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# Triangle PQR is right angled at Q. QT is perpendicular to PR

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Manager
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Triangle PQR is right angled at Q. QT is perpendicular to PR  [#permalink]

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Updated on: 24 May 2014, 09:09
5
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Difficulty:

65% (hard)

Question Stats:

69% (01:49) correct 31% (02:38) wrong based on 405 sessions

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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

A. 3/5
B. $$6\sqrt{5}$$
C. $$\frac{6}{\sqrt{5}}$$
D. 4
E. None

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Originally posted by mailnavin1 on 12 Dec 2010, 05:28.
Last edited by Bunuel on 24 May 2014, 09:09, edited 2 times in total.
Renamed the topic and edited the question.
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Joined: 02 Sep 2009
Posts: 52285

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12 Dec 2010, 06:23
6
3
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 52645 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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Joined: 03 Sep 2006
Posts: 780

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13 Dec 2010, 04:51
1
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
Manager
Joined: 28 Aug 2010
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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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Joined: 02 Sep 2009
Posts: 52285

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24 Jan 2011, 12:39
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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Joined: 14 Feb 2011
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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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Joined: 09 Feb 2011
Posts: 226
Concentration: General Management, Social Entrepreneurship
Schools: HBS '14 (A)
GMAT 1: 770 Q50 V47

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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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Joined: 16 Nov 2010
Posts: 1420
Location: United States (IN)
Concentration: Strategy, Technology

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04 May 2011, 06:59
1/2 * QT * PR = 1/2 * PQ * QR

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

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Manager
Joined: 15 Aug 2013
Posts: 247

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24 May 2014, 08:58
Bunuel wrote:
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

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: $$PR=3\sqrt{5}$$ --> $$QT=\frac{6}{\sqrt{5}}$$.

Hi Bunuel,

I'm a little confused here. let me try to explain:

If I equate the ratios:

I get: QT/6 = 5/3Root5
I'm equating the ratios as such: the height of the QTR over the height of the the main triangle which equals, the hypotenuse of QTR to the hypotenuse of the main, which i 3 Root 5.

Can you please clarify what the ratios actually are(height of one vs. base of another etc). It's a little confusing since the variables are repeated between the 3 triangles. My worry comes in because if I did this same ratio and I equated it to QTP, i would've gotten a completely different answer. I'm surely doing something wrong -- perhaps in how i'm setting up the ratios?
Math Expert
Joined: 02 Sep 2009
Posts: 52285

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24 May 2014, 09:23
russ9 wrote:
Bunuel wrote:
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

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: $$PR=3\sqrt{5}$$ --> $$QT=\frac{6}{\sqrt{5}}$$.

Hi Bunuel,

I'm a little confused here. let me try to explain:

If I equate the ratios:

I get: QT/6 = 5/3Root5
I'm equating the ratios as such: the height of the QTR over the height of the the main triangle which equals, the hypotenuse of QTR to the hypotenuse of the main, which i 3 Root 5.

Can you please clarify what the ratios actually are(height of one vs. base of another etc). It's a little confusing since the variables are repeated between the 3 triangles. My worry comes in because if I did this same ratio and I equated it to QTP, i would've gotten a completely different answer. I'm surely doing something wrong -- perhaps in how i'm setting up the ratios?

Corresponding sides are opposite the corresponding (equal) angles. Corresponding (equal) angles are marked in the diagram below:

Similar questions to practice:
in-the-diagram-to-the-right-triangle-pqr-has-a-right-angle-169506.html
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Manager
Joined: 15 Aug 2013
Posts: 247

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24 May 2014, 11:10
Bunuel wrote:
russ9 wrote:

Hi Bunuel,

I'm a little confused here. let me try to explain:

If I equate the ratios:

I get: QT/6 = 5/3Root5
I'm equating the ratios as such: the height of the QTR over the height of the the main triangle which equals, the hypotenuse of QTR to the hypotenuse of the main, which i 3 Root 5.

Can you please clarify what the ratios actually are(height of one vs. base of another etc). It's a little confusing since the variables are repeated between the 3 triangles. My worry comes in because if I did this same ratio and I equated it to QTP, i would've gotten a completely different answer. I'm surely doing something wrong -- perhaps in how i'm setting up the ratios?

Corresponding sides are opposite the corresponding (equal) angles. Corresponding (equal) angles are marked in the diagram below:

Similar questions to practice:
in-the-diagram-to-the-right-triangle-pqr-has-a-right-angle-169506.html
in-the-figure-above-the-circles-are-centered-at-o-0-0-and-162390.html
in-the-figure-above-segments-rs-and-tu-represent-two-positi-167496.html
in-triangle-pqr-the-angle-q-9-pq-6-cm-qr-8-cm-x-is-161440.html
in-triangle-abc-to-the-right-if-bc-3-and-ac-4-then-88061.html
if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html
in-triangle-abc-if-bc-3-and-ac-4-then-what-is-the-126937.html
what-are-the-lengths-of-sides-no-and-op-in-triangle-nop-130657.html
in-the-diagram-what-is-the-length-of-ab-127098.html
length-of-ac-119652.html

I see where I was going wrong. I was only accounting to align the hypotenuse but I would switch the other two angles. Thanks for clarifying.

One question to that extent: While aligning the ratios of the 3 triangles (2 inner and 1 outer), Side A/B of Inner #1 = Side A/B of Outer = Side A/B of Inner #2 right? What I mean by that is, I can equate any of the triangles to each other - not just inner/outer but also the two inner triangles because they ALL are similar triangles. Correct?

Math Expert
Joined: 02 Sep 2009
Posts: 52285

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24 May 2014, 13:35
russ9 wrote:
Bunuel wrote:
russ9 wrote:

Hi Bunuel,

I'm a little confused here. let me try to explain:

If I equate the ratios:

I get: QT/6 = 5/3Root5
I'm equating the ratios as such: the height of the QTR over the height of the the main triangle which equals, the hypotenuse of QTR to the hypotenuse of the main, which i 3 Root 5.

Can you please clarify what the ratios actually are(height of one vs. base of another etc). It's a little confusing since the variables are repeated between the 3 triangles. My worry comes in because if I did this same ratio and I equated it to QTP, i would've gotten a completely different answer. I'm surely doing something wrong -- perhaps in how i'm setting up the ratios?

Corresponding sides are opposite the corresponding (equal) angles. Corresponding (equal) angles are marked in the diagram below:

Similar questions to practice:
in-the-diagram-to-the-right-triangle-pqr-has-a-right-angle-169506.html
in-the-figure-above-the-circles-are-centered-at-o-0-0-and-162390.html
in-the-figure-above-segments-rs-and-tu-represent-two-positi-167496.html
in-triangle-pqr-the-angle-q-9-pq-6-cm-qr-8-cm-x-is-161440.html
in-triangle-abc-to-the-right-if-bc-3-and-ac-4-then-88061.html
if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html
in-triangle-abc-if-bc-3-and-ac-4-then-what-is-the-126937.html
what-are-the-lengths-of-sides-no-and-op-in-triangle-nop-130657.html
in-the-diagram-what-is-the-length-of-ab-127098.html
length-of-ac-119652.html

I see where I was going wrong. I was only accounting to align the hypotenuse but I would switch the other two angles. Thanks for clarifying.

One question to that extent: While aligning the ratios of the 3 triangles (2 inner and 1 outer), Side A/B of Inner #1 = Side A/B of Outer = Side A/B of Inner #2 right? What I mean by that is, I can equate any of the triangles to each other - not just inner/outer but also the two inner triangles because they ALL are similar triangles. Correct?

Yes, you can equate the ratios of corresponding sides in all three triangles.
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Re: Triangle PQR is right angled at Q. QT is perpendicular to PR  [#permalink]

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21 Oct 2018, 13:23
Hey guys,

What I don't understand, however, is why can we assume that the perpendicular segment QT cuts through angle PQR, could it not also be anywhere else along QR? And would the theorem (that a perpendicular to the hypotenuse creates three similar triangles) still work even if the line segment does not cut through the right angle (is not the height of the big triangle)?
Re: Triangle PQR is right angled at Q. QT is perpendicular to PR &nbs [#permalink] 21 Oct 2018, 13:23
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