Question

In the given figure $△ABC$ and $△AMP$ are right angled at $B$ and $M$ respectively.
Given $AB=10cm,AP=15cm$ and $PM=12cm$.
(i) Prove $△ABC\sim △AMP$
(ii) Find $AB$ and $BC$.

Answer 1

(i) In $\Delta ABC$ and $\Delta AMP$

$\angle ABC=\angle AMP$ (Both are equal to $90˚$)

$\angle BAC=\angle PAM$ (Common Angles)

$\angle APM=\angle ACB$ (Common Angles)

Therefore

By AAA corollary, $\Delta ABC\sim \Delta AMP$

(ii) Now,

$AP=15cm,PM=12cm$

Using Pythagoras theorem in $\Delta APM$, we get

$AM=\sqrt{{AP}^2-{PM}^2}$

$AM=\sqrt{225-144}=9cm$

$\frac{AB}{BP}=\frac{AM}{MC}$

$\frac{10}{5}=\frac{9}{MC}$

$\Rightarrow MC=4.5cm$

Also,

$\frac{AB}{BC}=\frac{AM}{MP}$

$\frac{10}{BC}=\frac{9}{12}$

$\Rightarrow BC=\frac{120}{9}cm$

$AB=10cm$