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

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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
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Difficulty:

65% (hard)

Question Stats:

67% (02:17) correct 33% (02:55) wrong based on 338 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

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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Posts: 61439
Re: Triangles  [#permalink]

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12 Dec 2010, 06:23
7
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 59953 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: $$PR=3\sqrt{5}$$ --> $$QT=\frac{6}{\sqrt{5}}$$.

Answer: C.
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Re: Triangles  [#permalink]

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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
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Re: Triangles  [#permalink]

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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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Posts: 61439
Re: Triangles  [#permalink]

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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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Re: Triangles  [#permalink]

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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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Re: Triangles  [#permalink]

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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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Re: Triangles  [#permalink]

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

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

Answer - C
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Re: Triangles  [#permalink]

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

Answer: C.

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?
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Joined: 02 Sep 2009
Posts: 61439
Re: Triangles  [#permalink]

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

Answer: C.

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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Re: Triangles  [#permalink]

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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-ad-4-ab-3-and-cd-168366.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
in-the-above-circle-ab-4-bc-6-ac-5-and-ad-6-what-106009.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?

About to start work on the links above. Thanks again.
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Posts: 61439
Re: Triangles  [#permalink]

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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-ad-4-ab-3-and-cd-168366.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
in-the-above-circle-ab-4-bc-6-ac-5-and-ad-6-what-106009.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?

About to start work on the links above. Thanks again.

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)?
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Re: Triangle PQR is right angled at Q. QT is perpendicular to PR  [#permalink]

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Re: Triangle PQR is right angled at Q. QT is perpendicular to PR   [#permalink] 18 Jan 2020, 14:03
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