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July 28th, 2014, 10:34 PM   #1
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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.
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July 28th, 2014, 11:06 PM   #2
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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.
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July 28th, 2014, 11:47 PM   #3
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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.
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July 28th, 2014, 11:49 PM   #4
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Exclamation asdad

Quote:
Originally Posted by tahirimanov View Post
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.
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July 29th, 2014, 12:16 AM   #5
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Quote:
Originally Posted by klafmn View Post
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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