Question

In the given figure, $\Delta ABC$ and $\Delta AMP$ are right-angled at B and M respectively. [3 MARKS]
Given AC = 10 cm, AP = 15 cm and PM = 12 cm
i) Prove $\Delta ABC\sim \Delta AMP$
ii) Find AB and BC

Answer 1

Part (i) : 1.5 Marks
Part (ii): 1.5 Marks

i) In triangles AMP and ABC, we have:
$\angle AMP=\angle ABC={90}^{\circ }$ [Given]
$\angle MAP=\angle CAB$ [Common]
\therefore \Delta AMP \sim \Delta ABC\) [By AA similarity]

ii) Now, triangle AMP $\sim$ triangle ABC
$\therefore \frac{AP}{AC}=\frac{MP}{BC}=\frac{AM}{AB}$
$\Rightarrow \frac{15}{10}=\frac{12}{BC}\Rightarrow BC=8cm$
$\therefore AB=\sqrt{{10}^2-8^2}=\sqrt{100-64}=\sqrt{36}=6cm$