Mensuration | Class 8 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 3 min read
Mensuration – this guide gives you a concise, exam-ready overview of Mensuration from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
9.2 Area of a Polygon
The area of polygons, especially quadrilaterals and pentagons, can be found by dividing them into simpler shapes such as triangles and trapeziums. For example, a pentagon can be divided by drawing diagonals from one vertex to non-adjacent vertices, thus splitting it into triangles. The total area of the polygon is then the sum of the areas of these triangles. Alternatively, perpendiculars can be drawn from vertices to a diagonal to divide the polygon into right-angled triangles and trapeziums, whose areas can be calculated using known formulas. This method simplifies the calculation by breaking complex polygons into manageable parts. The chapter provides examples and exercises to practice this method, including dividing polygons like pentagons and hexagons into triangles and trapeziums and calculating their areas using base and height measurements.
📊 Diagram: See figure_3 and figure_4: Fig 9.1 and Fig 9.2 showing a pentagon divided into triangles and trapeziums; figure_6, figure_7, figure_8, figure_9: polygons divided into parts for area calculation; figure_10, figure_11, figure_12, figure_13, figure_14: hexagon and trapezium examples.
🧪 Activity: Try These: Divide given polygons into triangles and trapeziums to find their areas using given dimensions.
🔗 Connection: Prepares for understanding surface area and volume of three-dimensional solids.
Frequently asked questions
1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are $1 \, \text{m}$ and $1.2 \, \text{m}$ and perpendicular distance between them is $0.8 \, \text{m}$.
Given: Parallel sides a = 1 m, b = 1.2 m, height h = 0.8 m.
Area of trapezium = \( \frac{1}{2} \times (a + b) \times h \) = \( \frac{1}{2} \times (1 + 1.2) \times 0.8 = \frac{1}{2} \times 2.2 \times 0.8 = 0.88 \text{ m}^2 \).
Hence, the area of the trapezium is 0.88 square meters.
2. The area of a trapezium is $34~\mathrm{cm}^2$ and the length of one of the parallel sides is $10~\mathrm{cm}$ and its height is $4~\mathrm{cm}$. Find the length of the other parallel side.
Given: Area = 34 cm², one parallel side a = 10 cm, height h = 4 cm.
Area of trapezium = \( \frac{1}{2} (a + b) h \), where b is the other parallel side.
Substitute values: 34 = \( \frac{1}{2} (10 + b) \times 4 \) => 34 = 2 (10 + b) => 10 + b = 17 => b = 17 - 10 = 7 \text{ cm} \).
Hence, the length of the other parallel side is 7 cm.
3. Length of the fence of a trapezium shaped field ABCD is $120\mathrm{m}$. If $\mathrm{BC} = 48\mathrm{m}$, $\mathrm{CD} = 17\mathrm{m}$ and $\mathrm{AD} = 40\mathrm{m}$, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.
Given: Perimeter = 120 m BC = 48 m, CD = 17 m, AD = 40 m AB is perpendicular to AD and BC.
Find AB: Perimeter = AB + BC + CD + DA 120 = AB + 48 + 17 + 40 => AB = 120 - 105 = 15 m
Area of trapezium = \( \frac{1}{2} (AD + BC) \times AB \) = \( \frac{1}{2} (40 + 48) \times 15 = \frac{1}{2} \times 88 \times 15 = 44 \times 15 = 660 \text{ m}^2 \).
Hence, the area of the field is 660 square meters.
4. The diagonal of a quadrilateral shaped field is $24\mathrm{m}$ and the perpendiculars dropped on it from the remaining opposite vertices are $8\mathrm{m}$ and $13\mathrm{m}$. Find the area of the field.
Given: Diagonal = 24 m Perpendiculars from opposite vertices = 8 m and 13 m
Area of quadrilateral = sum of areas of two triangles formed by the diagonal = \( \frac{1}{2} \times 24 \times 8 + \frac{1}{2} \times 24 \times 13 \) = 12 \times 8 + 12 \times 13 = 96 + 156 = 252 \text{ m}^2 \).
Hence, the area of the field is 252 square meters.
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