Mensuration Test

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Mensuration Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

The area of a trapezium is $$28 cm^{2}$$ and one of its parallel sides $$6 $$cm. If its altitude is $$4 $$cm then its other parallel side is

A
$$8$$ cm

B
$$4$$ cm

C
$$6$$ cm

D
none of these

Solution

Area of trapezium = $$\displaystyle \frac {1}{2}\times (sum\ of\ parallel \ sides) \times altitude$$

Let other parallel side be $$x$$ cm.

Then,

$$\displaystyle \dfrac {1}{2}\, (x\, +\, 6)\, \times\, 4\, =\, 28$$

$$\Rightarrow 2(x+6)=28$$

$$\Rightarrow x=14-6$$

$$\Rightarrow x\, =\, 8 cm$$
Question 2
1 / -0

What shape are the faces of a cube?

A
Circle

B
Triangle

C
Square

D
Pentagon

Solution

We know, a cuboid whose length, breadth and height are all equal is called a cube.
Since, in a cube, all the edges of are equal, we can surely say that the faces of a cube are identical with equal edges and hence each face will be a square

Therefore, option $$C$$ is correct.
Question 3
1 / -0

The volume of air in a room $$10 m$$ long, $$6.5 m$$ wide, and $$5 m$$ height is:

A
$$425$$ cu m

B
$$225$$ cu m

C
$$325$$ cu m

D
None of the above

Solution

$$l=10m,b=6.5m,h=5m$$
Volume of air$$=l\times b\times h=10\times 6.5\times 5=325m^3$$
Question 4
1 / -0

A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape It is $$30\ cm$$ long, $$25\ cm$$ wide, and $$25\ cm$$ high The area of glass is:

A
$$5000\ cm^{2}$$

B
$$ 4800\ cm^{2}$$

C
$$4250\ cm^{2}$$

D
$$4500\ cm^{2}$$

Solution

Since the greenhouse is cuboidal in shape, we need to calculate its total surface area by applying the formula.
Total surface area of a cuboid $$ = 2(l \times b + b\times h + l \times h) $$
Given:-
$$l=30\ cm$$
$$b=25\ cm$$
$$h=25\ cm$$

Hence, the total surface area of the herbarium will be,
$$S =2(30 \times 25 + 25 \times 25 + 30 \times 25) $$
$$= 2(750+625+750) $$
$$=2(2125)$$
$$= 4250 \ cm^2$$
Question 5
1 / -0

Find the volume of a cuboid of length $$20$$ cm, breadth $$15$$ cm and height $$10$$ cm.

A
$$8794$$ $$cm^3$$

B
$$7680$$ $$cm^3$$

C
$$3000$$ $$cm^3$$

D
$$5760$$ $$cm^3$$

Solution

Given, length of the cuboid$$ = 20$$ cm
Breadth of the cuboid $$= 15$$ cm
Height of the cuboid$$ = 10$$ cm
$$\therefore$$ Volume of the cuboid $$=$$ length $$\times$$ breadth $$\times$$ height
$$=3000 cm^3$$
Question 6
1 / -0

A cuboid is $$3$$ cm high, $$4$$cm wide and $$5$$cm long. What is its volume?

A
$$60 cm^3$$

B
$$50 cm^3$$

C
$$150 cm^3$$

D
$$100 cm^3$$

Solution

$$Volume\ of\ cuboid=length*breadth*height$$
$$length=5\ cm$$
$$breadth=4\ cm$$
$$height=3\ cm$$
$$Volume=5*4*3=60\ cm^3$$
Question 7
1 / -0

A plastic box $$1.5$$ m long, $$1.25$$ m wide, and $$65$$ cm deep is to be made. It is to be opened at the top. Ignoring the thickness of the plastic, the cost of the sheet for covering it, if a sheet measuring $$1$$$$\displaystyle m^{2}$$ costs Rs. $$20$$ is:

A
Rs. $$100$$

B
Rs. $$109$$

C
Rs. $$115$$

D
Rs. $$110$$

Solution

Length of the box, $$l=1.5m$$. Breadth of box,$$b=1.25m$$

Depth of box, $$h=65cm=0.65m$$

Area of sheet required$$=$$ Area of cuboid$$-$$Area of open top

$$=2(lb+bh+hl)-lb$$

$$=lb+2h(b+l)$$

$$=1.5\times 1.25+2\times 0.65(1.5+1.25)$$

$$=5.45m^2$$

Cost of sheet of $$5.45m^2=$$Rs. $$(5.45\times 20)=$$Rs. $$109$$
Question 8
1 / -0

The radius of the cylinder whose lateral surface area is $$704\, cm^2$$ and height $$8$$ cm is

A
$$6$$ cm

B
$$4$$ cm

C
$$8$$ cm

D
$$14$$ cm

Solution

Lateral surface area of a cylinder $$=2 \pi r h$$
Given, $$\, 2\, \pi\, rh\, =\, 704\, cm^2$$.
$$\therefore\, r\, =\, \displaystyle \frac {704}{2\pi h}\, $$
$$=\, \displaystyle \frac {704}{2\, \times\, \displaystyle \frac {22}{7}\, \times\, \times\, 8}\, $$
$$=\, 14$$ cm
Question 9
1 / -0

Given an isosceles trapezium $$ABCD$$ in order with $$AB = 6, CD = 12$$ and area $$36$$ sq units. Find the length of the side $$BC$$.

A
$$6$$

B
$$5$$

C
$$4.5$$

D
$$5.5$$

Solution

The trapezium can be divided in $$3$$ parts, a rectangle of $$6 $$ units, $$2$$ equal triangles.
Let the height be $$x$$,
Area of trapezium $$=$$ Area of rectangle $$+2$$(Area of triangle)
$$36 = 6\times x + 2(\dfrac 12\times 3\times x)$$ , base$$=3$$ units, the sides are equally divided.
$$36 = 9x,$$
$$x = 4 units$$,
$$BC = 5 units $$ (Pythagoras theorem)
Question 10
1 / -0

The area of a trapezium is $$34\;cm^2$$ and the length of one of the parallel sides is $$10\;cm$$ and its height is $$4\;cm$$. Find the length of the other parallel side.

A
$$7\;cm$$

B
$$4\;cm$$

C
$$7\;m$$

D
$$0.7\;cm$$

Solution

Let the required side be $$x\;cm$$.

We know, the area of the trapezium

$$=\begin{bmatrix}\displaystyle\frac{1}{2}\times(a+b)\times height\end{bmatrix}$$ {where $$a$$ and $$b$$ are length of parallel sides}

$$=\begin{bmatrix}\displaystyle\frac{1}{2}\times(10+x)\times4\end{bmatrix}cm^2=2(10+x)\;cm^2$$

Given, the area of the trapezium $$=34\;cm^2$$ (given)

$$\implies \;2(10+x)=34$$

$$\Rightarrow\;10+x=17$$

$$\Rightarrow\;x=17-10=7$$

Hence, the other parallel side $$=7\;cm$$.

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Mensuration Test - 20

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Mensuration Test - 20

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SHARING IS CARING

Question 1

$$8$$ cm

$$4$$ cm

$$6$$ cm

none of these

Solution

Question 2

Circle

Triangle

Square

Pentagon

Solution

Question 3

$$425$$ cu m

$$225$$ cu m

$$325$$ cu m

None of the above

Solution

Question 4

$$5000\ cm^{2}$$

$$ 4800\ cm^{2}$$

$$4250\ cm^{2}$$

$$4500\ cm^{2}$$

Solution

Question 5

$$8794$$ $$cm^3$$

$$7680$$ $$cm^3$$

$$3000$$ $$cm^3$$

$$5760$$ $$cm^3$$

Solution

Question 6

$$60 cm^3$$

$$50 cm^3$$

$$150 cm^3$$

$$100 cm^3$$

Solution

Question 7

Rs. $$100$$

Rs. $$109$$

Rs. $$115$$

Rs. $$110$$

Solution

Question 8

$$6$$ cm

$$4$$ cm

$$8$$ cm

$$14$$ cm

Solution

Question 9

$$6$$

$$5$$

$$4.5$$

$$5.5$$

Solution

Question 10

$$7\;cm$$

$$4\;cm$$

$$7\;m$$

$$0.7\;cm$$

Solution

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