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Geometry

Mensuration

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CAT 2025 Lesson : Mensuration - 2-D Figures

1. Introduction

Mensuration, meaning measurement, is a branch of mathematics which deals with the calculation of measurements such as distance, area, volume, etc. In CAT and other management entrance tests, we are tested on 2-dimensional (2-D) and 3-dimensional (3-D) figures only, which are detailed below.

Figure	2-D Figures	3-D Figures
Definition	A shape that can be drawn on a flat piece of paper or any other flat plane.	A solid object which occupies space and has volume.
Dimensions	Length and Breadth	Length, Breadth and Height
Example	Triangle, Quadrilateral, Pentagon, etc.	Cube, Cylinder, Sphere, etc.,

Mensuration pertaining to 2-D figures has been covered in the previous lessons in geometry, i.e., Lines & Triangles, Quadrilaterals and Circles. This lesson would serve as a recap for mensuration pertaining to 2-D figures and will provide mensuration for 3-D figures in detail.

1.1 Basic Conversions of units

Knowing these conversions will be helpful while solving mensuration questions.

Length / Distance	Area	Volume
1 Kilometre (km) = 1000 m		1 Kilolitre (kl) = 1000 l
1 Hectometre (hm) = 100 m	1 Hectare (ha) = 100 a	1 Hectolitre (hl) = 100 l
1 Decametre (dam) = 10 m	1 Decare (daa) = 10 a	1 Decalitre (dal) = 10 l
1 Metre (m)	1 Are (a) = 100 $\bold{m^2}$	1 Litre (l) = 1000 $\bold{cm^3}$
1 Decimetre (dm) = 0.1 m	1 Deciare (da) = 0.1 a	1 Decilitre (dl) = 0.1 l
1 Centimetre (cm) = 0.01 m	1 Centiare (ca) = 0.01 a	1 Centilitre (cl) = 0.01 l
1 Millimetre (mm) = 0.001 m		1 Millilitre (ml) = 0.001 l

Note: Please note the are and litre equivalents provided in the table above.

When area or volume is provided as a unit of length (say metres), it can be converted to a different unit by squaring or cubing their respective length conversions. The following table contains a few examples.

Area	Volume
$13 \ m^{2} = 13 \times (100 \ cm)^2 = 130,000 \ cm^2$	$5 \ m^{3} = 5 \times (100 \ cm)^3 = 5,000,000 \ cm^3$
$24 \ km^{2} = 24 \times (1000 \ m)^2 = 24,000,000 \ m^2$	$8 \ km^{3} = 8 \times (1000 \ m)^3 = 8,000,000,000 \ m^3$
$5,000 \ mm^{2} = 5,000 \times (0.1 \ cm)^2 = 50 \ cm^2$	$7,000 \ mm^{3} = 7,000 \times (0.1 \ cm)^3 = 7 \ cm^3$

Inches and feet are other measures of length wherein 1 inch = 2.54 cm and 1 feet = 12 inches.

With feet being a measure of length, areas are also mentioned with square feet (sqft) as a unit.

Acre is another unit used for area, wherein 1 acre = 43,560 sqft

2. 2-D Figures

A shape with only two dimensions (length and breadth) and no thickness is a 2D figure. We can calculate only the perimeter and area for these figures. Some examples of 2-D figures are square, rectangle, triangle, etc. Each of the figures listed below have been covered in the three preceding lessons under Geometry.

Shape	Perimeter	Area
Scalene Triangle	$a + b + c$	$\dfrac{1}{2} \times b \times h = \dfrac{1}{2} \times ab \times sin\theta$ = $\sqrt {s(s - a) (s - b) (s - c)}$ = $\dfrac{abc}{4R} = s \times r$ (where $s = \dfrac{a+ b + c}{2}$ , $R$ is the circumradius and $r$ is the inradius )
Isosceles Triangle	$2a + b$	$\dfrac{b}{4} \times \sqrt {4 a^2 - b^2}$
Equilateral Triangle	$3a$	$\dfrac{\sqrt3}{4} a^2$
Quadrilateral	AB + BC + CD + DA	$\dfrac{1}{2} \times$ AC $\times (h_1 + h_2)$
Parallelogram	$2(a + b)$	$b \times h$
Rhombus	$4a$	$\dfrac{1}{2} \times d_1 \times d_2$

Shape	Perimeter	Area
Rectangle	$2(l + b)$	$lb$
Square	$4a$	$a^2$
Trapezium	AB + BC + CD + AD	$\dfrac{1}{2} h (a + b)$
Circle	$2 \pi r$	$\pi r^2$
Semicircle	$\pi r + 2r$	$\dfrac{1}{2} \pi r^2$
Sector	$\left(\dfrac{\theta}{360 \degree} \times 2 \pi r \right) + 2r$	$\left(\dfrac{\theta}{360 \degree} \times \pi r^2\right)$

Example 1

Two pathways, each with a width of 2m, run through the centre of an 80m

\times

60m rectangular park. One of the pathways is along the length of the park, while the other is along the breadth of the park. What is the area of the park that is occupied by the pathways?

(1) 136

m^2

(2) 140

m^2

(3) 276

m^2

(4) 280

m^2

Solution


The pathways are in the form of a rectangle, with their lengths being 80m and 60m and their breadth being 2m each. Sum of areas of 2 pathways = $80 \times 2 + 60 \times 2 = 280 \ m^2$ The two pathways overlap at the centre. The overlapping part is a square of side 2m. The area of this square when subtracted from the sum above provides the required area. Area occupied by pathways = $280 - 2^2 = 276 \ m^2$

The pathways are in the form of a rectangle, with their lengths being 80m and 60m and their breadth being 2m each.

Sum of areas of 2 pathways =

80 \times 2 + 60 \times 2 = 280 \ m^2

The two pathways overlap at the centre. The overlapping part is a square of side 2m. The area of this square when subtracted from the sum above provides the required area.

Area occupied by pathways =

280 - 2^2 = 276 \ m^2

Answer: (3) 276

m^2

Example 2

In a rectangular paper of dimensions 75 cm

\times

50 cm, a margin of 2 cm from each of the edges is to be painted. If the cost of painting is Rs. 0.50/

cm^2

, then what is the total painting cost?

Solution

In the adjacent figure, the margin lies between the outer and inner rectangles. Therefore, to find its area, we subtract the area of the 2 rectangles. Margin area = $75 \times 50 - 71 \times 46$ = $3750$ - $3266$ = $484$ Cost of Painting = $484 \times 0.5 = Rs. 242$

Answer: Rs. 242

Example 3

A rectangular shed of dimensions

22 \ m \times 7 \ m

is constructed in the middle of a large farm. A cow is tied to one of the corners of the shed using a 14m long rope. If the area occupied by the shed is not grazable, then what is the area of the field that is available for the cow to graze?

Solution


OPQR is the rectangular shed, wherein the cow is tied to point O. The $270 \degree$ sector OAB, with radius of 14 cm, is fully accessible to the cow. After this, rope of 7m is stuck along OP and with P as the centre, the cow can graze the $90 \degree$ sector of PBC, whose radius is 7m. Grazable area = Area of sector OAB + Area of sector PBC. = $\dfrac{270}{360} \times \pi \times 14^2 + \dfrac{90}{360} \times \pi \times 7^2 = 462 + 38.5 = 500.5 \ m^2$

OPQR is the rectangular shed, wherein the cow is tied to point O. The

270 \degree

sector OAB, with radius of 14 cm, is fully accessible to the cow. After this, rope of 7m is stuck along OP and with P as the centre, the cow can graze the

90 \degree

sector of PBC, whose radius is 7m.

Grazable area = Area of sector OAB + Area of sector PBC.
=

\dfrac{270}{360} \times \pi \times 14^2 + \dfrac{90}{360} \times \pi \times 7^2 = 462 + 38.5 = 500.5 \ m^2

Answer:

500.5 \ m^2

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