Try this beautiful problem from Geometry: The area of trapezoid.
Quadrilateral ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid?
Geometry
Trapezoid
Area of Triangle
Answer:$750$
AMC-8(2011) Problem 20
Pre College Mathematics
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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