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