### How to find the area of a triangle in the following diagram?

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\bigtriangleup ABC$△`A``B``C`?

### 1 Answer

From the information given, $\bigtriangleup ABC\sim\bigtriangleup BED$△`A``B``C`∼△`B``E``D` . The hypotenuse of right triangle BED is $\sqrt{1^2+2^2}=\sqrt{5}$√1^{2}+2^{2}=√5

$\frac{AC}{BC}=\frac{BD}{ED}=\frac{2}{1}$`A``C``B``C` =`B``D``E``D` =21 $AC=2BC$`A``C`=2`B``C`

$\frac{AC}{AB}=\frac{BD}{BE}$`A``C``A``B` =`B``D``B``E` => $\frac{AC}{1}=\frac{2}{\sqrt{5}}$`A``C`1 =2√5 => $AC=\frac{2}{\sqrt{5}}$`A``C`=2√5

Since AC=2BC, BC= $\frac{1}{\sqrt{5}}$1√5 , triangle ABC is right triangle os the area is just

$\frac{1}{2}\times AC\times BC=\frac{1}{2}\times\frac{2}{\sqrt{5}}\times\frac{1}{\sqrt{5}}=\frac{1}{5}$12 ×`A``C`×`B``C`=12 ×2√5 ×1√5 =15