Data:
H (height) and B (base)
Area = 80 in²
\(\frac{B}{H} = \frac{5}{2}\)
Knowing that the formula of the area of a triangle:
\(A = \frac{B*H}{2}\)
Solving:
We isolate one of the terms (B or H) to find its values
Product of extremes equals product of means:
\(\frac{B}{H} = \frac{5}{2}\to 2*B = 5*H\to 2B = 5H\to B = \frac{5H}{2}\)
Now substitute in the formula the data found:
\(A = \frac{B*H}{2}\)
\(80 = \frac{ \frac{5H}{2} *H}{2}\)
\(80*2 = \frac{5H^2}{2}\)
\(160 = \frac{5H^2}{2}\)
\(160*2 = 5H^2\)
\(320 = 5H^2\)
\(5H^2 = 320\)
\(H^2 = \frac{320}{5}\)
\(H^2 = 64\)
\(H = \sqrt{64}\)
\(\boxed{H = 8\:in}\longrightarrow\:\textbf{height} \end{array}}\qquad\quad\checkmark\)
Now find the value of base (B), if:
\(B = \frac{5H}{2}\)
Soon:
\(B = \frac{5*8}{2}\)
\(B = \frac{40}{2}\)
\(\boxed{B = 20\:in}\longrightarrow\:\textbf{base} \end{array}}\qquad\quad\checkmark\)
