NCERT Solutions for Class 8 Maths Chapter 9: Mensuration — Free PDF

Mensuration is all about measuring — areas of 2D shapes and surface areas and volumes of 3D solids. In Class 8, you extend your knowledge from rectangles and triangles to trapeziums, general quadrilaterals, and polygons. You also learn to compute the surface area and volume of cubes, cuboids, and cylinders.

This chapter is formula-heavy but highly scoring. Most questions follow a direct application of formulas, so memorising and practising them is the key to full marks. The real-world applications are everywhere: from calculating how much paint is needed to cover a room, to finding the capacity of a water tank.

The chapter has 2 exercises with a total of around 20 problems. Exercise 9.1 deals entirely with areas of 2D figures — trapezium, general quadrilateral, rhombus, and polygons. Exercise 9.2 covers surface area (TSA, LSA/CSA) and volume of three-dimensional solids: cubes, cuboids, and cylinders. This chapter often carries 8-10 marks in CBSE exams, making it one of the highest-scoring chapters in Class 8.

Perimeter: The total length of the boundary of a 2D figure. Measured in cm, m, etc.

Area: The amount of surface enclosed by a 2D figure. Measured in cm$^2$, m$^2$, etc.

Lateral Surface Area (LSA) / Curved Surface Area (CSA): The area of only the side surfaces of a 3D solid, excluding the top and bottom faces.

Total Surface Area (TSA): The area of all surfaces of a 3D solid, including the top and bottom faces.

Volume: The amount of space occupied by a 3D solid. Measured in cm$^3$, m$^3$, or litres.

Important Conversions:
- $1$ m$^2$ $= 10{,}000$ cm$^2$
- $1$ m$^3$ $= 1{,}000{,}000$ cm$^3$
- $1$ m$^3$ $= 1{,}000$ litres
- $1$ litre $= 1{,}000$ cm$^3$
- $1$ hectare $= 10{,}000$ m$^2$

2D Shapes — Areas:

• Trapezium: $A = \dfrac{1}{2}(a + b) \times h$, where $a, b$ are the parallel sides and $h$ is the perpendicular height between them.
• General Quadrilateral: Split into two triangles using a diagonal $d$:
  
  $$A = \dfrac{1}{2} \times d \times (h_1 + h_2)$$
  
  where $h_1$ and $h_2$ are the perpendicular distances from the opposite vertices to the diagonal.
• Rhombus: $A = \dfrac{1}{2} \times d_1 \times d_2$, where $d_1$ and $d_2$ are the diagonals.
• Regular Polygon: Divide into congruent triangles from the centre and sum their areas.

3D Solids — Surface Area and Volume:

• **Cuboid ($l, b, h$):**
  
  $$\text{TSA} = 2(lb + bh + lh), \quad \text{LSA} = 2h(l + b), \quad V = l \times b \times h$$
• **Cube (side $a$):**
  
  $$\text{TSA} = 6a^2, \quad \text{LSA} = 4a^2, \quad V = a^3$$
• **Cylinder ($r, h$):**
  
  $$\text{CSA} = 2\pi rh, \quad \text{TSA} = 2\pi r(r + h), \quad V = \pi r^2 h$$

**Q1. Find the area of a trapezium with parallel sides $12$ cm and $8$ cm, and height $5$ cm.**

Solution:

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**Q2. The diagonals of a rhombus are $24$ cm and $10$ cm. Find its area.**

Solution:

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**Q3. Find the area of a quadrilateral $ABCD$ where diagonal $AC = 14$ cm and the perpendicular distances from $B$ and $D$ to $AC$ are $4$ cm and $6$ cm respectively.**

Solution:

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**Q4. The area of a trapezium is $180$ cm$^2$, one parallel side is $16$ cm, and the height is $12$ cm. Find the other parallel side.**

Solution:

Let the other parallel side be $b$.

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**Q5. A field is in the shape of a pentagon. It is divided into a trapezium and a triangle by drawing a line parallel to one side. The parallel sides of the trapezium are $20$ m and $28$ m with height $15$ m. The triangle has base $28$ m and height $8$ m. Find the total area.**

Solution:

Area of trapezium $= \dfrac{1}{2}(20 + 28) \times 15 = \dfrac{1}{2} \times 48 \times 15 = 360$ m$^2$.

Area of triangle $= \dfrac{1}{2} \times 28 \times 8 = 112$ m$^2$.

Total area $= 360 + 112 = 472$ m$^2$.

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**Q6. The diagonals of a rhombus are in the ratio $3 : 4$. If the area is $96$ cm$^2$, find the diagonals.**

Solution:

Let the diagonals be $3k$ and $4k$.

Diagonals are $3 \times 4 = 12$ cm and $4 \times 4 = 16$ cm.

**Q1. Find the total surface area and volume of a cuboid with dimensions $10$ cm $\times$ $6$ cm $\times$ $4$ cm.**

Solution:

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**Q2. A cylindrical tank has radius $7$ m and height $10$ m. Find the capacity in litres.** (Use $\pi = 22/7$)

Solution:

Since $1 \text{ m}^3 = 1000$ litres: Capacity $= 1540 \times 1000 = 15{,}40{,}000$ litres.

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**Q3. The lateral surface area of a cube is $144$ cm$^2$. Find its volume.**

Solution:

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**Q4. A cuboid has a total surface area of $340$ cm$^2$. Its length is $10$ cm and breadth is $8$ cm. Find its height.**

Solution:

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**Q5. A hollow cylinder is open at both ends. Its outer radius is $7$ cm, inner radius is $3.5$ cm, and height is $10$ cm. Find the total surface area.**

Solution:

Outer CSA $= 2\pi R h = 2 \times \dfrac{22}{7} \times 7 \times 10 = 440$ cm$^2$.

Inner CSA $= 2\pi r h = 2 \times \dfrac{22}{7} \times 3.5 \times 10 = 220$ cm$^2$.

Area of two ring-shaped ends $= 2 \times \pi(R^2 - r^2) = 2 \times \dfrac{22}{7} \times (49 - 12.25) = 2 \times \dfrac{22}{7} \times 36.75 = 231$ cm$^2$.

TSA $= 440 + 220 + 231 = 891$ cm$^2$.

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**Q6. How many bricks, each measuring $25$ cm $\times$ $12.5$ cm $\times$ $7.5$ cm, are needed to build a wall $6$ m long, $5$ m high, and $0.5$ m thick?**

Solution:

Volume of wall $= 600 \times 500 \times 50 = 15{,}000{,}000$ cm$^3$ (converting m to cm).

Volume of one brick $= 25 \times 12.5 \times 7.5 = 2343.75$ cm$^3$.

Number of bricks $= \dfrac{15000000}{2343.75} = 6400$.

Here are more problems covering exam-level patterns.

**Example 1. A road roller has a diameter of $84$ cm and is $1$ m long. How much area does it cover in $500$ revolutions?**

Solution:

Radius $= 42$ cm $= 0.42$ m. Length $= 1$ m.

Area covered in one revolution $= 2\pi r \times l = 2 \times \dfrac{22}{7} \times 0.42 \times 1 = 2.64$ m$^2$.

Area in $500$ revolutions $= 500 \times 2.64 = 1320$ m$^2$.

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**Example 2. The dimensions of a cuboidal room are $12$ m $\times$ $8$ m $\times$ $4$ m. Find the cost of whitewashing the four walls and the ceiling at Rs $25$ per m$^2$. The room has $2$ doors ($2$ m $\times$ $1.5$ m each) and $3$ windows ($1.5$ m $\times$ $1$ m each).**

Solution:

Area of four walls $= 2h(l + b) = 2 \times 4 \times (12 + 8) = 160$ m$^2$.

Area of ceiling $= l \times b = 12 \times 8 = 96$ m$^2$.

Total area $= 160 + 96 = 256$ m$^2$.

Area of doors $= 2 \times (2 \times 1.5) = 6$ m$^2$.

Area of windows $= 3 \times (1.5 \times 1) = 4.5$ m$^2$.

Area to be whitewashed $= 256 - 6 - 4.5 = 245.5$ m$^2$.

Cost $= 245.5 \times 25 = \text{Rs } 6{,}137.50$.

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**Example 3. A cylinder and a cube have the same volume. The radius of the cylinder is $7$ cm and its height is $12$ cm. Find the side of the cube.**

Solution:

Volume of cylinder $= \pi r^2 h = \dfrac{22}{7} \times 49 \times 12 = 1848$ cm$^3$.

Let the side of the cube be $a$. Then $a^3 = 1848$.

$a = \sqrt[3]{1848} \approx 12.27$ cm.

Mistake 1: Confusing TSA and LSA/CSA.
TSA includes the top and bottom faces. LSA/CSA includes only the side surfaces. Read the question carefully to know which one is needed. If a box is open at the top, the area is $\text{TSA} - \text{area of top face}$.

Mistake 2: Forgetting to convert units.
All dimensions must be in the same unit before calculating. If length is in metres and breadth in centimetres, convert both to the same unit first. Also remember: $1$ m$^3 \neq 1000$ cm$^3$. Actually $1$ m$^3 = 10^6$ cm$^3$.

Mistake 3: Using diameter instead of radius in cylinder formulas.
The formulas use radius $r$, not diameter $d$. If the question gives the diameter, divide by $2$ first.

Mistake 4: Forgetting units in the answer.
Area must be in squared units (cm$^2$, m$^2$) and volume in cubed units (cm$^3$, m$^3$). Marks are often deducted for missing units.

Mistake 5: Not splitting irregular shapes correctly.
For general quadrilaterals and polygons, you must draw the diagonal correctly and identify the perpendicular heights accurately. A poorly drawn diagram leads to wrong heights.

Q1. Find the area of a trapezium with parallel sides $18$ cm and $12$ cm, and height $8$ cm.

Answer: $A = \dfrac{1}{2}(18 + 12) \times 8 = \dfrac{1}{2} \times 30 \times 8 = 120$ cm$^2$.

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Q2. A cube has a volume of $512$ cm$^3$. Find its total surface area.

Answer: $a^3 = 512 \implies a = 8$ cm. TSA $= 6 \times 8^2 = 384$ cm$^2$.

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Q3. Find the volume and CSA of a cylinder with radius $3.5$ cm and height $20$ cm.

Answer: $V = \dfrac{22}{7} \times (3.5)^2 \times 20 = \dfrac{22}{7} \times 12.25 \times 20 = 770$ cm$^3$. CSA $= 2 \times \dfrac{22}{7} \times 3.5 \times 20 = 440$ cm$^2$.

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Q4. The diagonals of a rhombus are $16$ cm and $12$ cm. Find (a) its area and (b) the side of the rhombus.

Answer: (a) Area $= \dfrac{1}{2} \times 16 \times 12 = 96$ cm$^2$. (b) Half-diagonals are $8$ and $6$. Side $= \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ cm.

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Q5. A cuboid-shaped swimming pool is $50$ m long, $30$ m wide, and $2.5$ m deep. Find the cost of cementing its floor and four walls at Rs $30$ per m$^2$.

Answer: Floor area $= 50 \times 30 = 1500$ m$^2$. Wall area $= 2 \times 2.5 \times (50 + 30) = 400$ m$^2$. Total $= 1900$ m$^2$. Cost $= 1900 \times 30 = \text{Rs } 57{,}000$.

• Area of trapezium $= \dfrac{1}{2}(a+b) \times h$.
  - For any quadrilateral, split it into triangles using a diagonal.
  - Rhombus area $= \dfrac{1}{2} \times d_1 \times d_2$.
  - Cuboid volume $= l \times b \times h$; Cube volume $= a^3$; Cylinder volume $= \pi r^2 h$.
  - TSA includes all faces; LSA/CSA excludes the top and bottom.
  - Always include units in your answer — area in cm$^2$/m$^2$, volume in cm$^3$/m$^3$.
  - Convert all dimensions to the same unit before calculating.
  - For real-world problems (painting, tiling, filling), identify whether you need area or volume.

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