10.5 sa of pyramids 1
TRANSCRIPT
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.