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Let    AB  =  w

and   BE  =  x

and   EC  =  y

The area of rectangle ABCD   =   ( length )( width )   =   15w

By the Pythagorean theorem, we can make these two equations:

w² + y²  =  14²

w² + x²  =  13²

Since ABCD is a rectangle, BC must be the same length as AD, and so we can make this third equation:

x + y  =  15

Now we can begin solving this system of three equations. Here's one way to do that:

x + y  =  15

y  =  15 - x

w² + x²  =  13²

w²  =  13² - x²

w²  =  169 - x²

w² + y²  =  14²

(169 - x²) + (15 - x)²  =  14²

(169 - x²) + (15 - x)²  =  196

169 - x² + (15 - x)(15 - x)  =  196

169 - x² + 225 - 30x + x²  =  196

-x² + x² - 30x  =  196 - 169 - 225

-30x  =  -198

x  =  6.6

w²  =  169 - x²

w²  =  169 - (6.6)²

w²  =  125.44

w  =  11.2

We don't care what  y  is, we just needed to know what  w  is.

Area of rectangle ABCD   =   15w   =   15(11.2)   =   168   sq units

In rectangle ABCD, AD=15 and point E is on side BC such that EA = 13 and ED = 14.

What is the area of rectangle ABCD?

\(\begin{array}{|rcll|} \hline x^2+y^2 &=& 13^2 \\ \mathbf{x^2} &=& \mathbf{13^2 - y^2} \\ \hline x^2 +(15-y)^2 &=& 14^2 \quad | \quad \mathbf{x^2=13^2 - y^2}\\ 13^2 -y^2 + 15^2-30y+y^2 &=& 14^2 \\ 13^2 + 15^2-30y &=& 14^2 \\ 30y &=& 13^2+15^2 - 14^2 \\ y &=& \dfrac{13^2+15^2 - 14^2}{30} \\ \mathbf{y} &=& \mathbf{6.6} \\ \hline \mathbf{x^2} &=& \mathbf{13^2 - y^2} \\ x^2 &=& 13^2 - 6.6^2 \\ x^2 &=& 125.44 \\ \mathbf{x} &=& \mathbf{11.2} \\ \hline \end{array} \)

Area of rectangle ABCD:

\(\begin{array}{|rcll|} \hline A &=& 15x \\ &=& 15*11.2 \\ \mathbf{A} &=& \mathbf{168} \\ \hline \end{array}\)