ML Aggarwal Solution Class 9 Chapter 16 Mensuration MCQS

 MCQS

Choose the correct answer from the given four options (1 to 24):

Question 1

Area of a triangle is 30 cm2. If its base is 10 cm, then its height is

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cm

Sol :
Area of a triangle= 30 cm2

Base=10 cm

$\therefore$ Height $=\frac{\text { Area } \times 2}{\text { Base }}=\frac{30 \times 2}{10}$

=6 cm 

Ans (b)


Question 2

If the perimeter of a square is 80 cm, then its area is

(a) 800 cm2

(b) 600 cm2

(c) 400 cm2

(d) 200 cm2

Sol :
Perimeter of a square=80 cm
$\therefore$ Side $=\frac{P}{4}=\frac{80}{4}=20 \mathrm{~cm}$

$\therefore$ Area $=(\text { side })^{2}=20 \times 20=400 \mathrm{~cm}^{2}$

Ans (c)


Question 3

Area of a parallelogram is 48 cm2. If its height is 6 cm then its base is

(a) 8 cm

(b) 4 cm

(c) 16 cm

(d) None of these

Sol :
Area of parallelogram $=48 \mathrm{~cm}^{2}$

Height $=6 \mathrm{~cm}$

$\therefore$ Base $=\frac{\text { Area }}{\text { Height }}=\frac{48}{6}=8 \mathrm{~cm}$

Ans (a)


Question 4

If d is the diameter of a circle, then its area is

(a) $\pi d^{2}$
(b) $\frac{\pi d^{2}}{2}$
(c) $\frac{\pi d^{2}}{4}$
(d) $2 \pi d^{2}$

Sol :

Diameter of circle=d

$\therefore \text { area }=\pi r^{2}=\pi\left(\frac{d}{2}\right)^{2$

$=\frac{\pi d^{2}}{4}$

Ans (c)


Question 5

If the area of a trapezium is 64 cm2 and the distance between parallel sides is 8 cm, then sum of its parallel sides is

(a) 8 cm

(b) 4 cm

(c) 32 cm

(d) 16 cm

Sol :
Area of trapezium $=64 \mathrm{~cm}^{2}$
Distance between parallel $(h)=8 \mathrm{~cm}$

$\therefore$ Sum of its parallel sides $=\frac{\text { Area } \times 2}{h}$

$=\frac{64 \times 2}{8}=16 \mathrm{~cm}$

Ans (d)


Question 6

Area of a rhombus whose diagonals are 8 cm and 6 cm is

(a) 48 cm2

(b) 24 cm2

(c) 12 cm2

(d) 96 cm2

Sol :
Area of rhombus $=\frac{d_{1} \times d_{2}}{2}$
$=\frac{8 \times 6}{2}=24 \mathrm{~cm}^{2}$
Ans (b)


Question 7

If the lengths of diagonals of a rhombus is doubled, then area of rhombus will be
(a) doubled
(b) tripled
(c) four times
(d) remains same
Sol :
Let $d_{1}, d_{2}$ be the diagonals of a rhombus

Then area $=\frac{d_{1} \times d_{2}}{2}$

If diagonals are doubled, then
Area $=\frac{2 d_{1} \times 2 d_{2}}{2}=4 \frac{d_{1} d_{2}}{2}$
=four times 
Ans (c)

Question 8

If the length of a diagonal of a quadrilateral is 10 cm and lengths of the perpendiculars on it from opposite vertices are 4 cm and 6 cm, then area of quadrilateral is

(a) 100 cm2

(b) 200 cm2

(c) 50 cm2

(d) None of these

Sol :
Length of diagonal of a quadrilateral=10 cm

Length of perpendicular on it from opposite vertices are 4 cm and 6 cm

$\therefore$ Area $=\frac{1}{2} \times 10 \times(4+6) \mathrm{cm}^{2}$
$=5 \times 10=50 \mathrm{~cm}^{2}$
Ans (c)

Question 9

Area of a rhombus is 90 cm2. If the length of one diagonal is 10 cm then the length of other diagonal is

(a) 18 cm

(b) 9 cm

(c) 36 cm

(d) 4.5 cm

Sol :
Area of rhombus $=90 \mathrm{~cm}^{2}$
One diagonal $\left(d_{1}\right)=10 \mathrm{~cm}$
Then $d_{2}=\frac{\text { Area } \times 2}{d_{1}}=\frac{90 \times 2}{10}$
=18 cm 
Ans (a)


Question 10

In the given figure, OACB is a quadrant of a circle of radius 7 cm. The perimeter of the quadrant is

(a) 11 cm

(b) 18 cm

(c) 25 cm

(d) 36 cm










Sol :

OACB is a quadrant of a circle with radius 7 cm

$\therefore$ Perimeter of the quadrant $=r+r+\frac{1}{4} \times 2 \pi r$

$=2 r+\frac{1}{2} \pi r=2 \times 7+\frac{1}{2} \times \frac{22}{7} \times 7$

=14+11=25 cm

Ans (c)


Question 11

In the given figure, OABC is a square of side 7 cm. OAC is a quadrant of a circle with O as centre. The area of the shaded region is












(a) 10.5 cm2

(b) 38.5 cm2

(c) 49 cm2

(d) 11.5 cm2

Sol :

OABC is a square with side 7 cm .OAC is a quadrant. 

Area of shaded portion = area of square area of quadrant

$=(7)^{2}-\frac{1}{4} \pi r^{2}$

$=(7)^{2}-\frac{1}{4} \times \frac{22}{7} \times 7 \times 7$

$=49-\frac{77}{2}=49-38.5=10.5 \mathrm{~cm}^{2}$

Ans (a)


Question 12

The given figure shows a rectangle and a semicircle. The perimeter of the shaded region is

(a) 70 cm

(b) 56 cm

(c) 78 cm

(d) 46 cm











Sol :
In the figure, ABCD is a rectangle and a semicircle is drawn on its side CD as diameter 14 cm
Length of rectangle =14 cm and breadth =10 cm

∴Perimeter of the shaded region
=DA+AB+BC+𝛑r
(Radius of semicircle $\left.=\frac{14}{2}=7 \mathrm{~cm}\right)$

$=10+14+10+\frac{22}{7} \times \frac{14}{2}=56 \mathrm{~cm}$
Ans (b)


Question 13

The area of the shaded region shown in Q. 12 (above is

(a) 140 cm2

(b) 77 cm2

(c) 294 cm2

(d) 217 cm2

Sol :
Area of shaded portion of the figure of question (12)
=Area of rectangle + Area of semicircle 
$=l \times b+\frac{1}{2} \pi r^{2}=14 \times 10+\frac{1}{2} \times \frac{22}{7} \times 7 \times 7$

$=14 \times 10+\frac{1}{2} \times \frac{22}{7} \times 7 \times 7 \mathrm{~cm}^{2}$

(Radius of semicircle =7 cm)

=140+77=217 cm2

Ans (d)


Question 14

In the given figure, the boundary of the shaded region consists of semicircular arcs. The area of the shaded region is equal to

(a) 616 cm2

(b) 385 cm2

(c) 231 cm2

(d) 308 cm2









Sol :
Area of shaded portion
=Area of semicircle of 14 cm  radius - area of semicircle of 7 cm radius + area of semicircle of 7 cm radius

=Area of semicircle of radius 14 cm

$=\frac{1}{2} \pi r^{2}=\frac{1}{2} \times \frac{22}{7} \times 14 \times 14 \mathrm{~cm}^{2}$

$=308 \mathrm{~cm}^{2}$

Ans (d)


Question 15

The perimeter of the shaded region shown in Q. 14 (above) is

(a) 44 cm

(b) 88 cm

(c) 66 cm

(d) 132 cm

Sol :
Perimeter of shaded portion of the figure given in Q 14 
Perimeter of bigger semicircle + perimeter of two small semicircles

$=\pi r+2 \times \pi r$

$=\frac{22}{7} \times 14+2 \times \frac{22}{7} \times 7 \mathrm{~cm}$

$=44+44=88 \mathrm{~cm}$

Ans (b)


Question 16

In the given figure, ABC is a right angled triangle at B. A semicircle is drawn on AB as diameter. If AB = 12 cm and BC = 5 cm, then the area of the shaded region is

(a) (60 + 18π) cm2

(b) (30 + 36π) cm2

(c) (30+18π) cm2

(d) (30 + 9π) cm2












Sol :

In the given figure, $\mathrm{ABC}$ is a right angled triangle right angle at $\mathrm{B}, \mathrm{AB}$ is diameter $=12 \mathrm{~cm}, \mathrm{BC}=5 \mathrm{~cm}$

Area of shaded portion = area of semicircle+ area of right $\Delta \mathrm{ABC}$

$=\frac{1}{2} \pi r^{2}+\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}$

$=\frac{1}{2} \pi(6)^{2}+\frac{1}{2} \times 12 \times 5 \mathrm{~cm}^{2}$

$=(18 \pi+30) \mathrm{cm}^{2}$

Ans (c)


Question 17

The perimeter of the shaded region shown in Q. 16 (above) is

(a) (30 + 6π) cm

(b) (30 + 12π) cm

(c) (18 + 12π) cm

(d) (18 + 6π) cm

Sol :

Perimeter of the shaded region in $Q .16$

$=\frac{1}{2} \times 2 \pi r+5 \mathrm{~cm}+\mathrm{AC}$

But $A C=\sqrt{A B^{2}+B C^{2}}$

(Pythagoras Theorem)

$=\sqrt{12^{2}+5^{2}}=\sqrt{144+25} \mathrm{~cm}$

$=\sqrt{169}=13 \mathrm{~cm}$

$\therefore$ Perimeter $=\pi \times 6+13 \mathrm{~cm}$

$=(18+6 \pi) \mathrm{cm}$

And (d)


Question 18

If the volume of a cube is 729 m3, then its surface area is

(a) 486 cm2

(b) 324 cm2

(c) 162 cm2

(d) None of these

Sol :
Volume of cube $=729 \mathrm{~m}^{3}$
$\therefore$ Side $=\sqrt[3]{\text { Volume }}=\sqrt[3]{729}=\sqrt[3]{9^{3}}$
$=9^{3 \times \frac{1}{3}}=9 \mathrm{~m}$

$\therefore$ Surface area $=6(9)^{2}=6 \times 81$
$=486 \mathrm{~m}^{2}$
Ans (a) 

Question 19

If the total surface area of a cube is 96 cm2, then the volume of the cube is
(a) 8 cm3
(b) 512 cm3
(c) 64 cm3
(d) 27 cm3

Sol :
Surface area of cube =96 cm2
$\therefore$ Side $=\sqrt{\frac{\text { Area }}{6}}$

$\sqrt{\frac{96}{6}}=\sqrt{16}=4 \mathrm{~cm}$

$\therefore$ Volume $=(\text { side })^{3}=(4)^{3} \mathrm{~cm}^{3}$

$=4 \times 4 \times 4=64 \mathrm{~cm}^{3}$

Ans (c)


Question 20

The length of the longest pole that can be put in a room of dimensions (10 m x 10 m x 5 m) is

(a) 15 m

(b) 16 m

(c) 10 m

(d) 12 m

Sol :

Longest pole in a room $=\sqrt{l^{2}+b^{2}+h^{2}}$

$=\sqrt{10^{2}+10^{2}+5^{2}}=\sqrt{100+100+25}$

$=\sqrt{225}=15 \mathrm{~m}$

Ans (a)


Question 21

The lateral surface area of a cube is 256 m2. The volume of the cube is

(a) 512 m3

(b) 64 m3

(c) 216 m3

(d) 256 m3

Sol :
Lateral surface area of a cube $=256 \mathrm{~m}^{2}$

$\therefore$ Side $=\sqrt{\frac{\text { Area }}{4}}=\sqrt{\frac{256}{4}}=\sqrt{64}=8 \mathrm{~m}$

$\therefore$ Volume of cube $=(\text { side })^{3}=(8)^{3} \mathrm{~m}^{3}$

$=8 \times 8 \times 8=512 \mathrm{~m}^{3}$


Question 22

If the perimeter of one face of a cube is 40 cm, then the sum of lengths of its edge is

(a) 80 cm

(b) 120 cm

(c) 160 cm

(d) 240 cm

Sol :

Perimeter of one face of cube =40 cm

∴Side $=\frac{40}{4}=10 \mathrm{~cm}$

No. of edges =12

$\therefore$ Sum of its edges $=12 \times 10=120 \mathrm{~cm}$

Ans (b)


Question 23

A cuboid container has the capacity to hold 50 small boxes. If all the dimensions of the container are doubled, then it can hold (small boxes of same size)

(a) 100 boxes

(b) 200 boxes

(c) 400 boxes

(d) 800 boxes

Sol :

In a cuboid containes, number of boxes =50

If the dimensions of the container are doubled then its volume $2 \times 2 \times 2=8$ times

∴Number of boxes in it will be =50×8

=400 boxes

Ans (c)


Question 24

The number of planks of dimensions (4 m x 50 cm x 20 cm) that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is

(a) 1900

(b) 1920

(c) 1800

(d) 1840

Sol :
Size of one planks=4 \mathrm{~m} \times 50 \mathrm{~cm} \times 20 \mathrm{~cm}

$=4 \mathrm{~m} \times \frac{1}{2} \times \frac{1}{5} \mathrm{~m}^{3}$

$=\frac{2}{5} \mathrm{~m}^{3}$

Volume of the pit $=16 \times 12 \times 4 \mathrm{~m}$ $=768 \mathrm{~m}^{3}$

$=768 \mathrm{~m}^{3}$

$\therefore$ Number of planks kept in it $=768 \div \frac{2}{5}$

$=768 \times \frac{5}{2}=1920$

Ans (b)

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