DEFINITION

Three-dimensional (3D) figures are solid shapes that have length, breadth (width), and height (or depth). They are different from plane (2D) shapes because they:

1. Occupy space.


2. Have volume (space inside).


3. Have surface area (area of all surfaces).


4. Have faces, edges, and vertices.

1. Cube

1. All edges are equal.


2. 6 faces (all squares).


3. 12 edges.


4. 8 vertices.

+------+
|      |\
|      | +
+------+ |
 \      \|
  +------+

Surface Area (SA) = 6s²
Volume (V) = s³

2. Cuboid

1. Opposite faces are equal rectangles.
2. 6 faces.
3. 12 edges.
4. 8 vertices.

+---------+
|         |\
|         | +
+---------+ |
 \         \|
  +---------+

SA = 2(lb + bh + lh)

V = l × b × h

3. Pyramid

A pyramid has a polygonal base and triangular faces meeting at a point (apex).

Examples: triangular pyramid, square pyramid.

    /\
   /  \
  /____\
  |    |
  |____|

Volume = ⅓ × Area of base × height

4. Cone

1. One circular base.


2. Curved surface.


3. One vertex (apex) and one edge.

   /\
  /  \
 /    \
/______\

SA = πr² + πrl (where l = slant height)
V = ⅓πr²h

5. Cylinder

Two equal circular bases joined by one curved surface.

   ****
  *    *
  *    *   (curved surface)
   ****
   ****
  *    *
  *    *
   ****

SA = 2πr² + 2πrh

V = πr²h

6. Sphere

1. Perfectly round solid.


2. Only curved surface.


3. No edges, no vertices.

   ****
 *      *
*        *
 *      *
   ****

SA = 4πr²

V = 4/3 πr³

Cube

Example 1:

Find the volume of a cube with side 10 cm.

V = s³ = 10 × 10 × 10 = 1000 cm³.

Answer: 1000 cm³

Example 2:
Find the surface area of a cube with side 7 cm.

SA = 6s² = 6 × 7 × 7 = 294 cm².

Answer: 294 cm²

Example 3:

A cube has volume 512 cm³. Find its side.

s³ = 512 → s = ∛512 = 8 cm.

Answer: 8 cm

Cuboid

Example 4:

Find the volume of a cuboid with length 12 cm, breadth 8 cm, height 5 cm.

V = l × b × h = 12 × 8 × 5 = 480 cm³.

Answer: 480 cm³

Example 5:

Find the surface area of a cuboid of length 10 cm, breadth 6 cm, height 4 cm.

SA = 2(lb + bh + lh) = 2(10×6 + 6×4 + 10×4) = 248 cm².

Answer: 248 cm²

Example 6:

A cuboid tank measures 5 m × 4 m × 3 m. Find its volume in litres (1 m³ = 1000 L).

V = 5 × 4 × 3 = 60 m³ = 60,000 L.

Answer: 60,000 L

Pyramid

Example 7:

Find the volume of a square pyramid with base side 10 cm and height 12 cm.

Base area = 10 × 10 = 100 cm².

V = ⅓ × 100 × 12 = 400 cm³.

Answer: 400 cm³

Example 8:

Find the volume of a triangular pyramid (tetrahedron) with base area 24 cm² and height 9 cm.

V = ⅓ × 24 × 9 = 72 cm³.

Answer: 72 cm³

Example 9:

A square pyramid has base 6 cm and height 9 cm. Find its volume.

Base area = 36 cm².

V = ⅓ × 36 × 9 = 108 cm³.

Answer: 108 cm³

Cone

Example 10:

Find the volume of a cone with radius 7 cm and height 12 cm (π = 22/7).

V = ⅓πr²h = ⅓ × 22/7 × 7 × 7 × 12 = 616 cm³.

Answer: 616 cm³

Example 11:

Find the curved surface area (CSA) of a cone with radius 5 cm, slant height 13 cm.

CSA = πrl = 22/7 × 5 × 13 = 286 cm².

Answer: 286 cm²

Example 12:

Find the total surface area of a cone with radius 7 cm, slant height 25 cm (π = 22/7).

TSA = πr² + πrl = 154 + 550 = 704 cm².

Answer: 704 cm²

Cylinder

Example 13:

Find the volume of a cylinder with radius 7 cm and height 10 cm (π = 22/7).

V = πr²h = 22/7 × 7 × 7 × 10 = 1540 cm³.

Answer: 1540 cm³

Example 14:

Find the curved surface area of a cylinder of radius 14 cm, height 20 cm.

CSA = 2πrh = 2 × 22/7 × 14 × 20 = 1760 cm².

Answer: 1760 cm²

Example 15:

Find the total surface area of a cylinder with radius 7 cm, height 24 cm.

TSA = 2πr² + 2πrh = 308 + 1056 = 1364 cm².

Answer: 1364 cm²

Sphere

Example 16:

Find the surface area of a sphere of radius 14 cm.

SA = 4πr² = 2464 cm².

Answer: 2464 cm²

Example 17:

Find the volume of a sphere of radius 7 cm.

V = 4/3πr³ = 1437⅓ cm³.

Answer: 1437⅓ cm³

Example 18:

A spherical ball has diameter 28 cm. Find its surface area.

r = 14 cm.

SA = 4πr² = 2464 cm².

Answer: 2464 cm²

SUMMARY

1. Cube: SA = 6s², V = s³


2. Cuboid: SA = 2(lb + bh + lh), V = l × b × h


3. Pyramid: V = ⅓ × Area of base × height


4. Cone: SA = πr² + πrl, V = ⅓πr²h


5. Cylinder: SA = 2πr² + 2πrh, V = πr²h


6. Sphere: SA = 4πr², V = 4/3πr³