Lesson 7: Surface Area and Volume. Pyramid and Cone
Lesson 7: Surface Area and Volume. Pyramid and Cone
Calculate surface area and volume of pyramids and cones.
In this lesson you will learn how to calculate surface area and volume of the given 3D shapes. Volume is how much space the shape has inside (its capacity). Surface Area is the total outside surface of the shape.
In order to calculate the volume of a square-based pyramid, we multiply the area of the base by the height of the pyramid and divide by 3: V = ⅓ × (area of base) × height. A cone works the same way: V = ⅓πr2h. Sometimes, the height is unknown and needs to be determined using the slant height of the lateral face and Pythagorean Theorem.
The process for finding the volume of a cone is similar.
In order to calculate the surface area of any 3D shape, we need to determine the area of all 2D shapes that make up the 3D figure and add them together. Thus, the surface area of a square-based pyramid would be the sum of the areas of 4 triangular faces and the area of a square base.
Depending on the situation, we may need to exclude some of the faces from our calculations when determining surface area and volume (for example, the floor inside the house when determining the area of the walls to be painted, etc.)
Angle Relationships Short Animation
Key ideas: Surface Area and Volume. Pyramid and Cone
Volume is the space inside a 3D shape (measured in cubic units, such as cm3). Surface area is the total area of all its outside faces (square units, such as cm2).
A pyramid or cone holds exactly one third of the prism or cylinder with the same base and height.
- Square-based pyramid (base side b, height h, slant height s): V = ⅓b2h and SA = b2 + 2bs. The base is b2, and the 4 triangular faces are 4 × ½bs = 2bs.
- Cone (radius r, height h, slant height s): V = ⅓πr2h and SA = πr2 + πrs. The base is πr2 and the curved side is πrs.
- The height h is measured straight up from the centre of the base. The slant height s runs along the face. They are connected by the Pythagorean theorem: for a pyramid, s2 = h2 + (b/2)2; for a cone, s2 = h2 + r2.
Worked examples
Example 1
A square-based pyramid has base side 10 cm and height 12 cm. Find its volume and surface area.
- Slant height: s2 = 122 + 52 = 169, so s = 13 cm.
- V = ⅓(102)(12) = 400 cm3.
- SA = 102 + 2(10)(13) = 100 + 260 = 360 cm2.
Example 2
A cone has radius 5 cm and height 12 cm. Find its volume and surface area.
- Slant height: s2 = 122 + 52 = 169, so s = 13 cm.
- V = ⅓π(52)(12) = 100π ≈ 314.2 cm3.
- SA = π(52) + π(5)(13) = 25π + 65π = 90π ≈ 282.7 cm2.
Example 3
A waffle cone (no lid) has radius 3 cm and slant height 10 cm. How much waffle is needed to make it?
- There is no base, so only the curved surface counts: πrs.
- π(3)(10) = 30π ≈ 94.2 cm2. Leaving out a face that is not there is common in real problems.
Common mistakes
- Forgetting the ⅓ in the volume formula.
- Using the slant height in the volume formula; volume uses the perpendicular height h.
- Using the diameter instead of the radius.
- Writing volume in cm2 or surface area in cm3.
Questions students ask
Why is there a ⅓ in the formulas?
Three pyramids with the same base and height as a prism fill it exactly, so each holds one third of the prism’s volume. The same is true for a cone and a cylinder.
When do I need the Pythagorean theorem?
When you know the height but need the slant height for surface area, or the reverse.
Check your understanding
Try the quiz for this lesson. You’ll see right away if each answer is correct.