Heptagonal pyramid

A hardwood for a column is in the form of a frustum of a regular heptagonal pyramid. The lower base edge is 18 cm, and the upper base of 14 cm. The altitude is 30 cm. Determine the weight in kg if the wood density is 10 grams/cm3.

Correct answer:

m =  17.3972 kg

Step-by-step explanation:

s1=18 cm s2=14 cm h=30 cm  n=7  S1=41 n s1 cotg(π/n)=41 7 18 cotg(3.1416/7)65.4104 cm2 S2=41 n s2 cotg(π/n)=41 7 14 cotg(3.1416/7)50.8748 cm2  V=31 h (S1+S2+S1 S2)=31 30 (65.4104+50.8748+65.4104 50.8748)1739.7177 cm3  ρ=10 g/cm3  m1=ρ V=10 1739.717717397.177 g  m=m1 kg=m1:1000  kg=17397.177:1000  kg=17.397 kg=17.3972 kg

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Showing 1 comment:
Math student
Problem #1] The aluminum frustum of a right circular cone has an altitude of 15 inches and, whose upper and lower bases have a radii 10 inches and 5 inches respectively; find the following;

a) The lateral surface area in ft2 .

b) The total surface area in ft2 .

c) The volume of the frustum of the cone in ft3 .

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