Question

In aright angle triangle ABC, E and D are the mid-points of AB and BC respectively Also $\angle B={90}^{\circ }AD=\sqrt{292}$ cm and $CE=\sqrt{208}$ cm. Find AC (in cm).

Answer 1

In $\Delta ABD,A{D}^2=A{B}^2+B{D}^2$
$\Rightarrow A{D}^2=A{B}^2+{\left(\frac{BC}{2}\right)}^2$ .....(1) ....($\because$ D is the mid-point of BC)

In $\Delta BEC,C{E}^2=B{C}^2+B{E}^2$

$\Rightarrow C{E}^2=B{C}^2+{\left(\frac{AB}{2}\right)}^2$ ...(2) ...($\because$ E is mid-point of AB)

Adding (1) and (2), we get
$\Rightarrow 292+208=\frac{5}{4}(A{B}^2+B{C}^2)$

$\Rightarrow 400=A{B}^2+B{C}^2\Rightarrow 400=A{C}^2$

$\Rightarrow AC=20$ cm