Question

In a pentagon ABCDE with AC = 5 cm, a line through B parallel to AC meets DC produced at F. The altitude of triangle ABC, perpendicular to AC, is 6 cm. Find the area of triangle ACF.

Answer 1

Let BG be perpendicular to AC.
From the given figure:
AC $\parallel$ BF (given)
AC = 5cm (given)
Length of the altitude BG, perpendicular to AC = 6 cm (given)

$△ACB\text{and}△ACF\text{lie on the same}\text{base AC and are between the same}\text{parallels AC and BF.}$

$\therefore \text{Area}△ACB=\text{Area}△ACF$

$\text{Area}△ACB=\frac{1}{2}\times \text{base}\times \text{Height}=\frac{1}{2}\times 5\times 6=15c{m}^2$

$\therefore \text{Area}△ACF=15c{m}^2$