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| Math Central | Quandaries & Queries |
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Question from Ahamed: In rectangle ABCD,angle BCD is trisected. CE,CF meet the sides AB,AD at E,F. BE=6cm,AF=2cm so, find the area of the rectangle ABCD? |
Hi,
Here is my diagram,
First consider triangle $EBC.$ Angle $EBC$ is a right angle and angle $BCE$ is one third of angle $BCD$ so the measure of angle $BCE$ is $30^{o}.$ What is the length of $BC?$
Now consider triangle $FCD.$
Penny
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