Question

In the given figure, $\Delta$ABC and $\Delta$AMP are right angled at B and M respectively.
Given AC$=10$cm, AP$=15$cm and PM$=12$cm. Find : AB and BC.

Answer 1

In $△ABC,△AMP$

$\angle AMP=\angle ABC={90}^{\circ }$

$\angle MAP=\angle BAC=\angle A$

As two angles are equal automatically third angle will be equal since sum of angles in a triangle is constant

By $AAA$ similarity

$△ABC\sim △AMP$

Given $AC=10cm,AP=15cm,PM=12cm$

As $\frac{AP}{PM}=\frac{AC}{CB}⟹\frac{10}{BC}=\frac{15}{12}$

$⟹BC=8cm$

In $△ABC,\angle B={90}^{\circ }$

$⟹A{C}^2=A{B}^2+B{C}^2⟹BC=\sqrt{A{C}^2-B{C}^2}=\sqrt{{10}^2-8^2}=6cm$