Try this beautiful problem from Geometry based on Area of Trapezoid.
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?
Geometry
Triangle
Trapezoid
Answer:\(3\)
AMC-8 (2002) Problem 20
Pre College Mathematics
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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