  # Mensuration Practice Set 2 5 Steps - 3 Clicks

# Mensuration Practice Set 2

### Introduction

Mensuration Practice set 2 is a topic in Geometry which is a branch of mathematics. Mensuration deals with length, area, and volume of different kinds of shape- both 2D and 3D. The article Mensuration Practice Quiz provides information about Mensuration, an important topic of Mathematics Consists of different types of Mensuration questions with solutions useful for candidates preparing for different competitive examinations like RRB, RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC CGL, SSC CHSL, IBPS, SBI PO, SBI Clerks, CAT and etc.

### Quiz

Q1. The radius of a circle is 4 m. What is the radius of another circle whose area is 16 times of that first?

A. 16 m
B. 64 m
C. 256 m
D. 400 m

Explanation:
Ratio of areas = (ratio of radii)²

$$\frac {16}{1}$$ = (ratio of radii)²

ratio of radii = $$\frac {4}{1}$$

The required radius = 16 m

Q2. What is the radius of the circle whose area is equal to the sum of the areas of two circles whose radii are 20 cm and 21 cm?

A. 27m
B. 28m
C. 29m
D. 25m

Explanation:
$$\pi R² = \pi r1² + \pi r2²$$

$$\pi R² = \pi (r1² + r2²)$$

R² = (400 + 441)

R = 29

Q3. Find the area of a white sheet required to prepare a cone with a height of 21cm amd the radius of 20cm.

A. 3080 cm²
B. 2300 cm²
C. 3460 cm²
D. 3600 cm²

Explanation:
r = 20 ; h = 21

l = $$\sqrt {(400 + 441)}$$= 29 cm

Total surface area = $$\pi rl + \pi r²$$

= $$\pi r(l + r)$$

= $$\frac {22}{7} \times 20(49)$$

= 3080 cm²

Q4. A hollow cylindrical tube open at both ends is made of plastic 4 cm thick. If the external diameter by 54 cm and the length of the tube be 490 cm, find the volume of plastic.

A. 320000 cm³
B. 340000 cm³
C. 306300 cm³
D. 308000 cm³

Explanation:

Volume of Plastic = $$\pi h(R² – r²) = \frac {22}{7} \times 490(27² – 23²) = 308000 cm³$$

Q5. A solid metallic cylinder of base radius 5 cm and height 7 cm is melted to form cones, each of height 1 cm and base radius 1 mm. Find the number of cones?

A. 53500
B. 49500
C. 51500
D. 52500

Explanation:
Number of cones = $$\frac {Volume of Cylinder}{Volume of one cone}$$

= $$\frac {\pi \times 5 \times 5 \times 7} {( \frac {1}{3} \pi \times \frac {1}{10} \times \frac {1}{10} \times 1)}$$

= 52500

Q1. The length of a rectangle wall is 3/2 times of its height. If the area of the wall is 600m². What is the sum of the length and height of the wall?

A. 40
B. 60
C. 50
D. 70

Explanation:
length = $$2x$$

height = $$3x$$

Area of the wall = $$3x \times 2x = 6x²$$ = 600

$$x$$ = 10;

Sum of the length and height of the wall = 50

Q2. If the area of a circle is 616cm² whose diameter is equal to the radius of a semicircle. Find the perimeter of the semicircle?

A. 184 cm
B. 154 cm
C. 144 cm
D. 164 cm

Explanation:
Area of a circle = 616cm²

$$\pi r² = 616;$$

r = 14

D = Radius of Semi Circle = 28

Perimeter of the semicircle = $$\pi r + 2r$$= 144 cm

Q3. One side of the rectangular ground is 8m and its diagonal is 17m. Find the area of the ground?

A. 150
B. 140
C. 130
D. 120

Explanation:
d = $$\sqrt {(l² + b²)}$$

17 = $$\sqrt {(l² + 8²)}$$

l² = 17² – 8² $$\Rightarrow$$l = 289 – 64 = 225

l = 15

Area = $$15 \times 8$$= 120 m²

Q4. The side of a square-shaped garden is 8√2. Find the maximum possible distance between any two corners

A. 18 meter
B. 15 meter
C. 16 metre
D. 14 metre

Explanation:
d = a $$\sqrt {2}$$

a = 8$$\sqrt {2}$$

d = 16m

Q5. A rectangular ground 16m long and 10m breadth. It has a gravel path 2.5m wide all around it on the outside. What is the area of the path?

A. 159 m²
B. 155 m²
C. 187 m²
D. 183 m²

Explanation:
Area of ground = $$16 \times 10$$ = 160

Total area (ground + path) = $$(16 + 5) \times (10 + 5)$$ = 315

Area of path = 315 – 160 = 155 m²

Q1. If the side of the square is increased by 30%, then how much % does its area get increased?

A. 59%
B. 69%
C. 79%
D. 49%

Explanation:
Area of the plot = $$1.3 \times 1.3 = 1.69 = 69 %$$

Q2. A ladder is resting with one end in contact with the top of the wall of height 15m and the other end of the ground is at a distance 8m from the wall. The length of the ladder is?

A. 17m
B. 18m
C. 16m
D. 15m

Explanation:
Hypotenuse = $$\sqrt {(base)² + (altitude)²}$$

$$\sqrt {(8)² + (15)²} = \sqrt {289}$$ = 17 m

Q3. The perimeter of a square is equal to twice the perimeter of a rectangle of length 10 cm and breadth 4 cm. What is the circumference of a semi-circle whose diameter is equal to the side of the square?

A. 38 cm
B. 23 cm
C. 46 cm
D. 36 cm

Explanation:
Perimeter of square = 2(l + b)

= $$2 \times 2(10 + 4) = 2 \times 28$$ = 56 cm

Side of square = $$\frac {56}{4}$$ = 14 cm

Radius of semi circle = $$\frac {14}{2}$$ = 7cm

Circumference of the semi-circle = $$\frac {22}{7} \times 7 + 14$$ = 36 cm

Q4. The length of a rectangle is $$\frac {3}{{5}^{th}}$$ of the side of a square. The radius of a circle is equal to the side of the square. The circumference of the circle is 132 cm. What is the area of the rectangle, if the breadth of the rectangle is 15 cm?

A. 112 cm²
B. 149 cm²
C. 189 cm²
D. Cannot be determined

Explanation:
Circumference of the circle = 132

$$2 \pi R$$ = 132;

R = 21 cm

Side of square = 21 cm

Length of the rectangle = $$\frac {3}{5} \times 21 = \frac {63}{5}$$

Area of the rectangle = $$\frac {63}{5} \times 15$$ = 189 cm²

Q5. Smallest side of a right-angled triangle is 13 cm less than the side of a square of perimeter 72 cm. The second-largest side of the right-angled triangle is 2 cm less than the length of the rectangle of area 112 cm² and breadth 8 cm. What is the largest side of the right-angled triangle?

A. 20 cm
B. 12 cm
C. 10 cm
D. 13 cm

Explanation:
Side of square = $$\frac {72}{4}$$ = 18 cm

Smallest side of the right angled triangle = 18 – 13 = 5 cm

Length of rectangle = $$\frac {112}{8}$$ = 14 cm

Second side of the right angled triangle = 14 – 2 = 12 cm

Hypotenuse of the right angled triangle = $$\sqrt {(25 + 144)}$$ = 13cm

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