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

Try this beautiful problem on area of trapezoid from Geometry.

## 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

• $182$
• $195$
• $210$
• $234$
• $260$

Geometry

Trapezoid

Triangle

## Check the Answer

Answer: $210$

AMC-10A (2002) Problem 25

Pre College Mathematics

## Try with Hints

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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