Re: In the figure, ABCD is a rectangle. The area of quadrilatera
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20 Mar 2019, 09:31
In the triangle EBD, the base is 3 and the height is AB. It's area, then, is: \(\frac{3AB}{2}\)
Similarly, in the triangle BDF, the base is 4 and the height is AB. So it's area is \(\frac{4AB}{2}\)
Both of these together, \(\frac{3AB}{2}+\frac{4AB}{2}=\frac{7AB}{2}\) form the area of quadrilateral EBFD. This is one half of the area of the rectangle ABCD.
The area of ABCD is AB*AD. So:
\(\frac{7AB}{2} = \frac{AB*AD}{2}\)
Multiply both sides by 2:
\(7AB = AB*AD\)
Divide both sides by AB:
\(7 = AD\)