Surface Area of Pyramids (and Cones)

Section 10.5

P. 548 - 552

• In this section, you will practice finding the “surface” area of pyramids. These pyramids will include both triangular and square bases. However, the base will be “regular” polygon – so either squares or equilateral triangles.

• This is not true of all pyramids, however.

Is the height of a lateral face (any face that is NOT the base)

GUIDED PRACTICE for Example 1

Find the surface area. Round to the nearest tenth.

1.

Area of the B = 8 x 8 = 64 ft2

Area of one lateral side is (8 x 12) / 2 = 48

4 Triangles x 48 = 192

Add 192 + 64 = 256 ft 2 is the surface area

Area of the B = 6 x 6 = 36

Area of one lateral side triangle = (6x11) /2 = 33

33 x 4 = 132

Add 132 + 36 = 168 cm2 for the surface area

GUIDED PRACTICE for Example 1

Area of the base + Area of the three lateral sides

B = 62.4

One lateral Triangle (12 x 10) /2 = 60

60 x 3 = 180 (for the 3 Tri. Sides)

SA = 180 + 62.4 = 242.4 in2

Area of B is given = 84.9

One lateral Triangle – (14 x 16) / 2 = 112

Time 3 (for the three triangle) 3 x 112 = 336

Add 336 + 84.9 = 420.9 yd2 for the surface area

Look at the sphere formula for surface area S = 4r2

You will NOT need to memorize this, but you should be able to substitute in the values correctly and solve the problem.

diameter= 9.4

S = 4r2

S = 4 x 3.14 x 4.72

S = 277.5 in2

• Assignment: P 550 #2-4, 17-19

• Show all your steps – Draw the nets only if you want to

• Remember – this is still area and labels will be square units.