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]
