Area of trapezoid | AMC 8, 2011|Problem 20

Quadrilateral  ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid?

Draw altitudes from the top points A and B to CD at X and Y points

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

The area of the trapezoid is \(\frac{1}{2} \times (AB+CD) \times\) (height between AB and CD)

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

Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle .

Using The Pythagorean theorem on \(\triangle ADX and \triangle BYC\) ,

\((DX)^2+(AX)^2=(AD)^2\)

\(\Rightarrow (a)^2+(12)^2=(15)^2\)

\(\Rightarrow a=\sqrt{(15)^2-(12)^2}=\sqrt {81} =9\)

and

\((BY)^2+(YC)^2=(BC)^2\)

\(\Rightarrow (12)^2+(b)^2=(20)^2\)

\(\Rightarrow b=\sqrt{(20)^2-(12)^2}=\sqrt {256} =16\)

Now ABYX is a Rectangle so \(XY=AB=50\)

\(CD=DX+XY+YC=a+XY+b=9+50+16=75\)

The area of the trapezoid is \(\frac{1}{2} \times (AB+CD) \times (height between AB and CD)=\frac {1}{2} \times (AB+CD) \times 12=750\)

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