Area (Mensuration) questions with answers and concepts

Learn area (mensuration) with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

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Mensuration is the branch of mathematics that deals with the measurement of geometric figures, including their areas, perimeters, and volumes. It involves calculating the space occupied by 2D shapes (like triangles, rectangles, circles) and understanding the relationships between different measurements. Mensuration problems commonly appear in aptitude tests and require knowledge of standard formulas and their applications.

Key formulas and rules

12 rules

Rectangle: $Area = l \times b$ , $Perimeter = 2 (l + b)$ , $Diagonal = l^{2} + b^{2}$

Square: $Area = a^{2} = \frac{1}{2} d^{2}$ , $Perimeter = 4 a$ , $Diagonal = a 2$

Triangle: $Area = \frac{1}{2} \times base \times height = s (s - a) (s - b) (s - c)$ (where $s = \frac{a + b + c}{2}$ )

Equilateral Triangle: $Area = \frac{3}{4} a^{2}$ , $Height = \frac{3}{2} a$

Isosceles Triangle: $Area = \frac{b}{4} 4 a^{2} - b^{2}$ (where $a$ = equal side, $b$ = base)

Circle: $Area = π r^{2}$ , $Circumference = 2 π r = π d$

Semicircle: $Area = \frac{1}{2} π r^{2}$ , $Perimeter = π r + 2 r = r (π + 2)$

Sector of Circle: $Area = \frac{θ}{36 0 ^{\circ}} \times π r^{2}$ , $Arc length = \frac{θ}{36 0 ^{\circ}} \times 2 π r$

Ring (Annulus): $Area = π (R^{2} - r^{2})$ (where $R = outer radius, r = inner radius$ )

Parallelogram: $Area = base \times height$ , $Perimeter = 2 (a + b)$

Rhombus: $Area = \frac{1}{2} \times d_{1} \times d_{2}$ , $Perimeter = 4 a$ , $Side a = \frac{1}{2} d_{1}^{2} + d_{2}^{2}$

Trapezium: $Area = \frac{1}{2} \times (a + b) \times h$ (where $a, b$ are parallel sides, $h$ is height)

Key concepts

Area vs. Perimeter

Area measures the region enclosed within a shape (in square units), while perimeter measures the total boundary length (in linear units). For the same perimeter, a circle has the maximum area. Understanding when to use area versus perimeter formulas is crucial - area is needed for coverage problems (painting, carpeting) while perimeter is needed for boundary problems (fencing, framing).

Units and Conversions

Always ensure all measurements are in the same unit before calculating. Common conversions: 1 $m = 100$ $c m = 1000$ mm; 1 $k m = 1000$ m; 1 hectare = 10,000 $m^{2} = 100$ ares; 1 acre = 4046.86 $m^{2}$ . Area conversions use square factors: 1 $m^{2} = 10$ ,000 $c m^{2}$ , not 100 $c m^{2}$ . Volume conversions use cubic factors.

Composite Figures

Complex shapes can be broken down into simpler standard shapes. Calculate the required measurement (area/perimeter) for each component and combine appropriately. For area of composite figures: add areas of components that make up the shape. For shaded region problems: subtract the unshaded area from total area.

Pathway Problems

When a path runs around/inside a garden or field: Outer dimension minus path width gives inner dimension. For a path of uniform width 'w' around a rectangle of dimensions L x B: Inner rectangle dimensions are (L-2w) x (B-2w). Path area = Outer area - Inner area = 2w(L + B - 2w).

Cost Calculation Problems

To find cost of painting, fencing, or paving: First calculate the required area or perimeter, then multiply by the rate per unit. Example: Cost to paint a wall = (Area of wall) x (Rate per $m^{2})$ . For cylindrical tanks or circular paths, be careful whether you need curved surface area or total surface area.

Ratio of Areas

When comparing similar figures or figures with related dimensions: If linear dimensions are in ratio a:b, then areas are in ratio $a^{2}$ : $b^{2}$ , and perimeters are in ratio a:b. Example: If two circles have radii in ratio 2:3, their areas are in ratio 4:9, and their circumferences are in ratio 2:3.

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Worked examples

Example 1

The length and breadth of a rectangle are in the ratio $5 : 3$ . If its perimeter is 128 m, find its area.

Solution

Step 1: Express dimensions using scale factor $x$ .
- Length $l = 5 x$
- Breadth $b = 3 x$

Step 2: Use perimeter formula.

Perimeter = 2 (l + b) = 2 (5 x + 3 x) = 16 x = 128

x = \frac{128}{16} = 8 m

Step 3: Calculate dimensions and area.

l = 5 \times 8 = 40 m, b = 3 \times 8 = 24 m

Area = l \times b = 40 \times 24 = 960 m^{2}

Answer: $960 m^{2}$ .

Example 2

Find the area of a triangle with side lengths 13 cm, 14 cm, and 15 cm.

Solution

Step 1: Calculate semi-perimeter $s$ .

s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 cm

Step 2: Apply Heron's formula.

Area = s (s - a) (s - b) (s - c)

Area = 21 \times (21 - 13) \times (21 - 14) \times (21 - 15)

Area = 21 \times 8 \times 7 \times 6 = 7, 056 = 84 cm^{2}

Answer: $84 cm^{2}$ .

Example 3

The area of a rhombus is $216 cm^{2}$ . If one diagonal is 24 cm, find the length of the other diagonal and the length of each side.

Solution

Step 1: Calculate the second diagonal using the area formula.

Area = \frac{1}{2} d_{1} d_{2}

216 = \frac{1}{2} \times 24 \times d_{2} = 12 d_{2} ⟹ d_{2} = \frac{216}{12} = 18 cm

Step 2: Calculate side length using Pythagoras theorem.
Diagonals of a rhombus bisect each other at right angles:

Side = (\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2} = 1 2^{2} + 9^{2} = 144 + 81 = 225 = 15 cm

Answer: Diagonal = 18 cm; Side = 15 cm.

Example 4

A circular park has a circumference of 440 m. A surrounding circular road 7 m wide runs around the outside of the park. Find the area of the road.

Solution

Step 1: Find the inner radius $r$ of the park.

2 π r = 440 ⟹ r = \frac{440}{2 \times \frac{22}{7}} = \frac{440 \times 7}{44} = 70 m

Step 2: Find outer radius $R$ .

R = r + 7 = 70 + 7 = 77 m

Step 3: Compute the area of the ring.

Area of road = π (R^{2} - r^{2}) = π (R + r) (R - r)

Area = \frac{22}{7} \times (77 + 70) \times (77 - 70) = \frac{22}{7} \times 147 \times 7 = 22 \times 147 = 3, 234 m^{2}

Answer: $3, 234 m^{2}$ .

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

Find the circumference of a circle of radius 5 cm using $p i = 22/7$ .

31.428571428571427 cm
15.714285714285714 cm
10 cm
$25 c m^{2}$

Answer: A. 31.428571428571427 cm

Explanation

Circumference = 2pir = 2 x 22/7 x $5 = 31$ .428571428571427 cm.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

Find the circumference of a circle of radius 6 cm using $p i = 22/7$ .

37.714285714285715 cm
12 cm
18.857142857142858 cm
$36 c m^{2}$

Answer: A. 37.714285714285715 cm

Explanation

Circumference = 2pir = 2 x 22/7 x $6 = 37$ .714285714285715 cm.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

Find the circumference of a circle of radius 7 cm using $p i = 22/7$ .

44 cm
22 cm
$49 c m^{2}$
14 cm

Answer: A. 44 cm

Explanation

Circumference = 2pir = 2 x 22/7 x $7 = 44$ cm.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Memorize all standard formulas for 2D shapes thoroughly. Create a formula sheet for quick revision.
Always verify units are consistent before calculating. Convert all measurements to the same unit first.
For composite figures, carefully identify which areas to add and which to subtract. Drawing a diagram helps visualize.
Remember that pi ~= 22/7 is a good approximation when calculations result in whole numbers, but use pi ~= 3.14 or calculator value for precision.
For pathway problems, be careful whether the path is inside (reducing dimensions) or outside (increasing dimensions) the main area.
When given ratios, use the technique of assuming actual values as ratio multiples (like 5x, 3x) and solve for x.
For cost-based problems, verify whether you're calculating cost based on area or perimeter before multiplying by the rate.
In right triangles, you can use x base x height, but for general triangles with three sides given, Heron's formula is essential.
For circle problems involving sectors and segments, always identify the central angle and use the appropriate fraction ( /360 deg) of the full circle values.
Double-check your final answer by estimating - if a rectangle has sides around 10 m each, the area should be around 100 m^2, not 1000 m^2.

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