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July 28th, 2014, 09:34 PM  #1 
Senior Member Joined: Jan 2012 Posts: 721 Thanks: 7  Similar triangles
How I wish I could show this problem in drawing. This problem has to deal with the concept of similar triangle. In a diagram, PQ//YZ. PQ is parallel to YZ, XP = 2cm, PY = 3cm, PQ= 6cm and the area of triangle $\displaystyle XPQ = 24cm^2$ Calculate the area of trapezium PQZY. Let me describe the shape. We have a large triangle XYZ. The large triange XYZ overllaps a small triangle XPQ I was able to find YZ using PX/YX = PQ/YZ i.e 2/5 =6/YZ YZ = 30/2 = 15 cm My greatest problem now is how to find the lenght XQ of the small triangle XPQ using the area $\displaystyle 24cm^2$ given. I find it hard because I can't figure out the height. Though am taking PQ as the base. Could XQ be the height? I don't think so. If you understand my problem could you please draw the shape so that I can confirm it. I have the diagram with me; is just that I don't have the means to show it. 
July 28th, 2014, 10:06 PM  #2 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry 
If XP=2 and PQ=6, then area of XPQ cannot be 24. Check problem again, and see if you did some mistake on post. And please change your profile image, I find it to offensive, and I have a right not be offended. 
July 28th, 2014, 10:47 PM  #3 
Newbie Joined: Jul 2014 From: India Posts: 2 Thanks: 0 
It's simple.. ar(XPQ)/ar(XYZ) = XP^2/XY^2 = 4 / 25 now, XPQ = 24 therefore, 24 / ar(XYZ) = 4 / 25 ar(XYZ) = 24 * 25/4 = 150 now, ar (PQZY) = ar (XYZ)  ar(XPQ) = 150  24 = 126 sq. cm and moreover you can't find the length of XQ, you can interpret this by using cosine formula because angle XPQ is not constant. 
July 28th, 2014, 10:49 PM  #4 
Newbie Joined: Jul 2014 From: India Posts: 2 Thanks: 0  asdad 
July 28th, 2014, 11:16 PM  #5 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry  Let $\displaystyle \angle XPQ$ be $\displaystyle \alpha$. $\displaystyle Area( \triangle XPQ ) = \frac{1}{2} \cdot XP \cdot PQ \cdot \sin \alpha = \frac{1}{2} \cdot 2 \cdot 6 \cdot \sin \alpha = 6 \cdot \sin \alpha$ And $\displaystyle max ( 6 \cdot \sin \alpha ) = 6$. 

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