An irregular pentagon has vertices A,B,C,D,E, with the angle at B being re-entrant and of 270 degrees, and those at C,E being right angles. These lengths are given: AB=3, BC=4, CD=5, AE=4.
(Stop this video
around 17 seconds in, and there is a diagram of the figure.)
Construct line AC.
Prove that AC and DE are parallel.
Since AB and CD are parallel, we can see AD as the hypotenuse of a right triangle with sides 4 and 8. Hence AD=sqrt(80)=4sqrt(5).
Triangle AED has right angle given, from which ED=8.
From here there are several ways to prove AC parallel to DE. One way is to use the law of cosines to show the alternate interior angles to transversal AD are equal:
cos(DCA)=[5^2+(4sqrt(5))^2-5^2]/[2*5*4sqrt(5)]=2/sqrt(5)
cos(ADE)=[8^2+(4sqrt(5))^2-4^2]/[2*8*4sqrt(5)]=2/sqrt(5)
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Posted by Jer
on 2019-08-24 15:23:06 |
One way
