In the figure, $ABCD$ is a rectangle, $AZ = WC = 6, AB = 12,$ and the area of trapezoid
$ZWCD$ is $120$ square units. What is the area of triangle $BQW$?
The height of this trapezoid is AB and the bases are ZD and WC
120 = (1/2) AB (ZD + WC)
120 = (1/2) 12 ( ZD + 6) multiply through by 2
240 = 12 (ZD + 6) divide both sides by 12
20 = ZD + 6 subtract 6 from both sides
14 = ZD
Now angle DBW = angle BDZ
And angle ZQD = angle BQW
And since AZ = WC
Then ZD = BW
So by AAS....triangle ZQD is congruent to triangle WQZ
And the altitude of trangle DQZ = altitude of triangle BQW
And since twice these altitudes = AB....then each altitude = 1/2(AB) = 6
So....the area of triangle BQW =
(1/2) BW * altitude of BQW =
(1/2) (14)(6) =
(1/2) (84) =
It is good that you like it Chris, it is your answer.
JoLink, do not take the credit for other people's answers!
Just say that you have found an answer by CPhill and give the link.
You will be given credit for your research skills and your willingness to help.
Linking a question to a good question will give you a good name.
Taking credit for other people's answers is plagiarism. It is immoral and very rude.
Taking credit for other people's work gives you a bad name.