Try this beautiful problem on area of trapezoid from Geometry.
In trapezoid \(ABCD\) with bases \(AB\) and \(CD\), we have \(AB = 52\), \(BC = 12\), \(CD = 39\), and \(DA = 5\). The area of \(ABCD\) is
Geometry
Trapezoid
Triangle
But try the problem first...
Answer: \(210\)
AMC-10A (2002) Problem 25
Pre College Mathematics
First hint
Given that \(ABCD\) is a Trapezium with bases \(AB\) and \(CD\), we have \(AB = 52\), \(BC = 12\), \(CD = 39\), and \(DA = 5\).we have to find out the area of the Trapezium.Normally the area of the trapezium is \(\frac{1}{2} (AD +BC) \times \)(height between CD & AB).but we don't know the height.So another way if we extend \(AD\) & \(BC\) ,they will meet a point \(E\).Now clearly Area of \(ABCD\)=Area of \(\triangle ABE\) - Area of \(\triangle EDC\).Can you find out the area of \(\triangle EAB\) & Area of \(\triangle EDC\)?
Can you now finish the problem ..........
Second Hint
Now \(AB||DC\) , Therefore \(\triangle EDC \sim \triangle EAB\)
\(\Rightarrow \frac{ED}{EA}=\frac{EC}{EB}=\frac{DC}{AB}\)
\(\Rightarrow \frac{ED}{ED+DA}=\frac{EC}{EC+BC}=\frac{DC}{AB}\)
\(\Rightarrow \frac{ED}{ED+5}=\frac{EC}{EC+12}=\frac{39}{52}\)
Now , \(\frac{ED}{ED+5}=\frac{39}{52}\)
\(\Rightarrow ED=15\)
And \( \frac{EC}{EC+12}=\frac{39}{52}\)
\(\Rightarrow CE=36\)
Therefore \(BE\)=\(12+36=48\) and \(AE=20\)
Notice that in the \(\triangle EDC\), \({ED}^2 +{EC}^2=(36)^2+(15)^2=(39)^2=(DC)^2\) \(\Rightarrow \triangle EDC\) is a Right-angle Triangle
Therefore Area of \(\triangle EDC=\frac{1}{2} \times 36 \times 15=270\)
Similarly In the \(\triangle EAB\), \({EA}^2 +{EB}^2=(48)^2+(20)^2=(52)^2=(AB)^2\) \(\Rightarrow \triangle EAB\) is a Right-angle Triangle
Therefore Area of \(\triangle EAB=\frac{1}{2} \times 48 \times 20=480\)
Now can you find out the area of \(ABCD\)?
can you finish the problem........
Final Step
Therefore Area of \(ABCD\)=Area of \(\triangle ABE\) - Area of \(\triangle EDC\)=\(480-270=210\)