2. Vertex Regular Pyramid – Slant Height - Altitude 1) Base is a regular polygon 2) Faces are congruent isosceles triangles 3) Altitude meets the base at its center Not the same as lateral edge

3. Lateral Area – area of one face multiplied by number of sides Lateral Area can also be found by taking half the perimeter of the base times the slant height. L.A. = ½p l T.A. = L.A. + B V = Bh

4. Find the L.A., T.A. and Volume. 6in. 8in. 6in. 8in. 10 L.A. = ½p l L.A. = ½(12 · 4)(10) L.A. = 240 in² T.A. = L.A. + B T.A. = 240 + (12 · 12) T.A. = 384 in² V = Bh= (12 · 12)(8) V = 384 in³

5. Find the L.A.,T.A. and Volume. 6in. 5in. 3 5 L.A. = ½p l L.A. = ½(6 · 4)(5) L.A. = 60 in² T.A. = L.A. + B T.A. = 60 + (6 · 6) T.A. = 96 in² V = Bh= (6 · 6)(4) V = 48 in³ 4

6. Slant Height L.A. = ½·2πr ·l V = πr²h T.A. = πr l + πr² Altitude Cone – Just like a pyramid, but with a circular base

7. 13m 24m Find the L.A., T.A. and Volume. 13m 12m 5 5 L.A. = ½·2πr ·l L.A. = π(12) · 13 L.A. = 156π T.A. = πr l + πr² T.A. = π(12 · 13) + π12² T.A. = 156π + 144π T.A. = 300π

8. 13m 24m 5 V = πr²h V = π(12)²(5) V = 240π

9. Find the L.A., T.A. and Volume. 6 ft 3ft L.A. = ½·2πr ·l 3²+ 6²= x²x²= 45 L.A. = π(3) · L.A. = 9π T.A. = πr l + πr² T.A. = 9π + 9π T.A. = π(3 · ) + π3² V = πr²h V = π(3)²(6) V = 18π

10. A solid metal cylinder with radius 6cm and height 18cm is melted down and recast as a solid cone with radius 9cm. Find the height of the cone. V = πr²h V = π(6)²(18) V = 648π V = πr²h 648π = π(9)²h h = 24cm 648π = 27πh

11. A soup can has a height of 5 inches and a diameter of 6 inches. Determine the area of the label for the can. If 1 oz = 11.5 cubic inches, about how many ounces are in the soup can? L.A. = 2πrh L.A. = 2π(3)·5 L.A. = 30π V = πr²h V = π(3)²·5 V = 45π = 141.3 in³ 12.3 oz.