Question

Question 18
Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangle drawn on the other two sides of the triangle.

Answer 1

Let a right triangle BAC in which $\angle A$ is a right angle and AC = y, AB = x.

Three equilateral triangles $\Delta AEC$, $\Delta AFB$ and $\Delta CBD$ are drawn on the three sides of $\Delta ABC$.
Again let the area of a triangle made on AC, AB and BC are $A_1,A_2$ and $A_3$ respectively.
To prove that $A_3=A_1+A_2$
Proof
In $\Delta CAB$, by using Pythagoras theorem,
$B{C}^2=A{C}^2+A{B}^2$
$\Rightarrow B{C}^2=y^2+x^2$
$\Rightarrow BC=\sqrt{y^2+x^2}$
We know that, area of an equilateral triangle = $\frac{\sqrt{3}}{4}(side{)}^2$
$\therefore$ Area of equaliteral $\Delta AEC,A_1=\frac{\sqrt{3}}{4}(AC{)}^2$
$\Rightarrow A_1=\frac{\sqrt{3}}{4}{y}^2$............(i)
And area of equilateral $\Delta AFB,A_2=\frac{\sqrt{3}}{4}(AB{)}^2$
$\Rightarrow A_2=\frac{\sqrt{3}}{4}{x}^2$............(ii)
And area of equilateral $\Delta AFB$, $A_3=\frac{\sqrt{3}}{4}(CB{)}^2$=$\frac{\sqrt{3}}{4}\times (y^2+x^2)$
$=\frac{\sqrt{3}}{4}(y^2+x^2)=\frac{\sqrt{3}}{4}{y}^2+\frac{\sqrt{3}}{4}{x}^2$
$=A_1+A_2$ [from Eq.(i) and (ii)]
Therefore $A_3=A_1+A_2$ ,hence proved.