Mensuration Test

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Mensuration Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Find the surface area of the following cubes with length of edge
$$4\;cm$$
$$3.2\;cm$$

A
$$96\;cm^2\;;\;61.44\;cm^2$$

B
$$90\;cm^2\;;\;61.44\;cm^2$$

C
$$96\;cm^2\;;\;60.44\;cm^2$$

D
$$92\;cm^2\;;\;61.44\;cm^2$$

Solution

Total surface area$$=6(4^2)=96 cm^2$$,for edge=4$$ cm^2$$
Total surface area$$=6(3.2^2)=61.44 cm^2$$, for edge=61.44 $$cm^2$$
Question 2
1 / -0

The area of a trapezium with height $$14\;cm$$ is $$504\;cm^2$$. If the parallel sides are in the ratio $$4\,\colon\,5$$, find their lengths.

A
$$30\;cm,\;42\;cm$$.

B
$$32\;cm,\;40\;cm$$.

C
$$32\;cm,\;42\;cm$$.

D
$$30\;cm,\;40\;cm$$.

Solution

Area of trapezium$$=\dfrac{1}{2}$$(sum of parallel opposite side)(height)
Let parallel side are $$4x,5x$$
$$504=\dfrac{1}{2}(4x+5x)(14)$$
$$\dfrac{(504)(2)}{9(14)}=x$$
$$x=8$$
side of one side$$=4(8)=32cm,5(8)=40cm$$
Question 3
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The sum of the radius of the base and the height of a solid cylinder is $$37 m$$. If the total surface area of the cylinder be $$1628 sq. m$$, then its volume is:

A
$$4620m^{3}$$

B
$$4630 m^{3}$$

C
$$4520 m^{3}$$

D
$$4830 m^{3}$$

Solution

Given, $$r + h = 37 m$$
Total Surface Area of a Cylinder $$ = 2\pi r(r + h)$$
Given, $$ 2\times \cfrac {7}{22} \times r \times 37 =1628 $$
$$\Rightarrow r= \cfrac {1628 \times 7}{22 \times 37 \times 2} =7 m $$
Height (H) $$ = 37 - r = 37 -7 = 30 m $$
Volume of a Cylinder $$ = \pi { R }^{ 2 }h $$ $$ = \cfrac {22}{7} \times 7 \times 7 \times 30 = 4620 {m}^{3} $$
Question 4
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The cross section of a canal is in the shape of trapezium. The canal is $$15 m$$ wide at the top and $$9 m$$ wide at the bottom. If the area of the cross section is $$720 m^{2}$$, then the depth of the canal is

A
$$55 m$$

B
$$50 m$$

C
$$58.5 m$$

D
$$60 m$$

Solution

Given,
The canal is $$15 m$$ wide at the top and $$9 m$$ wide at the bottom.
Area of the cross section$$=720 m^{2}$$
Let the depth of the canal be $$h$$
Area of a trapezium $$ = \cfrac { 1 }{ 2 } \times$$ Sum of parallel sides $$\times$$ height
$$\Rightarrow 720= \cfrac { 1 }{ 2 } \times (9 + 15) \times h$$
$$\Rightarrow 720= \cfrac { 24\times h }{ 2 }$$
$$\Rightarrow h= \cfrac { 720\times 2 }{ 24 }$$
$$\Rightarrow h= 60$$
Hence, depth of the canal is $$ 60\ m $$.
Question 5
1 / -0

The area of a field is in the shape of a trapezium measures $$1440 m^{2}$$. The perpendicular distance between its parallel sides is $$24 m$$. If the ratio of the parallel sides is $$5 : 3$$. What is the length of the longer parallel side?

A
$$75 m$$

B
$$70 m$$

C
$$80 m$$

D
$$85 m$$

Solution

Let sides of the field are $$5x$$ and $$3x$$.
Area of field $$=\cfrac{side_1 +side_2}{2}\times h $$
Area of field$$=\cfrac { 1 }{ 2 } \times \left( 5x+3x \right) \times 24$$
$$=96x\ m^2$$
$$\Rightarrow 96x=1440$$
$$\Rightarrow x=\cfrac { 1440 }{ 96 } =15m$$
Therefore, length of longer parallel side$$=15\times 5=75m$$
Question 6
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The shape of the top surface of a table is trapezium. Find its area if its parallel sides are $$1\;m\;and\;1.2\;m$$ and the perpendicular distance between them is $$0.8\;m$$.

A
$$0.88\;m^2$$

B
$$0.8\;m^2$$

C
$$8.8\;m^2$$

D
$$0.88\;cm^2$$
Question 7
1 / -0

The area of a trapezium is $$720\;cm^2$$. The ratio of the parallel sides is $$2\,\colon\,1$$. If the distance between the parallel sides is $$20\;cm$$ then find the length of the parallel sides.

A
$$20,\,30\;cm$$

B
$$24,\,48\;cm$$

C
$$42,\,46\;cm$$

D
None of these

Solution

Let length of parallel side$$ = 2x $$ and $$x$$ ($$\because$$ they are in ratio $$2:1$$)
Area of tapezium $$= \dfrac{1}{2}\times h \left(a+b \right) = 720 cm^2$$ (where $$h = 20 cm $$ given)
$$ = \dfrac{1}{2} \times 20 \left(2x+x \right) = 720$$
$$ = 10 \left(2x+x\right) = 720$$
$$30x = 720$$
$$ x = 24$$
Hence the length of the parallel sides $$= 24cm$$ and $$48cm$$
Question 8
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Find the area covered by a road roller of width $$80\;cm$$ and diameter $$140\;cm$$ in $$40$$ revolutions.

A
$$130.8\;m^2$$

B
$$143.6\;m^2$$

C
$$141.8\;m^2$$

D
$$140.8\;m^2$$

Solution

Curved surface area of cylinder$$=2 \pi rh=2 \pi(70)(80)=35200 cm^2=3.52 m^2
$$
Area of 40 revolutions$$=3.52 \times 40=140.8 m^2$$
Question 9
1 / -0

Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope. What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length $$25\space cm$$ with a $$3.5\space cm$$ radius? You may take $$\left (\pi = \dfrac{22}{7}\right )$$

A
$$540\space cm^2$$

B
$$520\space cm^2$$

C
$$550\space cm^2$$

D
$$560\space cm^2$$

Solution

Kaleidoscope is of cylindrical shape.
$$\therefore $$ Area of chart paper required $$=$$ CSA of cylinder
$$=$$ $$2\pi rh$$
$$ =$$ $$2\times \dfrac { 22 }{ 7 } \times 3.5\times 25$$
$$ =$$ $$550$$ $${ cm }^{ 2 }$$
Question 10
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From a metal sheet in the shape of trapezium with parallel sides $$64\; \text{cm}\;$$ and $$\;100\; \text{cm}$$ and height $$56\; \text{cm},$$ a circle of radius $$7\; \text{cm}$$ are cut out. Find the area of the metal sheet that it left.

A
$$230\; \text{cm}^2$$

B
$$260\; \text{cm}^2$$

C
$$4438\; \text{cm}^2$$

D
$$250\; \text{cm}^2$$

Solution

Area of sheet left $$=$$ Area of trapezium - Area of circle
$$=$$ $$\dfrac { 1 }{ 2 } \times \text{sum of parallel sides}\times \text{height}-\pi { r }^{ 2 }$$

$$ =$$ $$\left(\dfrac { 1 }{ 2 } \times \left( 64+100 \right) \times 56-\dfrac { 22 }{ 7 } \times 7\times 7\right)\ \text{cm}^2$$
$$=$$ $$(4592 - 154)\ \text{cm}^2$$
$$=$$ $$4438$$ $$\text{ cm }^{ 2 }$$

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Mensuration Test - 21

Result

Mensuration Test - 21

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Question 1

$$96\;cm^2\;;\;61.44\;cm^2$$

$$90\;cm^2\;;\;61.44\;cm^2$$

$$96\;cm^2\;;\;60.44\;cm^2$$

$$92\;cm^2\;;\;61.44\;cm^2$$

Solution

Question 2

$$30\;cm,\;42\;cm$$.

$$32\;cm,\;40\;cm$$.

$$32\;cm,\;42\;cm$$.

$$30\;cm,\;40\;cm$$.

Solution

Question 3

$$4620m^{3}$$

$$4630 m^{3}$$

$$4520 m^{3}$$

$$4830 m^{3}$$

Solution

Question 4

$$55 m$$

$$50 m$$

$$58.5 m$$

$$60 m$$

Solution

Question 5

$$75 m$$

$$70 m$$

$$80 m$$

$$85 m$$

Solution

Question 6

$$0.88\;m^2$$

$$0.8\;m^2$$

$$8.8\;m^2$$

$$0.88\;cm^2$$

Question 7

$$20,\,30\;cm$$

$$24,\,48\;cm$$

$$42,\,46\;cm$$

None of these

Solution

Question 8

$$130.8\;m^2$$

$$143.6\;m^2$$

$$141.8\;m^2$$

$$140.8\;m^2$$

Solution

Question 9

$$540\space cm^2$$

$$520\space cm^2$$

$$550\space cm^2$$

$$560\space cm^2$$

Solution

Question 10

$$230\; \text{cm}^2$$

$$260\; \text{cm}^2$$

$$4438\; \text{cm}^2$$

$$250\; \text{cm}^2$$

Solution

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