Surface Area of Pyramids and Cones

*Notes 46

VocabularyRegular pyramid- a pyramid whose

base is a regular polygon, and the faces are congruent isosceles triangles.

Slant height of a pyramid- the distance from the vertex to the midpoint of an edge of the base.

Slant height of a cone- the distance from the vertex to a point on the edge of the base.

The diagram shows a square pyramid. The blue dashed line labeled l is the slant height of the pyramid, the distance from the vertex to the midpoint of an edge of the base.

The base of a regular pyramid is a regular polygon, and the faces are congruent isosceles triangles.

Additional Example 1A: Finding the Surface Area of a Pyramid

Find the surface area of the pyramid.

S = 81 + 180

S = 261 m2

S = B + Pl12

S = (9 • 9) + (36)(10)12

Use the formula.

S = lw + Pl12

B = lw

Substitute. P = 4(9) = 36

Add.

The surface area is 261 square meters.

Additional Example 1B: Finding the Surface Area of a Pyramid

Find the surface area of the pyramid.

S = 62.28 + 108

S = 170.28

S = B + Pl12 Use the formula.

B= ½bh.S = bh + Pl12

12

S = (12)(10.38) + (36)(6)12

12

The surface area is 170.28 m2.

Check It Out: Example 1A

Find the surface area of each pyramid.

Check It Out: Example 1B

Find the surface area of the pyramid

The diagram shows a cone and its net. The blue dashed line is the slant height of the cone, the distance from the vertex to a point on the edge of the base.

Additional Example 2: Finding the Surface Area of a Cone

Find the surface area of the cone. Use 3.14 for .

Use the formula.

S ≈ 122.46

S = r2 + rl

S ≈ (3.14)(32) + (3.14)(3)(10)

S ≈ 28.26 + 94.2

The surface area is about 122.46 square centimeters.

Substitute.

Multiply.

Add.

Check It Out: Example 2A

Find the surface area of the cone. Use 3.14 for .

Check It Out: Example 2B

The dimensions of the cone from Exercise 2a quadrupled. Find the surface area of the cone. Use 3.14 for .