Objective – To find the surface area of pyramids and cones.

10 m10 m

13 m

12 m SA BA LA= +

SurfaceArea = Base

Area + LateralArea

SA l w= • ( )1 4 bh2+10 m ( )2

( )1SA 10 10 4 10 132= • + • •

( )1SA 100 4 1302= + •

SA 100 260= +2SA 360 m=

Find the surface area of the pyramid below.

6 mSquare Base

6 m

5 m

4 m

4 Lateral Triangles+

( )( ) 4 1 b h2 •l w•

6 6•36 ( )14 6 52 • •

60=

2

2 6 5• •

Total Surface Area = 36 + 60 296 m=

Equilateral Triangular Pyramid

8 8

8 10 8 8

8

8

4

x2 2 2a b c+ =2 2 2x 4 8+ =2x 16 64+ =

16 16

x

SA BA LA= +1SA bh2= ( )1 3 bh2+

16 16− −

x 48=x 6.93≈

2x 48=

( )1 1SA (8 6.93) 3 8 102 2≈ • + • •

SA 27.72 120≈ + 2147.72 un≈

Surface Area of a Cone

SA BA LA= +

SurfaceArea = Base

Area + LateralArea

5 m5 m2SA r r= π• + π• • l

3 m

5 m5 m2SA 3 3 5= π• + π• •

SA 9 15= π+ π

SA 24= π 24(3.14)≈2SA 75.36 m≈

Find the surface area of the cone below.

SA BA LA= +

SurfaceArea = Base

Area + LateralArea

2SA r r= π• + π• • l

13 in.12 in.

2SA 5 5 13= π• + π• •

SA 25 65= π+ π

SA 90= π 90(3.14)≈2SA 282.6 in≈

5 in.

Lesson 10-5

Pre-Algebra Slide Show: Teaching Made Easy As Pi, by James Wenk © 2011