vertex regular pyramid – slant height - altitude 1) base is a regular polygon 2) faces are...
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Vertex
Regular Pyramid –
Slant Height -Altitude
1) Base is a regular polygon2) Faces are congruent isosceles triangles3) Altitude meets the base at its center
Not the same as lateral edge
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. = ½pl T.A. = L.A. + B
31
Find the L.A., T.A. and Volume.
6in.
8in.6in.
8in.
10
10
L.A. = ½plL.A. = ½(12·4)(10)L.A. = 240 in²
T.A. = L.A. + BT.A. = 240 +
(12·12)T.A. = 384 in²31
31
V = 384 in³
Find the L.A.,T.A.
and Volume.
6in.
5in.
3
5
L.A. = ½plL.A. = ½(6·4)(5)
L.A. = 60 in²
T.A. = L.A. + BT.A. = 60 + (6·6)
T.A. = 96 in²
31
31
V = 48 in³
4
Slant Height
L.A. = ½·2πr·l
31
T.A. = πrl + πr²
Altitude
Cone – Just like a pyramid, but with a circular base
13m
24m
Find the L.A., T.A. and Volume.
13m
12m
55
L.A. = ½·2πr·l
L.A. = π(12)·13L.A. = 156π
T.A. = πrl + πr²T.A. = π(12·13) + π12²T.A. = 156π + 144π
T.A. = 300π
13m
24m
5
31
31
V = 240π
Find the L.A., T.A. and Volume.
6 ft
3ft
L.A. = ½·2πr·l
3²+ 6²= x² x²= 45
53
L.A. = π(3)· 53
L.A. = 9π
5
T.A. = πrl + πr²
5
T.A. = π(3· ) + π3²
53
31
31
V = 18π
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
31
648π = π(9)²h31
h = 24cm
648π = 27πh
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 = πr²hV = π(3)²·5
V = 45π =
141.3 in³
11.5141.3
12.3 oz.