Problema Solution
the edges of the bases of a frustum of a regular square pyramid are 6 cm and 12 cm, respectively and the altitude is 7 cm. Determine the slant height, lateral area and volume of the frustum.
Answer provided by our tutors
m1 = 12/2
m1 = 6 cm
m2 = 6/2
m2 = 3 cm
h = 7 cm
B1 = 12^2 the area of the bigger base
B1 = 144 cm^2
B2 = 6^2 the area of the smaller base
B2 = 36 cm^2
a = the slant height
A = lateral area
V = the volume of the flustrum
Using the Pythagorean Theorem we find the slant height:
a^2 = h^2 + (m1 - m2)^2
a^2 = 7^2 + (6 - 3)^2
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click here to see the equation solved for a
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a = 7.6 cm
The lateral area is:
A = B1 + B2 + 4*(1/2)(m1 + m2)*a
A = 144 + 36 + 4*(1/2)(6 + 3)*6.7
A = 300.6 cm^2
The volume is:
V = (h/3)(B1 + B2 + √(B1B2))
V = (7/3)(144 + 36 + √(144*36))
V = 588 cm^3