Chapter 11.3, Problem 33WE

To determine

Show that the area of square ABCD equals the area of rectangle EFGD.

Explanation of Solution

Given:

  

Calculation:

We have GD = GC in the right triangle DGC .

Also , C is the mid-point of GF.

So, by Pythagoras Theorem in right triangle DGC ,

  $\begin{array}{l}C{D}^2=G{D}^2+G{C}^2\\ \Rightarrow C{D}^2=2G{D}^2\mathrm{..............}[GD=GC]\\ \Rightarrow C{D}^2=2GD\times GD\\ \Rightarrow C{D}^2=2GD\times GC\mathrm{........}[GD=GC]\\ \Rightarrow C{D}^2=2GD\times \frac{GF}{2}\\ \Rightarrow C{D}^2=GD\times GF\end{array}$

Hence,

Area of square ABCD = Area of rectangle EFGD

Hence proved.