1. Surface Area of Cylinders and Pyramids
Sec 10-2C and 2E pg

2. Cylinders Axis – the segment with endpoints that are centers
of the circular bases
If the axis is also an altitude,
then it is a right cylinder
(it stands up straight!)

3. Surface Area Just like with prisms, surface
area combines area of faces
with the area of the bases. 20 7
S.A. = 2rh + 2r2
S.A. = 2rh + 2r2
S.A. = 2(3.14)(7)(20) + 2(3.14)(7)2
S.A. = u u2
S.A. = u2

4. Work This Problem Find the surface area S.A. = 2rh + 2r2
6 ft.
Find the surface area
7.5 ft.
S.A. = 2rh + 2r2
S.A. = 2(3.14)(3 ft.)(7.5 ft.) + 2(3.14)(3 ft.)2
S.A. = ft ft. 2
S.A. = ft. 2

5. Definitions Regular Pyramid – a pyramid where
- the base is a regular polygon, &
- the height from the vertex to the base is perpendicular
Height of the Pyramid (red line) the distance from the vertex to the base measured perpendicularly
Slant Height (blue line) the height from the vertex to the base of the face.

6. Surface Area S.A. = B + ½ Pl where S.A. = B + ½ Pl S.A.= bh +½ P l
½ Pl = Surface area of faces, and
B = area of the base
8 ft
10 ft
S.A. = B + ½ Pl
S.A.= bh +½ P l
S.A. = (8 ft) (8 ft) +½ (32 ft) (10 ft)
S.A. = 64 ft ft 2
S.A. = 224 ft 2