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July 28th, 2013, 12:18 AM   #1
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Similarity and possibly algebra oriented geometry question.

Hello, I would appreciate some help with the following question please.

In the diagram,

XP = 1/2 PQ,

QY = 1/3 PQ and

XR = 2/5 XZ.

Find the ratio of the area of triangle QRZ to the area of triangle QYZ.

The drawing is not drawn to scale.

I have attached a picture of the diagram.

Thanks for taking the time to respond!
Attached Images
 Math q.png (18.0 KB, 164 views)

 July 28th, 2013, 11:42 AM #2 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: Similarity and possibly algebra oriented geometry questi Let $XP=m$ , $XR=n$ , and angle YXZ=$\alpha$. Then $XP=\frac{1}{2} m$, $XR=\frac{1}{3} m$, $XZ=\frac{5}{2} n$ and $RZ=\frac{3}{2} n$. Area of triangle QRZ is $P(QRZ)=P(XQZ)-P(XQR)$ and area of triangle $P(QYZ)=P(XYZ)-P(XQZ)$. Let's express area of those triangles over m and n. $P(XQZ)=\frac{\frac{3}{2}m \cdot \frac{5}{2} n \cdot sin \alpha}{2}=\frac{15}{8}mnsin \alpha$ $P(XQR)=\frac{n \cdot \frac{3}{2} m \cdot sin \alpha}{2}=\frac{3}{4}nm sin\alpha$ $P(XYZ)=\frac{\frac{11}{6}m \cdot \frac{5}{2} n sin \alpha}{2}=\frac{55}{24}nm sin \alpha$ $P(QRZ)=P(XQZ)-P(XQR)=\frac{15}{8} mn sin\alpha - \frac{6}{8} mn sin \alpha =\frac{9}{8} mn sin \alpha$ $P(QYZ)=P(XYZ)-P(XQZ)=\frac{55}{24}mn sin \alpha -\frac{45}{24} mn sin \alpha=\frac{3}{8} mn sin \alpha$ Finally $\frac{P(QRZ)}{P(QYZ)}=3$ Thanks from ErumFawwad
 July 28th, 2013, 11:46 AM #3 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: Similarity and possibly algebra oriented geometry questi Since I can't edit my post , I meant $PQ=m$
July 28th, 2013, 11:55 AM   #4
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Re: Similarity and possibly algebra oriented geometry questi

Quote:
 Originally Posted by crom $P(QYZ)=P(XYZ)-P(XQZ)=\frac{55}{24}mn sin \alpha -\frac{45}{24} mn sin \alpha=\frac{3}{8} mn sin \alpha$
It's$P(QYZ)=\frac{5}{12}mn sin \alpha$, not $P(QYZ)=\frac{3}{8} mn sin \alpha$, so the final score is little different.
Really sorry for three posts in a row (if needed let admins deleted).

July 28th, 2013, 02:00 PM   #5
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It's simplest to post the entire corrected post, then ask a moderator to delete the inaccurate ones (if nobody has replied to them). Triangles with the same height have areas in proportion to the corresponding bases.

Attached Images
 AreaRatio.GIF (1.8 KB, 119 views)

Last edited by skipjack; April 29th, 2014 at 11:10 PM.

 July 28th, 2013, 03:03 PM #6 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: Similarity and possibly algebra oriented geometry questi Okay, I'll keep that in mind.
 July 28th, 2013, 09:33 PM #7 Newbie   Joined: Jul 2013 Posts: 4 Thanks: 0 Re: Similarity and possibly algebra oriented geometry questi Thanks for quick and detailed responses! It has helped clarify my problem.
 April 21st, 2014, 06:42 AM #8 Newbie   Joined: Apr 2014 From: Indonesia Posts: 2 Thanks: 0 RE:Help How to solve this problem without using trigonometry? How did skipjack arrived at his/her answer for the areas of the triangle?
 April 21st, 2014, 06:43 AM #9 Newbie   Joined: Apr 2014 From: Indonesia Posts: 2 Thanks: 0 How did you arrive at the areas of each triangle?
 April 21st, 2014, 10:56 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 882 It would be easier if you started your own thread, and ask your own question; hijacking an inactive thread (as you did) is a good way for your post to be missed...

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# Xp=1/2PQ and QY=1/3pq and XR=2/5RZ, find the numerical value of area of triangle QRZ and QYZ

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