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# Triangles

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Triangles [#permalink]  12 Dec 2010, 05:28
00:00

Difficulty:

45% (medium)

Question Stats:

49% (02:27) correct 50% (01:57) wrong based on 63 sessions
Triangle PQR is right angled at Q. QT is perpendicular to PR, where T is the point on PR. If PQ = 3cm and QR = 6cm. Find QT

1. 3/5
2. 6\sqrt{5}
3. 6/\sqrt{5}
4. 4
5. None
[Reveal] Spoiler: OA

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Re: Triangles [#permalink]  12 Dec 2010, 06:23
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Expert's post
mailnavin1 wrote:
Triangle PQR is right angled at Q. QT is perpendicular to PR, where T is the point on PR. If PQ = 3cm and QR = 6cm. Find QT

1. 3/5
2. 6\sqrt{5}
3. 6/\sqrt{5}
4. 4
5. None

Look at the diagram:
Attachment:

untitled.PNG [ 5.73 KiB | Viewed 2459 times ]

You can solve this question using the fact that: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle, which means that all 3 triangles PQT, QRT and PRQ are similar. Now, as in similar triangles, corresponding sides are all in the same proportion then QT/QR=PQ/PR, as PR=\sqrt{3^2+6^2}=\sqrt{45}=3\sqrt{5} then \frac{QT}{6}=\frac{3}{3\sqrt{5}} --> QT=\frac{6}{\sqrt{5}}.

Or you can equate the area: area_{PQR}=\frac{1}{2}*PQ*QR=\frac{1}{2}*QT*PR --> the same here: PR=3\sqrt{5} --> QT=\frac{6}{\sqrt{5}}.

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Re: Triangles [#permalink]  13 Dec 2010, 04:51
Find the hypotenuse by using the Pythagorean theorem.

3^2 + 6 ^2 = 3\sqrt{5}

Area of the right angled triangle = \frac{1}{2}*base*height

\frac{1}{2}*QR*PQ

Above can be equated with \frac{1}{2}*PR*QT

and QT can be found easily
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Re: Triangles [#permalink]  13 Dec 2010, 05:47
All great explanations.

+1 for c
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Re: Triangles [#permalink]  24 Jan 2011, 12:14
Hi Bunuel, I have a small doubt regarding similarity of triangles. Could you please explain the concept in little more detail or point me to a place where i can learn the concept.

eg lets say same diagram except now QT = 4 , PT = 3 .....we have to find TR = ?

Could you please explain this ?
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Re: Triangles [#permalink]  24 Jan 2011, 12:39
Expert's post
ajit257 wrote:
Hi Bunuel, I have a small doubt regarding similarity of triangles. Could you please explain the concept in little more detail or point me to a place where i can learn the concept.

eg lets say same diagram except now QT = 4 , PT = 3 .....we have to find TR = ?

Could you please explain this ?

If QT=4 and PT=3 --> corresponding legs in QTR and in PTQ are in the same ratio: TR/QT=QT/PT --> TR=16/3.

You can use area equation approach if you are not comfortable with similarity.
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Re: Triangles [#permalink]  22 Feb 2011, 15:27
Bunuel ,on your comment , corresponding legs in QTR and in PTQ are in the same ratio: TR/QT=QT/PT --> TR=16/3. When I list out propertions, its coming TR=TP, Can you explain wat did I do wrong? Thanks!
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Re: Triangles [#permalink]  04 May 2011, 01:51
ajit257 wrote:
Hi Bunuel, I have a small doubt regarding similarity of triangles. Could you please explain the concept in little more detail or point me to a place where i can learn the concept.

eg lets say same diagram except now QT = 4 , PT = 3 .....we have to find TR = ?

Could you please explain this ?

You might want to refer to this, i found it very comprehensive, like a high school chapter http://www.nos.org/Secmathcour/eng/ch-17.pdf
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Re: Triangles [#permalink]  04 May 2011, 06:59
1/2 * QT * PR = 1/2 * PQ * QR

QT = 18/root(45) = 6/root(5)

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Re: Triangles   [#permalink] 04 May 2011, 06:59
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