Question

ABC is an isosceles triangle, right-angled at B. similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ∆ABE and ∆ACD.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},\mathrm{ABC}\mathrm{as}\mathrm{an}\mathrm{isosceles}\mathrm{triangle},\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{B}. \\ \mathrm{Now},\mathrm{AB}=\mathrm{BC} \end{gathered}$$

Applying Pythagoras theorem in right-angled triangle ABC, we get:

${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2=2{\mathrm{AB}}^2(\because \mathrm{AB}=\mathrm{AC})$ ...(i)

$\because$ $△\mathrm{ACD}~△\mathrm{ABE}$

$\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{ratio}\mathrm{of}\mathrm{areas}\mathrm{of}2\mathrm{similar}\mathrm{triangles}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{squares}\mathrm{of}\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{their}\mathrm{corresponding}\mathrm{sides}.$

$$\begin{gathered} \therefore \frac{\mathrm{ar}(△\mathrm{ABE})}{\mathrm{ar}(△\mathrm{ACD})}=\frac{{\mathrm{AB}}^2}{{\mathrm{AC}}^2}=\frac{{\mathrm{AB}}^2}{2{\mathrm{AB}}^2}\left[\mathrm{from}\left(\mathrm{i}\right)\right] \\ =\frac{1}{2}=1:2 \end{gathered}$$