Surface Area and Volume - Grade 8

1. Volume of Cubes, Prisms, and Pyramids

Volume: The amount of space inside a three-dimensional object, measured in cubic units.

Volume Formulas:

Cube: \( V = s^3 \)

where \( s \) = side length

Rectangular Prism (Cuboid): \( V = l \times w \times h \)

where \( l \) = length, \( w \) = width, \( h \) = height

Prism (General): \( V = Bh \)

where \( B \) = area of the base, \( h \) = height

Triangular Prism: \( V = \frac{1}{2}bhl \)

where \( b \) = base of triangle, \( h \) = height of triangle, \( l \) = length (depth) of prism

Pyramid: \( V = \frac{1}{3}Bh \)

where \( B \) = area of the base, \( h \) = height of pyramid

Examples:

Example 1: Find the volume of a cube with side length 5 cm.

\( V = s^3 = 5^3 = 125 \) cm³

Example 2: Find the volume of a rectangular prism with length 8 m, width 3 m, and height 4 m.

\( V = l \times w \times h = 8 \times 3 \times 4 = 96 \) m³

Example 3: Find the volume of a square pyramid with base side 6 cm and height 9 cm.

Base area: \( B = 6^2 = 36 \) cm²

\( V = \frac{1}{3}Bh = \frac{1}{3} \times 36 \times 9 = 108 \) cm³

2. Surface Area of Cubes, Prisms, and Pyramids

Surface Area: The total area of all faces of a three-dimensional object.

Surface Area Formulas:

Cube: \( SA = 6s^2 \)

6 square faces, each with area \( s^2 \)

Rectangular Prism: \( SA = 2(lw + lh + wh) \)

Sum of areas of all 6 rectangular faces

Prism (General): \( SA = 2B + Ph \)

where \( B \) = area of base, \( P \) = perimeter of base, \( h \) = height

Pyramid: \( SA = B + \frac{1}{2}Pl \)

where \( B \) = area of base, \( P \) = perimeter of base, \( l \) = slant height

Key Terms:

Lateral Surface Area (LSA): Area of the sides only (excludes bases)
Total Surface Area (TSA): Area of all faces including bases
Slant Height (l): Distance from apex to midpoint of base edge (pyramids)

Examples:

Example 1: Find the surface area of a cube with side 4 cm.

\( SA = 6s^2 = 6 \times 4^2 = 6 \times 16 = 96 \) cm²

Example 2: Find the surface area of a rectangular prism with dimensions 5 m × 3 m × 4 m.

\( SA = 2(lw + lh + wh) = 2(5×3 + 5×4 + 3×4) \)

\( = 2(15 + 20 + 12) = 2(47) = 94 \) m²

3. Volume of Cylinders

Volume of Cylinder: \( V = \pi r^2 h \)

where \( r \) = radius of base, \( h \) = height of cylinder

Understanding the Formula:

A cylinder is like a prism with a circular base
Base area = \( \pi r^2 \) (area of circle)
Volume = Base area × height
Use \( \pi = 3.14 \) or \( \pi = \frac{22}{7} \) as specified

Examples:

Example 1: Find the volume of a cylinder with radius 7 cm and height 10 cm. (Use \( \pi = \frac{22}{7} \))

\( V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 10 = \frac{22}{7} \times 49 \times 10 \)

\( V = 22 \times 7 \times 10 = 1540 \) cm³

Example 2: A cylindrical tank has diameter 14 m and height 5 m. Find its volume. (Use \( \pi = 3.14 \))

Radius = \( \frac{14}{2} = 7 \) m

\( V = \pi r^2 h = 3.14 \times 7^2 \times 5 = 3.14 \times 49 \times 5 = 769.3 \) m³

4. Volume of Cones

Volume of Cone: \( V = \frac{1}{3}\pi r^2 h \)

where \( r \) = radius of base, \( h \) = height (perpendicular height from apex to base)

Key Relationship:

Volume of cone = \( \frac{1}{3} \) × Volume of cylinder with same base and height

Examples:

Example 1: Find the volume of a cone with radius 3 cm and height 7 cm. (Use \( \pi = \frac{22}{7} \))

\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 3^2 \times 7 \)

\( V = \frac{1}{3} \times \frac{22}{7} \times 9 \times 7 = \frac{1}{3} \times 22 \times 9 = 66 \) cm³

Example 2: A cone has diameter 10 cm and height 12 cm. Find its volume. (Use \( \pi = 3.14 \))

Radius = \( \frac{10}{2} = 5 \) cm

\( V = \frac{1}{3} \times 3.14 \times 5^2 \times 12 = \frac{1}{3} \times 3.14 \times 25 \times 12 \)

\( V = \frac{942}{3} = 314 \) cm³

5. Surface Area of Cylinders

Formulas:

Curved (Lateral) Surface Area: \( CSA = 2\pi rh \)

Area of the curved side only

Total Surface Area: \( TSA = 2\pi r^2 + 2\pi rh \) or \( TSA = 2\pi r(r + h) \)

Includes two circular bases and the curved surface

Components:

Two circular bases: \( 2 \times \pi r^2 = 2\pi r^2 \)
Curved surface: \( 2\pi rh \) (circumference × height)

Examples:

Example 1: Find the total surface area of a cylinder with radius 7 cm and height 10 cm. (Use \( \pi = \frac{22}{7} \))

\( TSA = 2\pi r(r + h) = 2 \times \frac{22}{7} \times 7 \times (7 + 10) \)

\( = 2 \times 22 \times 17 = 748 \) cm²

Example 2: Find the curved surface area of a cylinder with radius 5 m and height 8 m. (Use \( \pi = 3.14 \))

\( CSA = 2\pi rh = 2 \times 3.14 \times 5 \times 8 = 251.2 \) m²

6. Surface Area of Cones

Formulas:

Curved (Lateral) Surface Area: \( CSA = \pi rl \)

where \( l \) = slant height

Total Surface Area: \( TSA = \pi r^2 + \pi rl \) or \( TSA = \pi r(r + l) \)

Includes circular base and curved surface

Finding Slant Height:

If height \( h \) and radius \( r \) are known: \( l = \sqrt{r^2 + h^2} \)

(Uses Pythagorean theorem)

Examples:

Example 1: Find the total surface area of a cone with radius 7 cm and slant height 10 cm. (Use \( \pi = \frac{22}{7} \))

\( TSA = \pi r(r + l) = \frac{22}{7} \times 7 \times (7 + 10) \)

\( = 22 \times 17 = 374 \) cm²

Example 2: A cone has radius 5 cm and height 12 cm. Find its curved surface area. (Use \( \pi = 3.14 \))

Step 1: Find slant height: \( l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \) cm

Step 2: \( CSA = \pi rl = 3.14 \times 5 \times 13 = 204.1 \) cm²

7. Volume of Spheres

Volume of Sphere: \( V = \frac{4}{3}\pi r^3 \)

where \( r \) = radius of sphere

Key Points:

A sphere is a perfectly round 3D shape
All points on surface are equidistant from center
Radius is the only dimension needed
If diameter is given, find radius first: \( r = \frac{d}{2} \)

Examples:

Example 1: Find the volume of a sphere with radius 3 cm. (Use \( \pi = \frac{22}{7} \))

\( V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 3^3 \)

\( = \frac{4}{3} \times \frac{22}{7} \times 27 = \frac{4 \times 22 \times 27}{21} = \frac{2376}{21} \approx 113.14 \) cm³

Example 2: A spherical ball has diameter 14 cm. Find its volume. (Use \( \pi = \frac{22}{7} \))

Radius = \( \frac{14}{2} = 7 \) cm

\( V = \frac{4}{3} \times \frac{22}{7} \times 7^3 = \frac{4}{3} \times \frac{22}{7} \times 343 \)

\( = \frac{4 \times 22 \times 49}{3} = \frac{4312}{3} \approx 1437.33 \) cm³

8. Surface Area of Spheres

Surface Area of Sphere: \( SA = 4\pi r^2 \)

where \( r \) = radius of sphere

Relationship to Circle:

Surface area of sphere = 4 × Area of a great circle (circle with same radius)

Examples:

Example 1: Find the surface area of a sphere with radius 7 cm. (Use \( \pi = \frac{22}{7} \))

\( SA = 4\pi r^2 = 4 \times \frac{22}{7} \times 7^2 = 4 \times \frac{22}{7} \times 49 \)

\( = 4 \times 22 \times 7 = 616 \) cm²

Example 2: A spherical balloon has diameter 10 cm. Find the area of material needed. (Use \( \pi = 3.14 \))

Radius = \( \frac{10}{2} = 5 \) cm

\( SA = 4\pi r^2 = 4 \times 3.14 \times 5^2 = 4 \times 3.14 \times 25 = 314 \) cm²

9. Volume and Surface Area of Similar Solids

Similar Solids: 3D shapes with the same shape but different sizes. All corresponding linear dimensions are proportional.

Scale Factor Relationships:

If the scale factor (ratio of corresponding lengths) is \( k \), then:

Measure	Ratio	Formula
Linear dimensions	\( k : 1 \)	\( \frac{l_1}{l_2} = k \)
Surface Area	\( k^2 : 1 \)	\( \frac{SA_1}{SA_2} = k^2 \)
Volume	\( k^3 : 1 \)	\( \frac{V_1}{V_2} = k^3 \)

Key Principle:

Linear scale factor = k
Surface area scale factor = k²
Volume scale factor = k³

Examples:

Example 1: Two similar cylinders have heights 4 cm and 12 cm. If the smaller cylinder has surface area 100 cm², find the surface area of the larger cylinder.

Scale factor: \( k = \frac{12}{4} = 3 \)

Surface area ratio: \( k^2 = 3^2 = 9 \)

\( \frac{SA_{\text{large}}}{SA_{\text{small}}} = 9 \) → \( \frac{SA_{\text{large}}}{100} = 9 \)

\( SA_{\text{large}} = 900 \) cm²

Example 2: Two similar spheres have volumes 64 cm³ and 216 cm³. If the smaller sphere has radius 4 cm, find the radius of the larger sphere.

Volume ratio: \( \frac{V_1}{V_2} = \frac{64}{216} = \frac{8}{27} = \frac{2^3}{3^3} \)

Scale factor: \( k = \frac{3}{2} = 1.5 \)

\( \frac{r_{\text{large}}}{r_{\text{small}}} = 1.5 \) → \( \frac{r_{\text{large}}}{4} = 1.5 \)

\( r_{\text{large}} = 6 \) cm

Example 3: Two similar cubes have edge lengths 5 cm and 10 cm. Find the ratio of their volumes.

Scale factor: \( k = \frac{10}{5} = 2 \)

Volume ratio: \( k^3 = 2^3 = 8 \)

Answer: The volumes are in the ratio 8:1 (or larger volume is 8 times the smaller)

Quick Reference: All Formulas

Shape	Volume	Surface Area
Cube	\( V = s^3 \)	\( SA = 6s^2 \)
Rectangular Prism	\( V = lwh \)	\( SA = 2(lw+lh+wh) \)
Prism (General)	\( V = Bh \)	\( SA = 2B + Ph \)
Pyramid	\( V = \frac{1}{3}Bh \)	\( SA = B + \frac{1}{2}Pl \)
Cylinder	\( V = \pi r^2 h \)	\( SA = 2\pi r(r+h) \)
Cone	\( V = \frac{1}{3}\pi r^2 h \)	\( SA = \pi r(r+l) \)
Sphere	\( V = \frac{4}{3}\pi r^3 \)	\( SA = 4\pi r^2 \)