1. Surface Area of Cones

2. ADDITION TO DIAGRAM – NEW VOCAB
The slant height of a right cone is the distance from the vertex to any place on the perimeter of the circular base.
Add the slant height onto the sketch of the cone on the notes sheet.

3. Add the surface area formula onto the notes sheet for cones.

4. Example 1
Find the surface area of a right cone with radius 9 in and slant height 5 in. Leave your answer in terms of  .
S = *9*5 + 92
= 45 + (9)2 = 126 in2

5. Example 2
Find the surface area of a right cone with a diameter of 6 ft and a slant height of 41 ft. Round your answer to the nearest hundredth.
Divide the diameter in half to find the radius (3)
S = *3*41 + 32
= 123 + 9
= 132 ft2
= ft2

6. Example 3 Find the surface area of the right cone. 3m 5m r
Use the Pythagorean Theorem to solve for the radius (4).
S = *5*4 + 42
= 20 + 16
= 36 in2 r

7. Example 4 Work backwards to solve for the missing information.
Find the slant height of a right cone with a radius of 13 yards and surface area of yd2.
= *13^2+ *13*l
873.99=13 *l
21.4 yards = l

8. Example 5 Work backwards to solve for the missing information.
Find the radius of a right cone where the diameter is equal to the slant height and the surface area is 300  cm2.
First, the radius would be half the diameter, which is l/2.
300 = *(l/2)*l+ (l/2)^2
All of the  ‘s will cancel from the problem, since they are in every term.
300=(l^2)/2+ (l^2)/4
300=(3/4)l^2
Multiply both sides by 4/3
400=l^2
So l = 20.
Since the radius is l/2, the radius is 10 cm.