Mensuration | Class 8 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 4 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.
9.4 Surface Area of Cube, Cuboid and Cylinder
Surface area of a solid is the total area covered by all its faces. To find the surface area, we calculate the area of each face and sum them up. For a cuboid, which has six rectangular faces with three pairs of identical faces, the total surface area is twice the sum of the areas of three adjacent faces. The formula for the total surface area of a cuboid with length l, breadth b, and height h is 2(lb + bh + hl). The lateral surface area is the area of the four side faces excluding the top and bottom, given by 2h(l + b). For a cube, all six faces are squares with side length l, so the total surface area is 6l². For a right circular cylinder with radius r and height h, the total surface area includes the areas of two circular bases and the curved surface area. The curved surface area is the area of a rectangle with length equal to the circumference of the base (2πr) and width equal to the height h, so it is 2πrh. The total surface area of the cylinder is 2πr² + 2πrh = 2πr(r + h). These formulas help in practical problems such as painting, wrapping, or covering objects.
📊 Diagram: See figure_41 and figure_42: Net of a cuboidal box showing six faces; figure_45: lateral surface area of cuboid room; figure_49 and figure_51: net and faces of a cube; figure_61 and figure_66: cylinder and its rectangular strip net; figure_67: cylinder surface area components.
🧪 Activity: Cover lateral surface of a cuboidal object with a strip of paper and measure its area to understand lateral surface area.
🔗 Connection: Prepares for volume calculations of these 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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