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# Area of the Trapezoid | AMC 8, 2002 | Problem 20

Try this beautiful problem from AMC-8, 2002, (Problem-20) based on area of Trapezoid.You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on Area of Trapezoid.

## Area of the Trapezoid – AMC- 8, 2002 – Problem 20

The area of triangle XYZ is 8  square inches. Points  A and B  are midpoints of congruent segments  XY and XZ . Altitude XC bisects YZ.What is the area (in square inches) of the shaded region?

• $6$
• $4$
• $3$

### Key Concepts

Geometry

Triangle

Trapezoid

Answer:$3$

AMC-8 (2002) Problem 20

Pre College Mathematics

## Try with Hints

Given that Points  A and B  are midpoints of congruent segments  XY and XZ and Altitude XC bisects YZ

Let us assume that the length of YZ=$x$ and length of $XC$= $y$

Can you now finish the problem ……….

Therefore area of the trapezoid= $\frac{1}{2} \times (YC+AO) \times OC$

can you finish the problem……..

Let us assume that the length of YZ=$x$ and length of $XC$= $y$

Given that area of $\triangle xyz$=8

Therefore $\frac{1}{2} \times YZ \times XC$=8

$\Rightarrow \frac{1}{2} \times x \times y$ =8

$\Rightarrow xy=16$

Given that Points  A and B  are midpoints of congruent segments  XY and XZ and Altitude XC bisects YZ

Then by the mid point theorm we can say that $AO=\frac{1}{2} YC =\frac{1}{4} YZ =\frac{x}{4}$ and $OC=\frac{1}{2} XC=\frac{y}{2}$

Therefore area of the trapezoid shaded area = $\frac{1}{2} \times (YC+AO) \times OC$= $\frac{1}{2} \times ( \frac{x}{2} + \frac{x}{4} ) \times \frac{y}{2}$ =$\frac{3xy}{16}=3$ (as $xy$=16)

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