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.
9.3 Solid Shapes
Solid shapes are three-dimensional figures that have length, breadth, and height. Unlike plane figures, solids occupy space and have volume. Examples include cubes, cuboids, cylinders, cones, spheres, and hemispheres. Each solid has faces, edges, and vertices. Some solids have faces that are congruent (identical in shape and size). For instance, a cube has six square faces all congruent to each other, while a cuboid has six rectangular faces with opposite faces congruent. Cylinders have two circular congruent faces and one curved surface. Understanding the properties of these solids is essential to calculate their surface areas and volumes. The chapter focuses on right circular cylinders, where the axis is perpendicular to the base. Other types of cylinders exist but are not covered here. Recognizing the shapes and their nets (2D layouts of faces) helps in visualizing and calculating surface areas.
📊 Diagram: See figure_24 and figure_25: Fig 9.10 showing various solids including cube, cuboid, and cylinder; figure_28 and figure_29: nets and examples of cuboidal, cubical, and cylindrical boxes.
🧪 Activity: Cut out faces of different boxes and identify congruent faces by placing them on each other.
🔗 Connection: Leads to calculation of surface areas 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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