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AMC 8 Geometry Math Olympiad USA Math Olympiad

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?

aera of trapezoid
  • \( 6\)
  • \( 4\)
  • \( 3\)

Key Concepts


Geometry

Triangle

Trapezoid

Check the Answer


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……..

Triangle with centre O

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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