 # Mensuration Practice Set 4 5 Steps - 3 Clicks

# Mensuration Practice Set 4

### Introduction

Mensuration 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 Set 4 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

1. What is the are of an equilateral triangle of side 16 cm?

A. 48√3 $${cm}^{2}$$
B. 128√3 $${cm}^{2}$$
C. 9.6√3 $${cm}^{2}$$
D. 64√3 $${cm}^{2}$$
E. None of these

Explanation:

Area of an equilateral triangle = √$$\frac{3}{4}$$ $${cm}^{2}$$

If S = 16, Area of triangle = √$$\frac{3}{4}$$ × 16 × 16 = 64√3 $${cm}^{2}$$

2. If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area?

A. 120 $${cm}^{2}$$
B. 130 $${cm}^{2}$$
C. 312 $${cm}^{2}$$
D. 315 $${cm}^{2}$$
E. None of thesae

Explanation:

The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.

Area of the triangle = $$\frac{1}{2}$$ × 24 × 10 = 120 $${cm}^{2}$$

3. The perimeter of a triangle is 28 cm and the inradius of the triangle is 2.5 cm. What is the area of the triangle?

A. 25 $${cm}^{2}$$
B. 42 $${cm}^{2}$$
C. 49 $${cm}^{2}$$
D. 70 $${cm}^{2}$$
E. None of these

Explanation:

Area of a triangle = r × s

Where r is the inradius and s is the semi perimeter of the triangle.
Area of triangle = 2.5 × $$\frac{28}{2}$$ = 35 $${cm}^{2}$$

4. Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.

A. 225 $${cm}^{2}$$
B. 275 $${cm}^{2}$$
C. 285 $${cm}^{2}$$
D. 315 $${cm}^{2}$$
E. None of these

Explanation:

Area of a trapezium = $$\frac{1}{2}$$ (sum of parallel sides) × (perpendicular distance between them)
= $$\frac{1}{2}$$ (20 + 18) × (15) = 285 $${cm}^{2}$$

5. Find the area of a parallelogram with base 24 cm and height 16 cm.

A. 262 $${cm}^{2}$$
B. 384 $${cm}^{2}$$
C. 192 $${cm}^{2}$$
D. 131 $${cm}^{2}$$
E. None of these

Explanation:

Area of a parallelogram = base × height = 24 × 16 = 384 $${cm}^{2}$$

1. The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?

A. 1 : 96
B. 1 : 48
C. 1 : 84
D. 1 : 68
E. None of these

Explanation:

Let the length and the breadth of the rectangle be 4x cm and 3x respectively.

(4x) (3x) = 6912
12$${x}^{2}$$ = 6912
$${x}^{2}$$ = 576 = 4 × 144 = 22 × 122 (x > 0)
=> x = 2 × 12 = 24
Ratio of the breadth and the areas = 3x : 12$${x}^{2}$$ = 1 : 4x = 1: 96.

2. The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?

A. 8
B. 12
C. 6
D. 2
E. None of these

Explanation:

Let the sides of the rectangle be l and b respectively.

From the given data,
(√$${l}^{2} + {b}^{2}$$) = (1 + 108 $$\frac{1}{3}$$ %)lb
=> $${l}^{2} + {b}^{2}$$ = (1 + $$\frac{325}{3}$$ × $$\frac{1}{100}$$)lb
= (1 + $$\frac{13}{12}$$)lb
= $$\frac{25}{12}$$ lb
=> $${l}^{2}$$ + $$\frac{{b}^{2}}{lb}$$ = $$\frac{25}{12}$$

12$${l}^{2} + {b}^{2}$$ = 25lb
12$${l}^{2}$$ + 12$${b}^{2}$$ + 24lb = 49lb
12($${l}^{2} + {b}^{2}$$ + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12$${(l + b)}^{2}$$ = 49lb
=> 12$${(14)}^{2}$$ = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.

3. The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?

A. 8.5 m
B. 17 m
C. 34 m
D. 51 m
E. None of these

Explanation:

Let the breadth of the plot be b m.

Length of the plot = 3 b m
(3b)(b) = 867
3$${b}^{2}$$ = 867
$${b}^{2}$$ = 289 = 172 (b > 0)
b = 17 m.

4. The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor?

A. 27 m
B. 24 m
C. 18 m
D. 21 m
E. None of these

Explanation:

Let the length and the breadth of the floor be l m and b m respectively.

l = b + 200% of b = l + 2b = 3b
Area of the floor = $$\frac{324}{3}$$ = 108 sq m
l b = 108 i.e., l × $$\frac{1}{3}$$ = 108
l2 = 324 => l = 18.

5. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?

A. Rs. 3642.40
B. Rs. 3868.80
C. Rs. 4216.20
D. Rs. 4082.40
E. None of these

Explanation:

Length of the first carpet = (1.44) (6) = 8.64 cm

Area of the second carpet = 8.64 (1 + $$\frac{40}{100}$$) × 6 (1 + $$\frac{25}{100}$$)
= 51.84 (1.4) ($$\frac{5}{4}$$) sq m = (12.96) (7) sq m
Cost of the second carpet = (45) (12.96 × 7) = 315 (13 – 0.04) = 4095 – 12.6 = Rs. 4082.40

1. What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58?

A. Rs. 3944
B. Rs. 3828
C. Rs. 4176
D. Cannot be determined
E. None of these

Explanation:

Let the side of the square plot be a ft.

$${a}^{2}$$ = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 × 58 = Rs. 3944.

2. The area of a square is equal to five times the area of a rectangle of dimensions 125 cm × 64 cm. What is the perimeter of the square?

A. 600 cm
B. 800 cm
C. 400 cm
D. 1000 cm
E. None of these

Explanation:

Area of the square = s × s = 5(125 × 64)

=> s = 25 × 8 = 200 cm
Perimeter of the square = 4 × 200 = 800 cm.

3. A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangule, whose length and breadth are in the ratio of 6 : 5. What is the area of the rectangle?

A. 60 $${cm}^{2}$$
B. 30 $${cm}^{2}$$
C. 45 $${cm}^{2}$$
D. 15 $${cm}^{2}$$
E. None of these.

Explanation:

The circumference of the circle is equal to the permeter of the rectangle.

Let l = 6x and b = 5x 2(6x + 5x) = 2 × $$\frac{22}{7}$$ × 3.5
=> x = 1
Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 × 5 = 30 $${cm}^{2}$$

4. The area of a square is 4096 sq cm. Find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square.

A. 18 : 5
B. 7 : 16
C. 5 : 14
D. 5 : 32
E. None of these.

Explanation:

Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.

$${a}^{2}$$ = 4096 = 212
a = (212)$$\frac{1}{2}$$ = 26 = 64
L = 2a and b = a – 24
b : l = a – 24 : 2a = 40 : 128 = 5 : 16

5. The parameter of a square is double the perimeter of a rectangle. The area of the rectangle is 480 sq cm. Find the area of the square.

A. 200 sq cm
B. 72 sq cm
C. 162 sq cm
D. Cannot be determined
E. None of these

Explanation:

Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.

4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found.

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