Find the volume of a frustum of a right pyramid whose lower base is a square with a side 5 in. and whose upper base is a square
1 answer:
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Answer:
a. 85974 m²
b. 17,194,800 AED
c. 18,450 AED
Step-by-step explanation:
The sides of the quadrilateral are given as follows;
AB = 192 m
BC = 576 m
CD = 228 m
DA = 480 m
Length of a diagonal AC = 672 m
a. We note that the area of the quadrilateral consists of the area of the two triangles (ΔABC and ΔACD) formed on opposite sides of the diagonal
The semi-perimeter, s₁, of ΔABC is found as follows;
s₁ = (AB + BC + AC)/2 = (192 + 576 + 672)/2 = 1440/2 = 720
The area, A₁, of ΔABC is given as follows;
= 6912·√(55) m²
Similarly, area, A₂, of ΔACD is given as follows;
The semi-perimeter, s₂, of ΔABC is found as follows;
s₂ = (AC + CD + D)/2 = (672 + 228 + 480)/2 = 690 m
We therefore have;
Therefore, the area of the quadrilateral ABCD = A₁ + A₂ = 6912×√(55) + 1260·√(759) = 85973.71 m² ≈ 85974 m² to the nearest meter square
b. Whereby the cost of 1 meter square land = 200 AED, we have;
Total cost of the land = 200 × 85974 = 17,194,800 AED
c. Whereby the cost of fencing 1 m = 12.50 AED, we have;
Total perimeter of the land = 576 + 192 + 480 + 228 = 1,476 m
The total cost of the fencing the land = 12.5 × 1476 = 18,450 AED