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Chikis July 28th, 2014 09:34 PM

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.

tahirimanov July 28th, 2014 10:06 PM

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.

klafmn July 28th, 2014 10:47 PM

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.

klafmn July 28th, 2014 10:49 PM

asdad
 
Quote:

Originally Posted by tahirimanov (Post 201690)
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.

tahirimanov July 28th, 2014 11:16 PM

Quote:

Originally Posted by klafmn (Post 201695)
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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