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 July 28th, 2014, 10:34 PM #1 Senior Member     Joined: Jan 2012 Posts: 725 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, 11:06 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, 11:47 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, 11:49 PM   #4
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Quote:
 Originally Posted by tahirimanov 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.
why not possible, please explain with arguement.

July 29th, 2014, 12:16 AM   #5
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Quote:
 Originally Posted by klafmn why not possible, please explain with arguement.
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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