Determine the volume (in ${f t}^{3}$ ) of a right truncated triangular prism with the following definitions:
Let the corners of the triangular base be defined by $A, B$ and $C$ .
The length of $A B = 10 f t ., B C = 9 f t .$ and $C A = 12 f t$ .
The sides $A, B$ and $C$ are perpendicular to the triangular base and have the height of $8.6 f t ., 7.1 f t .$ and $5.5 f t .$ respectively.

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Solution

The volume truncated right triangle prism = $\frac{A (l_{1} + l_{2} + l_{3})}{3}$
A = area of right section
$l_{1}, l_{2}, l_{3}$ = length of lateral edges
A = $\sqrt{s (s - a) (s - b) (s - c)}, s = \frac{a + b + c}{2} = \frac{10 + 9 + 12}{2} = 15.5 f t$
= $\sqrt{15.5 \times 5.5 \times 6.5 \times 3.5} = 44.04 {f t}^{2}$
Volume = $\frac{44.04 (8.6 + 7.1 + 5.5)}{3}$
= $\frac{44.04 \times 21.2}{3}$
$\approx$ $311 {f t}^{3}$

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The volume truncated right triangle prism = A(l1+l2+l3)3A = area of right sectionl1,l2,l3 = length of lateral edgesA = √s(s−a)(s−b)(s−c),s=a+b+c2=10+9+122=15.5ft = √15.5×5.5×6.5×3.5=44.04ft2Volume = 44.04(8.6+7.1+5.5)3 = 44.04×21.23 ≈ 311ft3