1. Surface Area of Pyramids and Cones

2. Vocabulary Regular 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.

3. 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.

5. Additional Example 1A: Finding the Surface Area of a Pyramid Find the surface area of the pyramid. S = 81 + 180 S = 261 m 2 S = B + Pl 1212 S = (9 9) + (36)(10) ‏ 1212 Use the formula. S = lw + Pl 1212 B = lw Substitute. P = 4(9) = 36 Add. The surface area is 261 square meters.

6. 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 + Pl 1212 Use the formula. B= ½bh. S = bh + Pl 1212 1212 S = (12)(10.38) + (36)(6) ‏ 1212 1212 The surface area is 170.28 m 2.

7. Check It Out: Example 1A Find the surface area of each pyramid.

8. Check It Out: Example 1B Find the surface area of the pyramid

9. 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.

11. 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 = r 2 + rl S ≈ (3.14)(3 2 ) + (3.14)(3)(10) ‏ S ≈ 28.26 + 94.2 The surface area is about 122.46 square centimeters. Substitute. Multiply. Add.

12. Check It Out: Example 2A Find the surface area of the cone. Use 3.14 for .

13. 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 .