Question
$ABC$ is an isosceles triangle, right-angled at $B$. Similar triangle $ACD$ and $ABE$ are constructed on sides $AC$ and $AB$. Find the ratio between the areas of $\triangle\text{ABE}$ and $\triangle\text{ACD}.$
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Answer
ABC is an isosceles triangle right angled at B, Let $AB = BC = x cm$
By Pythagoras theorem, $AC^2 = AB^2 + BC^2 = x^2 + x^2 AC^2 = 2x^2$
$\text{AC}=\sqrt{2}\text{x}$ $\triangle\text{ACD}\approx\triangle\text{ABE}$
(Given) $\therefore\frac{\text{ar}\triangle\text{ABE}}{\text{ar}\triangle\text{ACD}}=\frac{\text{AB}^2}{\text{AC}^2}=\frac{\text{x}^2}{\big(\sqrt{2}\text{x}\big)^2}$
$=\frac{\text{x}^2}{2\text{x}^2}=\frac{1}{2}=1:2$
By Pythagoras theorem, $AC^2 = AB^2 + BC^2 = x^2 + x^2 AC^2 = 2x^2$
$\text{AC}=\sqrt{2}\text{x}$ $\triangle\text{ACD}\approx\triangle\text{ABE}$
(Given) $\therefore\frac{\text{ar}\triangle\text{ABE}}{\text{ar}\triangle\text{ACD}}=\frac{\text{AB}^2}{\text{AC}^2}=\frac{\text{x}^2}{\big(\sqrt{2}\text{x}\big)^2}$
$=\frac{\text{x}^2}{2\text{x}^2}=\frac{1}{2}=1:2$
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