Ssc Gd Mensuration 3D

Jagdeep Singh

Published: 16 Dec, 2025

SSC GD Math Notes

Introduction to 3D Shapes
Understanding Volume and Surface Area
Cube
Cuboid
Cylinder
Basic Cone and Sphere

Introduction to 3D Shapes

Welcome to your study notes for ssc gd mensuration! We are going to learn about 3D shapes.

What is a 3D shape?

3D means “Three Dimensions.”
It means the shape has length, width, and height.
These shapes are solid, like a ball or a box.
2D shapes (like squares) are flat, but 3D shapes take up space in the real world.
Learning these concepts is key to solving ssc gd mensuration questions.

Understanding Volume and Surface Area

When we study 3D shapes for mensuration ssc gd, we look for two main things: Volume and Surface Area.

Volume (How much fits inside?)

Volume tells us how much space is inside the shape.
Imagine you have a big water tank. The volume is how much water the tank can hold.
Volume is always measured in cubic units (like $c m^{3}$ or $m^{3}$ ).

Surface Area (How much paint is needed?)

Surface Area tells us the total area of all the outside faces (sides) of the shape.
Imagine you want to paint a box. The surface area is the total amount of paint you need to cover the entire box.
Surface Area is always measured in square units (like $c m^{2}$ or $m^{2}$ ).

Cube

A Cube is a very special 3D shape. Think of a standard dice or a sugar cube.

All the sides (faces) of a cube are squares.
All the edges (length, width, and height) are exactly the same size. We call this size ‘a’.
Understanding the cube is essential for solving basic ssc gd mensuration questions formula.

Formulas for Cube

Measurement	Formula (where ‘a’ is the side length)
Volume (V)	$V = a \times a \times a = a^{3}$
Total Surface Area (TSA)	$T S A = 6 \times a^{2}$ (Because there are 6 square faces)
Lateral Surface Area (LSA)	$L S A = 4 \times a^{2}$ (Area of the 4 side walls, excluding top and bottom)

4 Solved Examples (Cube)

Example 1: Finding Volume

A cube has a side length of 5 cm. What is its Volume? This is a common type of ssc gd mensuration question.

Solution:

First, let’s understand what we need to find. We need the Volume (how much space is inside).
Now, write down the formula for Volume of a Cube: $V = a^{3}$ .
Put the numbers in. The side length ( $a$ ) is 5 cm. $V = 5 \times 5 \times 5$
Calculate it. $V = 25 \times 5 = 125$
So, the answer is $125 {cm}^{3}$ .

Example 2: Finding Total Surface Area

If the side of a cube is 10 meters, what is the Total Surface Area (TSA)?

Solution:

First, we need the Total Surface Area (the area of all 6 sides).
Now, write down the formula for TSA: $T S A = 6 a^{2}$ .
Put the numbers in. The side length ( $a$ ) is 10 m. $T S A = 6 \times (10)^{2}$
Calculate the square first: $T S A = 6 \times (10 \times 10) = 6 \times 100$
Calculate the final area: $T S A = 600$
So, the answer is $600 m^{2}$ .

Example 3: Finding Lateral Surface Area

A room is shaped like a cube with a side of 4 meters. We only want to paint the four walls (LSA). How much area needs painting? This helps practice mensuration ssc gd questions formula.

Solution:

We need the Lateral Surface Area (LSA), which means the 4 walls only.
Write down the formula for LSA: $L S A = 4 a^{2}$ .
Put the numbers in. The side length ( $a$ ) is 4 m. $L S A = 4 \times (4)^{2}$
Calculate the square: $L S A = 4 \times 16$
Calculate the final area: $L S A = 64$
So, the answer is $64 m^{2}$ .

Example 4: Finding Side from Volume

The volume of a cube is $27 {cm}^{3}$ . What is the length of its side? This is a tricky ssc gd mensuration question.

Solution:

We know the Volume ( $V = 27$ ) and the formula $V = a^{3}$ .
We need to find the number ( $a$ ) that, when multiplied by itself three times, gives 27. $a^{3} = 27$
We need to find the cube root of 27. $3 \times 3 \times 3 = 9 \times 3 = 27$
So, the side length ( $a$ ) must be 3.
The answer is 3 cm.

Cuboid

A Cuboid is like a stretched cube. Think of a matchbox, a book, or a standard brick.

A Cuboid has 6 faces, but they are rectangles, not necessarily squares.
It has three different measurements: Length ( $l$ ), Breadth (width, $b$ ), and Height ( $h$ ).
These mensuration ssc gd questions often involve packing or storage.

Formulas for Cuboid

Measurement	Formula
Volume (V)	$V = l \times b \times h$
Total Surface Area (TSA)	$T S A = 2 (l b + b h + h l)$
Lateral Surface Area (LSA)	$L S A = 2 h (l + b)$ (Area of the 4 walls)

4 Solved Examples (Cuboid)

Example 1: Finding Volume

A box has a length of 8 cm, a breadth of 3 cm, and a height of 2 cm. Find the volume of the box.

Solution:

We need the Volume (V), which is how much the box holds.
Write down the formula: $V = l \times b \times h$ .
Put the numbers in: $l = 8$ , $b = 3$ , $h = 2$ . $V = 8 \times 3 \times 2$
Calculate it: $V = 24 \times 2 = 48$
So, the answer is $48 {cm}^{3}$ .

Example 2: Finding Total Surface Area

Calculate the Total Surface Area (TSA) of a cuboid with $l = 5$ m, $b = 4$ m, and $h = 3$ m. This is a standard ssc gd mensuration questions pdf problem.

Solution:

We need the Total Surface Area (TSA) of all 6 faces.
Write down the formula: $T S A = 2 (l b + b h + h l)$ .
Calculate the three parts inside the bracket first:
- $l b = 5 \times 4 = 20$
- $b h = 4 \times 3 = 12$
- $h l = 3 \times 5 = 15$
Add them up: $T S A = 2 (20 + 12 + 15)$ $T S A = 2 (47)$
Calculate the final area: $T S A = 94$
So, the answer is $94 m^{2}$ .

Example 3: Finding Height from Volume

A water tank has a volume of $100 m^{3}$ . Its length is 10 m and its breadth is 5 m. What is the height of the tank?

Solution:

We know the Volume ( $V = 100$ ) and the formula $V = l \times b \times h$ .
Put the known numbers into the formula: $100 = 10 \times 5 \times h$
Multiply the length and breadth: $100 = 50 \times h$
To find $h$ , we divide the volume by the area of the base: $h = 100 / 50$
Calculate it: $h = 2$
So, the height is 2 meters.

Example 4: Finding Lateral Surface Area

A hall has dimensions $l = 12$ m, $b = 8$ m, and $h = 5$ m. Find the cost of painting the four walls (LSA) if the cost is ₹2 per square meter. This is a practical mensuration ssc gd problem.

Solution:

First, find the Lateral Surface Area (LSA) of the four walls. Formula: $L S A = 2 h (l + b)$ .
Put the numbers in: $L S A = 2 \times 5 \times (12 + 8)$
Calculate the bracket first: $L S A = 10 \times (20)$ $L S A = 200 m^{2}$
Now, find the total cost. Cost = Area $\times$ Rate. Cost = $200 \times 2$
Calculate the cost: Cost = ₹400
The total cost of painting is ₹400.

Cylinder

A Cylinder is a shape like a soda can, a pipe, or a drum.

It has two flat circular bases (top and bottom).
It has one curved surface connecting the bases.
We use the radius ( $r$ ) of the circle and the height ( $h$ ) of the cylinder.
Remember that $π$ (pi) is a constant number, usually taken as $22 / 7$ or 3.14. These ssc gd mensuration questions formula rely heavily on $π$ .

Formulas for Cylinder

Measurement	Formula
Volume (V)	$V = π r^{2} h$
Curved Surface Area (CSA)	$C S A = 2 π r h$ (The area of the side label only)
Total Surface Area (TSA)	$T S A = 2 π r h + 2 π r^{2}$ (CSA + Area of top circle + Area of bottom circle)

4 Solved Examples (Cylinder)

Example 1: Finding Volume

A cylindrical pipe has a radius ( $r$ ) of 7 cm and a height ( $h$ ) of 10 cm. Find its volume. (Use $π = 22 / 7$ ).

Solution:

We need the Volume (V).
Write down the formula: $V = π r^{2} h$ .
Put the numbers in: $r = 7$ , $h = 10$ , $π = 22 / 7$ . $V = (22 / 7) \times (7)^{2} \times 10$
Simplify the calculation: $V = (22 / 7) \times (7 \times 7) \times 10$ $V = 22 \times 7 \times 10$ (One 7 cancels out the 7 in the denominator)
Calculate it: $V = 154 \times 10 = 1540$
The volume is $1540 {cm}^{3}$ .

Example 2: Finding Curved Surface Area

A cylindrical pillar has a radius of 1 meter and a height of 5 meters. Find the Curved Surface Area (CSA). (Use $π = 3.14$ ). This is a typical ssc gd mensuration question.

Solution:

We need the Curved Surface Area (CSA), the area of the side.
Write down the formula: $C S A = 2 π r h$ .
Put the numbers in: $r = 1$ , $h = 5$ , $π = 3.14$ . $C S A = 2 \times 3.14 \times 1 \times 5$
Multiply the simple numbers first: $C S A = (2 \times 5) \times 3.14$ $C S A = 10 \times 3.14$
Calculate it: $C S A = 31.4$
The CSA is $31.4 m^{2}$ .

Example 3: Finding Total Surface Area

A closed drum has $r = 7$ cm and $h = 5$ cm. Find the Total Surface Area (TSA). (Use $π = 22 / 7$ ). This is a common mensuration ssc gd questions formula application.

Solution:

We need the Total Surface Area (TSA). Formula: $T S A = 2 π r h + 2 π r^{2}$ .
Calculate the CSA part ( $2 π r h$ ): $C S A = 2 \times (22 / 7) \times 7 \times 5$ $C S A = 2 \times 22 \times 5 = 220 {cm}^{2}$
Calculate the area of the two circles ( $2 π r^{2}$ ): $2 π r^{2} = 2 \times (22 / 7) \times 7 \times 7$ $2 π r^{2} = 2 \times 22 \times 7 = 308 {cm}^{2}$
Add the parts together: $T S A = 220 + 308$ $T S A = 528$
The TSA is $528 {cm}^{2}$ .

Example 4: Finding Height from Volume

The volume of a cylinder is $440 {cm}^{3}$ . If the radius is 7 cm, what is the height? (Use $π = 22 / 7$ ). This is a key ssc gd mensuration questions pdf type.

Solution:

We know $V = 440$ and the formula $V = π r^{2} h$ .
Put the known numbers in: $440 = (22 / 7) \times (7)^{2} \times h$
Simplify the radius part: $440 = (22 / 7) \times 49 \times h$ $440 = 22 \times 7 \times h$ (Since $49 / 7 = 7$ )
Multiply the constants: $440 = 154 \times h$
Solve for $h$ : $h = 440 / 154$
Divide (both numbers are divisible by 22): $h = 20 / 7$
The height is $20 / 7$ cm (or approximately 2.86 cm).

Basic Cone and Sphere

For the ssc gd mensuration syllabus, you need to know the basic formulas for cones and spheres.

Cone

A Cone is a shape like an ice cream cone or a birthday hat. It has a circular base and comes to a point (apex).

$r$ = radius
$h$ = height (straight up)
$l$ = slant height (the slanted side)

Measurement	Formula
Volume (V)	$V = (1 / 3) π r^{2} h$ (It is 1/3 the volume of a cylinder)
Curved Surface Area (CSA)	$C S A = π r l$

Sphere

A Sphere is a perfectly round 3D object, like a football or a tennis ball. It only has a radius ( $r$ ).

Measurement	Formula
Volume (V)	$V = (4 / 3) π r^{3}$
Surface Area (SA)	$S A = 4 π r^{2}$ (Total Surface Area and Curved Surface Area are the same)

4 Solved Examples (Cone and Sphere)

Example 1 (Cone Volume)

A cone has a radius of 3 cm and a height of 7 cm. Find its Volume. (Use $π = 22 / 7$ ).

Solution:

We need the Volume (V).
Write down the formula: $V = (1 / 3) π r^{2} h$ .
Put the numbers in: $r = 3$ , $h = 7$ , $π = 22 / 7$ . $V = (1 / 3) \times (22 / 7) \times (3)^{2} \times 7$
Simplify the calculation: $V = (1 / 3) \times (22 / 7) \times 9 \times 7$
Cancel out the 7s and simplify the 3: $V = (1 / 3) \times 22 \times 9$ $V = 22 \times 3$ (Since $9 / 3 = 3$ )
Calculate it: $V = 66$
The volume is $66 {cm}^{3}$ .

Example 2 (Cone CSA)

A cone has a radius of 5 m and a slant height ( $l$ ) of 10 m. Find its Curved Surface Area (CSA). (Use $π = 3.14$ ). This is a basic ssc gd mensuration question.

Solution:

We need the Curved Surface Area (CSA).
Write down the formula: $C S A = π r l$ .
Put the numbers in: $r = 5$ , $l = 10$ , $π = 3.14$ . $C S A = 3.14 \times 5 \times 10$
Multiply the simple numbers first: $C S A = 3.14 \times 50$
Calculate it: $C S A = 157$
The CSA is $157 m^{2}$ .

Example 3 (Sphere Surface Area)

Find the Surface Area of a sphere with a radius of 7 meters. (Use $π = 22 / 7$ ). This is a common type of mensuration ssc gd calculation.

Solution:

We need the Surface Area (SA).
Write down the formula: $S A = 4 π r^{2}$ .
Put the numbers in: $r = 7$ , $π = 22 / 7$ . $S A = 4 \times (22 / 7) \times (7)^{2}$
Simplify the calculation: $S A = 4 \times (22 / 7) \times 49$ $S A = 4 \times 22 \times 7$ (Since $49 / 7 = 7$ )
Calculate it: $S A = 88 \times 7 = 616$
The Surface Area is $616 m^{2}$ .

Example 4 (Sphere Volume)

Calculate the volume of a sphere if the radius is 3 cm. (Use $π = 3.14$ ). This helps practice ssc gd mensuration questions formula.

Solution:

We need the Volume (V).
Write down the formula: $V = (4 / 3) π r^{3}$ .
Put the numbers in: $r = 3$ , $π = 3.14$ . $V = (4 / 3) \times 3.14 \times (3)^{3}$
Calculate $r^{3}$ : $(3)^{3} = 3 \times 3 \times 3 = 27$
Substitute back: $V = (4 / 3) \times 3.14 \times 27$
Simplify the fraction: $V = 4 \times 3.14 \times 9$ (Since $27 / 3 = 9$ )
Multiply the constants: $V = 36 \times 3.14$
Calculate the final volume: $V = 113.04$
The volume is $113.04 {cm}^{3}$ .

Introduction to 3D Shapes
Understanding Volume and Surface Area
Cube
Cuboid
Cylinder
Basic Cone and Sphere

Introduction to 3D Shapes

Welcome to your study notes for ssc gd mensuration! We are going to learn about 3D shapes.

What is a 3D shape?

3D means “Three Dimensions.”
It means the shape has length, width, and height.
These shapes are solid, like a ball or a box.
2D shapes (like squares) are flat, but 3D shapes take up space in the real world.
Learning these concepts is key to solving ssc gd mensuration questions.

Understanding Volume and Surface Area

When we study 3D shapes for mensuration ssc gd, we look for two main things: Volume and Surface Area.

Volume (How much fits inside?)

Volume tells us how much space is inside the shape.
Imagine you have a big water tank. The volume is how much water the tank can hold.
Volume is always measured in cubic units (like $c m^{3}$ or $m^{3}$ ).

Surface Area (How much paint is needed?)

Surface Area tells us the total area of all the outside faces (sides) of the shape.
Imagine you want to paint a box. The surface area is the total amount of paint you need to cover the entire box.
Surface Area is always measured in square units (like $c m^{2}$ or $m^{2}$ ).

Cube

A Cube is a very special 3D shape. Think of a standard dice or a sugar cube.

All the sides (faces) of a cube are squares.
All the edges (length, width, and height) are exactly the same size. We call this size ‘a’.
Understanding the cube is essential for solving basic ssc gd mensuration questions formula.

Formulas for Cube

Measurement	Formula (where ‘a’ is the side length)
Volume (V)	$V = a \times a \times a = a^{3}$
Total Surface Area (TSA)	$T S A = 6 \times a^{2}$ (Because there are 6 square faces)
Lateral Surface Area (LSA)	$L S A = 4 \times a^{2}$ (Area of the 4 side walls, excluding top and bottom)

4 Solved Examples (Cube)

Example 1: Finding Volume

A cube has a side length of 5 cm. What is its Volume? This is a common type of ssc gd mensuration question.

Solution:

First, let’s understand what we need to find. We need the Volume (how much space is inside).
Now, write down the formula for Volume of a Cube: $V = a^{3}$ .
Put the numbers in. The side length ( $a$ ) is 5 cm. $V = 5 \times 5 \times 5$
Calculate it. $V = 25 \times 5 = 125$
So, the answer is $125 {cm}^{3}$ .

Example 2: Finding Total Surface Area

If the side of a cube is 10 meters, what is the Total Surface Area (TSA)?

Solution:

First, we need the Total Surface Area (the area of all 6 sides).
Now, write down the formula for TSA: $T S A = 6 a^{2}$ .
Put the numbers in. The side length ( $a$ ) is 10 m. $T S A = 6 \times (10)^{2}$
Calculate the square first: $T S A = 6 \times (10 \times 10) = 6 \times 100$
Calculate the final area: $T S A = 600$
So, the answer is $600 m^{2}$ .

Example 3: Finding Lateral Surface Area

A room is shaped like a cube with a side of 4 meters. We only want to paint the four walls (LSA). How much area needs painting? This helps practice mensuration ssc gd questions formula.

Solution:

We need the Lateral Surface Area (LSA), which means the 4 walls only.
Write down the formula for LSA: $L S A = 4 a^{2}$ .
Put the numbers in. The side length ( $a$ ) is 4 m. $L S A = 4 \times (4)^{2}$
Calculate the square: $L S A = 4 \times 16$
Calculate the final area: $L S A = 64$
So, the answer is $64 m^{2}$ .

Example 4: Finding Side from Volume

The volume of a cube is $27 {cm}^{3}$ . What is the length of its side? This is a tricky ssc gd mensuration question.

Solution:

We know the Volume ( $V = 27$ ) and the formula $V = a^{3}$ .
We need to find the number ( $a$ ) that, when multiplied by itself three times, gives 27. $a^{3} = 27$
We need to find the cube root of 27. $3 \times 3 \times 3 = 9 \times 3 = 27$
So, the side length ( $a$ ) must be 3.
The answer is 3 cm.

Cuboid

A Cuboid is like a stretched cube. Think of a matchbox, a book, or a standard brick.

A Cuboid has 6 faces, but they are rectangles, not necessarily squares.
It has three different measurements: Length ( $l$ ), Breadth (width, $b$ ), and Height ( $h$ ).
These mensuration ssc gd questions often involve packing or storage.

Formulas for Cuboid

Measurement	Formula
Volume (V)	$V = l \times b \times h$
Total Surface Area (TSA)	$T S A = 2 (l b + b h + h l)$
Lateral Surface Area (LSA)	$L S A = 2 h (l + b)$ (Area of the 4 walls)

4 Solved Examples (Cuboid)

Example 1: Finding Volume

A box has a length of 8 cm, a breadth of 3 cm, and a height of 2 cm. Find the volume of the box.

Solution:

We need the Volume (V), which is how much the box holds.
Write down the formula: $V = l \times b \times h$ .
Put the numbers in: $l = 8$ , $b = 3$ , $h = 2$ . $V = 8 \times 3 \times 2$
Calculate it: $V = 24 \times 2 = 48$
So, the answer is $48 {cm}^{3}$ .

Example 2: Finding Total Surface Area

Calculate the Total Surface Area (TSA) of a cuboid with $l = 5$ m, $b = 4$ m, and $h = 3$ m. This is a standard ssc gd mensuration questions pdf problem.

Solution:

We need the Total Surface Area (TSA) of all 6 faces.
Write down the formula: $T S A = 2 (l b + b h + h l)$ .
Calculate the three parts inside the bracket first:
- $l b = 5 \times 4 = 20$
- $b h = 4 \times 3 = 12$
- $h l = 3 \times 5 = 15$
Add them up: $T S A = 2 (20 + 12 + 15)$ $T S A = 2 (47)$
Calculate the final area: $T S A = 94$
So, the answer is $94 m^{2}$ .

Example 3: Finding Height from Volume

A water tank has a volume of $100 m^{3}$ . Its length is 10 m and its breadth is 5 m. What is the height of the tank?

Solution:

We know the Volume ( $V = 100$ ) and the formula $V = l \times b \times h$ .
Put the known numbers into the formula: $100 = 10 \times 5 \times h$
Multiply the length and breadth: $100 = 50 \times h$
To find $h$ , we divide the volume by the area of the base: $h = 100 / 50$
Calculate it: $h = 2$
So, the height is 2 meters.

Example 4: Finding Lateral Surface Area

A hall has dimensions $l = 12$ m, $b = 8$ m, and $h = 5$ m. Find the cost of painting the four walls (LSA) if the cost is ₹2 per square meter. This is a practical mensuration ssc gd problem.

Solution:

First, find the Lateral Surface Area (LSA) of the four walls. Formula: $L S A = 2 h (l + b)$ .
Put the numbers in: $L S A = 2 \times 5 \times (12 + 8)$
Calculate the bracket first: $L S A = 10 \times (20)$ $L S A = 200 m^{2}$
Now, find the total cost. Cost = Area $\times$ Rate. Cost = $200 \times 2$
Calculate the cost: Cost = ₹400
The total cost of painting is ₹400.

Cylinder

A Cylinder is a shape like a soda can, a pipe, or a drum.

It has two flat circular bases (top and bottom).
It has one curved surface connecting the bases.
We use the radius ( $r$ ) of the circle and the height ( $h$ ) of the cylinder.
Remember that $π$ (pi) is a constant number, usually taken as $22 / 7$ or 3.14. These ssc gd mensuration questions formula rely heavily on $π$ .

Formulas for Cylinder

Measurement	Formula
Volume (V)	$V = π r^{2} h$
Curved Surface Area (CSA)	$C S A = 2 π r h$ (The area of the side label only)
Total Surface Area (TSA)	$T S A = 2 π r h + 2 π r^{2}$ (CSA + Area of top circle + Area of bottom circle)

4 Solved Examples (Cylinder)

Example 1: Finding Volume

A cylindrical pipe has a radius ( $r$ ) of 7 cm and a height ( $h$ ) of 10 cm. Find its volume. (Use $π = 22 / 7$ ).

Solution:

We need the Volume (V).
Write down the formula: $V = π r^{2} h$ .
Put the numbers in: $r = 7$ , $h = 10$ , $π = 22 / 7$ . $V = (22 / 7) \times (7)^{2} \times 10$
Simplify the calculation: $V = (22 / 7) \times (7 \times 7) \times 10$ $V = 22 \times 7 \times 10$ (One 7 cancels out the 7 in the denominator)
Calculate it: $V = 154 \times 10 = 1540$
The volume is $1540 {cm}^{3}$ .

Example 2: Finding Curved Surface Area

A cylindrical pillar has a radius of 1 meter and a height of 5 meters. Find the Curved Surface Area (CSA). (Use $π = 3.14$ ). This is a typical ssc gd mensuration question.

Solution:

We need the Curved Surface Area (CSA), the area of the side.
Write down the formula: $C S A = 2 π r h$ .
Put the numbers in: $r = 1$ , $h = 5$ , $π = 3.14$ . $C S A = 2 \times 3.14 \times 1 \times 5$
Multiply the simple numbers first: $C S A = (2 \times 5) \times 3.14$ $C S A = 10 \times 3.14$
Calculate it: $C S A = 31.4$
The CSA is $31.4 m^{2}$ .

Example 3: Finding Total Surface Area

A closed drum has $r = 7$ cm and $h = 5$ cm. Find the Total Surface Area (TSA). (Use $π = 22 / 7$ ). This is a common mensuration ssc gd questions formula application.

Solution:

We need the Total Surface Area (TSA). Formula: $T S A = 2 π r h + 2 π r^{2}$ .
Calculate the CSA part ( $2 π r h$ ): $C S A = 2 \times (22 / 7) \times 7 \times 5$ $C S A = 2 \times 22 \times 5 = 220 {cm}^{2}$
Calculate the area of the two circles ( $2 π r^{2}$ ): $2 π r^{2} = 2 \times (22 / 7) \times 7 \times 7$ $2 π r^{2} = 2 \times 22 \times 7 = 308 {cm}^{2}$
Add the parts together: $T S A = 220 + 308$ $T S A = 528$
The TSA is $528 {cm}^{2}$ .

Example 4: Finding Height from Volume

The volume of a cylinder is $440 {cm}^{3}$ . If the radius is 7 cm, what is the height? (Use $π = 22 / 7$ ). This is a key ssc gd mensuration questions pdf type.

Solution:

We know $V = 440$ and the formula $V = π r^{2} h$ .
Put the known numbers in: $440 = (22 / 7) \times (7)^{2} \times h$
Simplify the radius part: $440 = (22 / 7) \times 49 \times h$ $440 = 22 \times 7 \times h$ (Since $49 / 7 = 7$ )
Multiply the constants: $440 = 154 \times h$
Solve for $h$ : $h = 440 / 154$
Divide (both numbers are divisible by 22): $h = 20 / 7$
The height is $20 / 7$ cm (or approximately 2.86 cm).

Basic Cone and Sphere

For the ssc gd mensuration syllabus, you need to know the basic formulas for cones and spheres.

Cone

A Cone is a shape like an ice cream cone or a birthday hat. It has a circular base and comes to a point (apex).

$r$ = radius
$h$ = height (straight up)
$l$ = slant height (the slanted side)

Measurement	Formula
Volume (V)	$V = (1 / 3) π r^{2} h$ (It is 1/3 the volume of a cylinder)
Curved Surface Area (CSA)	$C S A = π r l$

Sphere

A Sphere is a perfectly round 3D object, like a football or a tennis ball. It only has a radius ( $r$ ).

Measurement	Formula
Volume (V)	$V = (4 / 3) π r^{3}$
Surface Area (SA)	$S A = 4 π r^{2}$ (Total Surface Area and Curved Surface Area are the same)

4 Solved Examples (Cone and Sphere)

Example 1 (Cone Volume)

A cone has a radius of 3 cm and a height of 7 cm. Find its Volume. (Use $π = 22 / 7$ ).

Solution:

We need the Volume (V).
Write down the formula: $V = (1 / 3) π r^{2} h$ .
Put the numbers in: $r = 3$ , $h = 7$ , $π = 22 / 7$ . $V = (1 / 3) \times (22 / 7) \times (3)^{2} \times 7$
Simplify the calculation: $V = (1 / 3) \times (22 / 7) \times 9 \times 7$
Cancel out the 7s and simplify the 3: $V = (1 / 3) \times 22 \times 9$ $V = 22 \times 3$ (Since $9 / 3 = 3$ )
Calculate it: $V = 66$
The volume is $66 {cm}^{3}$ .

Example 2 (Cone CSA)

A cone has a radius of 5 m and a slant height ( $l$ ) of 10 m. Find its Curved Surface Area (CSA). (Use $π = 3.14$ ). This is a basic ssc gd mensuration question.

Solution:

We need the Curved Surface Area (CSA).
Write down the formula: $C S A = π r l$ .
Put the numbers in: $r = 5$ , $l = 10$ , $π = 3.14$ . $C S A = 3.14 \times 5 \times 10$
Multiply the simple numbers first: $C S A = 3.14 \times 50$
Calculate it: $C S A = 157$
The CSA is $157 m^{2}$ .

Example 3 (Sphere Surface Area)

Find the Surface Area of a sphere with a radius of 7 meters. (Use $π = 22 / 7$ ). This is a common type of mensuration ssc gd calculation.

Solution:

We need the Surface Area (SA).
Write down the formula: $S A = 4 π r^{2}$ .
Put the numbers in: $r = 7$ , $π = 22 / 7$ . $S A = 4 \times (22 / 7) \times (7)^{2}$
Simplify the calculation: $S A = 4 \times (22 / 7) \times 49$ $S A = 4 \times 22 \times 7$ (Since $49 / 7 = 7$ )
Calculate it: $S A = 88 \times 7 = 616$
The Surface Area is $616 m^{2}$ .

Example 4 (Sphere Volume)

Calculate the volume of a sphere if the radius is 3 cm. (Use $π = 3.14$ ). This helps practice ssc gd mensuration questions formula.

Solution:

We need the Volume (V).
Write down the formula: $V = (4 / 3) π r^{3}$ .
Put the numbers in: $r = 3$ , $π = 3.14$ . $V = (4 / 3) \times 3.14 \times (3)^{3}$
Calculate $r^{3}$ : $(3)^{3} = 3 \times 3 \times 3 = 27$
Substitute back: $V = (4 / 3) \times 3.14 \times 27$
Simplify the fraction: $V = 4 \times 3.14 \times 9$ (Since $27 / 3 = 9$ )
Multiply the constants: $V = 36 \times 3.14$
Calculate the final volume: $V = 113.04$
The volume is $113.04 {cm}^{3}$ .

Introduction to 3D Shapes

Understanding Volume and Surface Area

Volume (How much fits inside?)

Surface Area (How much paint is needed?)

Cube

Formulas for Cube

4 Solved Examples (Cube)

Example 1: Finding Volume

Example 2: Finding Total Surface Area

Example 3: Finding Lateral Surface Area

Example 4: Finding Side from Volume

Cuboid

Formulas for Cuboid

4 Solved Examples (Cuboid)

Example 1: Finding Volume

Example 2: Finding Total Surface Area

Example 3: Finding Height from Volume

Example 4: Finding Lateral Surface Area

Cylinder

Formulas for Cylinder

4 Solved Examples (Cylinder)

Example 1: Finding Volume

Example 2: Finding Curved Surface Area

Example 3: Finding Total Surface Area

Example 4: Finding Height from Volume

Basic Cone and Sphere

Cone

Sphere

4 Solved Examples (Cone and Sphere)

Example 1 (Cone Volume)

Example 2 (Cone CSA)

Example 3 (Sphere Surface Area)

Example 4 (Sphere Volume)

Introduction to 3D Shapes

Understanding Volume and Surface Area

Volume (How much fits inside?)

Surface Area (How much paint is needed?)

Cube

Formulas for Cube

4 Solved Examples (Cube)

Example 1: Finding Volume

Example 2: Finding Total Surface Area

Example 3: Finding Lateral Surface Area

Example 4: Finding Side from Volume

Cuboid

Formulas for Cuboid

4 Solved Examples (Cuboid)

Example 1: Finding Volume

Example 2: Finding Total Surface Area

Example 3: Finding Height from Volume

Example 4: Finding Lateral Surface Area

Cylinder

Formulas for Cylinder

4 Solved Examples (Cylinder)

Example 1: Finding Volume

Example 2: Finding Curved Surface Area

Example 3: Finding Total Surface Area

Example 4: Finding Height from Volume

Basic Cone and Sphere

Cone

Sphere

4 Solved Examples (Cone and Sphere)

Example 1 (Cone Volume)

Example 2 (Cone CSA)

Example 3 (Sphere Surface Area)

Example 4 (Sphere Volume)

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