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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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12 Dec 2010, 06:28
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 24 May 2014, 10:09, edited 2 times in total.
Renamed the topic and edited the question.
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12 Dec 2010, 07:23
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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 6400 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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13 Dec 2010, 05:51
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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13 Dec 2010, 06:47
All great explanations.

+1 for c
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24 Jan 2011, 13:14
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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24 Jan 2011, 13:39
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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22 Feb 2011, 16:27
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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04 May 2011, 02:51
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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04 May 2011, 07:59
1/2 * QT * PR = 1/2 * PQ * QR

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

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24 May 2014, 09:58
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?
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24 May 2014, 10:23
Expert's post
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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
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
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in-the-diagram-what-is-the-length-of-ab-127098.html
length-of-ac-119652.html
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24 May 2014, 12:10
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?

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24 May 2014, 14:35
Expert's post
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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26 Oct 2015, 15:00
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Re: Triangle PQR is right angled at Q. QT is perpendicular to PR   [#permalink] 26 Oct 2015, 15:00
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