$ where as surface area of the cylinder is

A
$$\pi r(r+h)$$

B
$$\pi rl(r+h)$$

C
$$2\pi r(r+h)$$

D
$$2\pi r^2l(r+h)$$

Solution

The curved surface area of the cylinder is $$2\pi r h$$ where as surface area of the cylinder is $$2\pi r(r+h)$$.
Surface area includes curved surface area and two circular bases of the cylinder.
Question 8
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The parallel sides of a trapezium are $$15\ cm$$ and $$9\ cm.$$ The distance between them is $$5\ cm$$. A rectangle has the same area as the trapezium. If its length is $$10\ cm,$$ what is its breadth?

A
$$6\ cm$$

B
$$3\ cm$$

C
$$5\ cm$$

D
$$10\ cm$$

Solution

Area of trapezium $$=\dfrac { 1 }{ 2 } \times \left( AB+CD \right) \times AM$$

$$=\dfrac { 1 }{ 2 } \times \left( 9+15 \right) \times 5$$

$$=\dfrac { 1 }{ 2 } \times 24\times 5=60{ cm }^{ 2 }$$
Now, area of rectangle $$=$$ area of trapezium
$$\Rightarrow \quad l\times b=60{ cm }^{ 2 }\Rightarrow 10\times b=60{ cm }^{ 2 }\Rightarrow b=\dfrac { 60 }{ 10 } =6cm$$

$$\therefore $$ $$Breadth = 6$$ cm
Question 9
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Calculate the total surface area of the right circular cylinder shown in the above figure (in square units).

A
$$\displaystyle 300\pi $$

B
$$\displaystyle 400\pi $$

C
$$\displaystyle 500\pi $$

D
$$\displaystyle 600\pi $$

Solution

From the given figure, we can see that:
The diameter of the cylinder $$(d)=20$$ and
The height of the cylinder $$(h)=20$$
Hence, the radius of the cylinder will be $$r=\dfrac{20}{2}=10$$

We know that, the total surface area of a cylinder is $$2\pi r^2+2\pi r h$$
Here, $$r=10,\ h=20$$
$$\therefore \ $$ The total surface area $$=2\pi (10)^{2}+2\pi(10\times20)$$
$$=2\pi(100+200)$$
$$=\pi(2\times 300)$$
$$=600\pi\ square\ units$$

Hence, the total surface area of the given cylinder is $$600\pi \ square\ units$$.
Question 10
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In the following figure, if the dimensions of the trapezoid are as shown and the area of the trapezoid is $$144$$, what is the value of $$x$$?

A
$$2$$

B
$$3$$

C
$$4$$

D
$$6$$

E
$$8$$

Solution

Given, area of trapezoid $$=144, h=4x, a=6x, c=12x$$
We know, area of trapezoid is $$A=\dfrac { h }{ 2 } \left( a+c \right) $$
$$\Rightarrow 144=\dfrac { 4x }{ 2 } \left( 6x+12x \right) $$
$$\Rightarrow 144=\dfrac { 4x }{ 2 } \left( 18x \right) $$
$$\Rightarrow 288={ 72x }^{ 2 }$$
$$\Rightarrow { x }^{ 2 }=\dfrac { 288 }{ 72 } =4$$
$$\Rightarrow x=2$$