HI

aqeelnaqvi10,

I think you can use 45-45-90 triangle here. In fact this formula - \(Area = \frac{(diagonal)^2}{2}\) is got after using the phythagoras theorem.

Consider a square of side 'a'. The area of the square = \(a^2\).

In a 45-45-90 triangle, formed by the diagonal and the sides of the square we get -

\(a^2+a^2 = (diagonal)^2\)

\(2a^2 = (diagonal)^2\)

\(a^2 = \frac{(diagonal)^2}{2}\)

Area of square \(= a^2 = \frac{(diagonal)^2}{2}\)

Bunuel has done this directly. I think this formula is a definite time saver!

aqeelnaqvi10 wrote:

I would like to know if we can use the 45-45-90 triangle properties here

_________________

My Best is yet to come!