Question

In △ ABC right-angled at B, a perpendicular is drawn from B which meets AC at M. If the ratio of areas of △ ABC and △ AMB is 9:4, find $\frac{AC}{AB}$ .

• A. $\frac{9}{4}$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{4}{9}$

Answer 1

The correct option is B $\frac{3}{2}$

‘If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other’.

By the above theorem, △ ABC is similar to △ AMB

Its given that $\frac{ar\left(ABC\right)}{ar\left(AMB\right)}$ =$\frac{9}{4}$

So, by using areas of similar triangles relation we can write,

$\frac{A{C}^2}{A{B}^2}$=$\frac{9}{4}$

$\frac{AC}{AB}$=$\frac{3}{2}$

Option b is correct.