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Right Angle

Date: 04/09/97 at 20:53:33
From: Raman Verma
Subject: Right Angle

I sure hope you can help me with this question. It goes like this:

          B                     C
          |\                   /|	
          | \                /  | 	
          |  \             /    |	
          |   \         /       |	
          |    \     /          |	
          A  P          D

The rectangle above has AB = 12cm and BC = 25cm. What value(s) for AP 
will make angle BPC a right angle?

I pretty sure that you have to prove the two triangles similar (I 
think), but I don't know what to do after that.

Date: 04/14/97 at 18:41:55
From: Doctor Wilkinson
Subject: Re: Right Angle

You've got the right idea; you just need to run with it.

First of all, can you prove the triangles are similar?  They're both 
right triangles, so you just need to show they have another pair of 
corresponding angles equal.  All you have to go on is that BPC is a 
right angle. This suggests looking at the angles BPA and CPD, because 
these angles together with BPC make a straight angle. If you subtract 
out the right angle in the middle, you find that BPA and CPD add up to 
a right angle. From this you should be able to conclude that BPA and 
PCD are equal. Do you see how to do that?

Now if you've got that far, then you know that triangles ABP and CPD 
are similar. What do you know about similar triangles? One thing you 
know is that they have corrsponding angles equal. But you already 
know that. What you want to use is the fact that corresponding sides 
are proportional.

So let x be the length of AP. That makes the length of PD 25 - x, 

So one of the two similar triangles has sides 12 and x (we're not 
going to bother with the hypotenuses). The other one has sides 25 - x 
and 12. Clearly the 12 doesn't correspond to the 12, so it must be the 
other way around!  So we have an equation which you should be able to 

 x/12 = 12/(25 -x)

-Doctor Wilkinson,  The Math Forum
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Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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