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The trapezoid is divided into 5 parts horizontally with AB and CD are equally divided. The Trapezoid also divided into 5 parts vertically with AD and BC are devided equally. If the shaded regions K has area 35 and the shaded region has area 55. Determine the area of the trapezoid ABCD.

Feb 11, 2020

#1
+25532
+3

The trapezoid is divided into 5 parts horizontally with AB and CD are equally divided.
The Trapezoid also divided into 5 parts vertically with AD and BC are devided equally.
If the shaded regions K has area 35 and the shaded region L has area 55.
Determine the area of the trapezoid ABCD.

$$\begin{array}{|rcll|} \hline 5b+2*\frac{1}{5}x = 5a &\text{ or }& \boxed{b = a-\frac{2}{25}x} \\ 5c+2*\frac{2}{5}x = 5a &\text{ or }& \boxed{c = a-\frac{4}{25}x} \\ 5d+2*\frac{3}{5}x = 5a &\text{ or }& \boxed{d = a-\frac{6}{25}x} \\ 5e+2*\frac{4}{5}x = 5a &\text{ or }& \boxed{e = a-\frac{8}{25}x} \\ 5f+2x = 5a &\text{ or }& \boxed{f = a-\frac{2}{5}x} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \text{trapezoid } ABCD &=& \left(\dfrac{5a+5f}{2}\right)5h \quad | \quad \boxed{f = a-\frac{2}{5}x} \\ &=& \left(\dfrac{5a+5(a-\frac{2}{5}x)}{2}\right)5h \\ &=& \left(\dfrac{10a-2x}{2}\right)5h \\ &=& (5a-x)5h \\ \mathbf{\text{trapezoid } ABCD} &=& \mathbf{25ah-5xh} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{K:} \\ \hline 35 &=& \left(\frac{d+e}{2}\right)h + 2 \left( \frac{e+f}{2}\right)h \\ 35 &=& \left(\frac{d+e}{2}\right)h + ( e+f )h \\ 35 &=& \left(\frac{a-\frac{6}{25}x+a-\frac{8}{25}x}{2}\right)h + ( a-\frac{8}{25}x+a-\frac{2}{5}x )h \\ \ldots \\ \mathbf{35} &=& \mathbf{3ah-xh} \quad (1) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{L:} \\ \hline 55 &=& 2 \left( \frac{a+b}{2}\right)h + \left(\frac{b+c}{2}\right)h \\ 55 &=& ( a+b )h + \left(\frac{b+c}{2}\right)h \\ 55 &=& ( a+a-\frac{2}{25}x )h + \left( \frac{ a-\frac{2}{25}x + a-\frac{4}{25}x }{2} \right)h \\ \ldots \\ \mathbf{55} &=& \mathbf{3ah-\frac{5}{25}xh} \quad (2) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (2)-(1): & 3ah-\frac{5}{25}xh -(3ah-xh) &=& 55-35 \\ & 3ah-\frac{5}{25}xh -3ah+ xh &=& 20 \\ & -\frac{5}{25}xh + xh &=& 20 \\ & \frac{20}{25}xh &=& 20 \\ & \frac{xh}{25} &=& 1 \\ & \mathbf{xh} &=& \mathbf{25} \\\\ (1): & 3ah &=& 35+xh \\ & 3ah &=& 35+25 \\ & 3ah &=& 60 \\ & \mathbf{ah} &=& \mathbf{20} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\text{trapezoid } ABCD} &=& \mathbf{25ah-5xh} \\ &=& 25*20-5*25 \\ &=& 500-125 \\ \mathbf{\text{trapezoid } ABCD} &=& \mathbf{375} \\ \hline \end{array}$$

Feb 13, 2020
edited by heureka  Feb 14, 2020
#2
+111433
+1

Wow!!!!....that's impressive !!!!!!

I'm going to have to study this one   !!!!

CPhill  Feb 13, 2020
#3
+25532
+2

Thank you, CPhill !

heureka  Feb 14, 2020