Question

Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.
OR
if the area of two similar triangles are equal, prove that they are congruent.

Answer 1

Let ABCD ba a square with side 'a'.

In $\Delta$ ABC,
$A{C}^2=A{B}^2+B{C}^2$
$=a^2+a^2$
$=2a^2$
AC = $\sqrt{2a^2}=\sqrt{2}a.$
Area of equilateral $\Delta$ BEC (formed on side BC of square ABCD)
$=\frac{\sqrt{3}}{4}\times (side{)}^2$
$=\frac{\sqrt{3}}{4}{a}^2$ .........(i)
Area of equilateral $\Delta$ ACF (formed on diagonal AC of square ABCD)
$=\frac{\sqrt{4}}{4}(\sqrt{2}a{)}^2$
$=\frac{\sqrt{3}}{4}\left(2a^2\right)$
$=2\frac{\sqrt{3}}{4}{a}^2$ .........(ii)

From eq. (i) and (ii),
ar $\Delta ACF=2\times ar\Delta BCE$
or
ar $\left(\Delta BCE\right)=\frac{1}{2}ar\left(\Delta ACF\right)$
i.e., area of triangle described on one side of square is half the area of triangle described on its diagonal.
OR

Given, $\Delta ABC\sim \Delta PQR$
And ar $\left(\Delta ABC\right)=ar\left(\Delta PQR\right)$
To prove :
$\Delta ABC\cong \Delta PQR$
$\therefore \frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}$
(Ratio of area of similar triangles is equal to the square of correspoding sides)
But $\frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=1$ (Given)
$\therefore \frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}=1$
So, $A{B}^2=P{Q}^{2}orAB=PQ$
$B{C}^2=Q{R}^{2}orBC=QR$
$A{C}^2=Q{R}^{2}orAC=PR$
By SSS congruency axiom
$\Delta ABC\cong \Delta PQR$ Hence Proved.