Problem on Area of Trapezoid | AMC-10A, 2002 | Problem 25

In trapezoid \(ABCD\) with bases \(AB\) and \(CD\), we have \(AB = 52\), \(BC = 12\), \(CD = 39\), and \(DA = 5\). The area of \(ABCD\) is

Given that \(ABCD\) is a Trapezium with bases \(AB\) and \(CD\), we have \(AB = 52\), \(BC = 12\), \(CD = 39\), and \(DA = 5\).we have to find out the area of the Trapezium.Normally the area of the trapezium is \(\frac{1}{2} (AD +BC) \times \)(height between CD & AB).but we don't know the height.So another way if we extend \(AD\) & \(BC\) ,they will meet a point \(E\).Now clearly Area of \(ABCD\)=Area of \(\triangle ABE\) - Area of \(\triangle EDC\).Can you find out the area of \(\triangle EAB\) & Area of \(\triangle EDC\)?

Can you now finish the problem ..........

Now \(AB||DC\) , Therefore \(\triangle EDC \sim \triangle EAB\)

\(\Rightarrow \frac{ED}{EA}=\frac{EC}{EB}=\frac{DC}{AB}\)

\(\Rightarrow \frac{ED}{ED+DA}=\frac{EC}{EC+BC}=\frac{DC}{AB}\)

\(\Rightarrow \frac{ED}{ED+5}=\frac{EC}{EC+12}=\frac{39}{52}\)

Now , \(\frac{ED}{ED+5}=\frac{39}{52}\)

\(\Rightarrow ED=15\)

And \( \frac{EC}{EC+12}=\frac{39}{52}\)

\(\Rightarrow CE=36\)

Therefore \(BE\)=\(12+36=48\) and \(AE=20\)

Notice that in the \(\triangle EDC\), \({ED}^2 +{EC}^2=(36)^2+(15)^2=(39)^2=(DC)^2\) \(\Rightarrow \triangle EDC\) is a Right-angle Triangle

Therefore Area of \(\triangle EDC=\frac{1}{2} \times 36 \times 15=270\)

Similarly In the \(\triangle EAB\), \({EA}^2 +{EB}^2=(48)^2+(20)^2=(52)^2=(AB)^2\) \(\Rightarrow \triangle EAB\) is a Right-angle Triangle

Therefore Area of \(\triangle EAB=\frac{1}{2} \times 48 \times 20=480\)

Now can you find out the area of \(ABCD\)?

can you finish the problem........

Therefore Area of \(ABCD\)=Area of \(\triangle ABE\) - Area of \(\triangle EDC\)=\(480-270=210\)

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