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# help

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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?

Dec 16, 2019

#1
+1 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

Dec 16, 2019
edited by hectictar  Dec 16, 2019
#2
+2

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:

$$\begin{array}{|rcll|} \hline A &=& 15x \\ &=& 15*11.2 \\ \mathbf{A} &=& \mathbf{168} \\ \hline \end{array}$$ Dec 16, 2019
#3
+1

THX, hectictar  and heureka    !!!!!   CPhill  Dec 16, 2019