Triangle PQR is right angled at Q. QT is perpendicular to PR : GMAT Problem Solving (PS)
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 04 Dec 2016, 19:07

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Senior Manager
Status: swimming against the current
Joined: 24 Jul 2009
Posts: 252
Location: Chennai, India
Followers: 4

Kudos [?]: 89 [3] , given: 30

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

### Show Tags

12 Dec 2010, 05:28
3
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

60% (02:58) correct 40% (02:20) wrong based on 263 sessions

### HideShow timer Statistics

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

_________________

Gonna make it this time

Last edited by Bunuel on 24 May 2014, 09:09, edited 2 times in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 35843
Followers: 6836

Kudos [?]: 89866 [4] , given: 10381

### Show Tags

12 Dec 2010, 06:23
4
KUDOS
Expert's post
2
This post was
BOOKMARKED
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 7445 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: $$PR=3\sqrt{5}$$ --> $$QT=\frac{6}{\sqrt{5}}$$.

_________________
Director
Joined: 03 Sep 2006
Posts: 879
Followers: 6

Kudos [?]: 740 [0], given: 33

### Show Tags

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
Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 406
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)
Followers: 7

Kudos [?]: 45 [0], given: 46

### Show Tags

13 Dec 2010, 05:47
All great explanations.

+1 for c
_________________
Senior Manager
Joined: 28 Aug 2010
Posts: 259
Followers: 6

Kudos [?]: 524 [0], given: 11

### Show Tags

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 ?
_________________

Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit

Math Expert
Joined: 02 Sep 2009
Posts: 35843
Followers: 6836

Kudos [?]: 89866 [0], given: 10381

### Show Tags

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.
_________________
Intern
Joined: 14 Feb 2011
Posts: 8
Followers: 0

Kudos [?]: 5 [0], given: 1

### Show Tags

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!
Senior Manager
Joined: 09 Feb 2011
Posts: 285
Concentration: General Management, Social Entrepreneurship
Schools: HBS '14 (A)
GMAT 1: 770 Q50 V47
Followers: 14

Kudos [?]: 181 [0], given: 13

### Show Tags

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
SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 504 [0], given: 36

### Show Tags

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

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

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 52 [0], given: 23

### Show Tags

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: 35843
Followers: 6836

Kudos [?]: 89866 [0], given: 10381

### Show Tags

24 May 2014, 09:23
Expert's post
3
This post was
BOOKMARKED
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
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
_________________
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 52 [0], given: 23

### Show Tags

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: 35843
Followers: 6836

Kudos [?]: 89866 [0], given: 10381

### Show Tags

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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12867
Followers: 559

Kudos [?]: 157 [0], given: 0

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

### Show Tags

26 Oct 2015, 14:00
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Triangle PQR is right angled at Q. QT is perpendicular to PR   [#permalink] 26 Oct 2015, 14:00
Similar topics Replies Last post
Similar
Topics:
1 In the figure to the right, the sum of the measures of angles PQR and 2 25 May 2016, 03:39
20 In the diagram to the right, triangle PQR has a right angle 13 28 Mar 2014, 23:17
12 In the diagram above, <PQR is a right angle, and QS is 13 05 May 2012, 18:37
71 In the diagram, triangle PQR has a right angle at Q and a 27 05 Feb 2012, 15:54
21 In the diagram to the right, triangle PQR has a right angle 6 31 Oct 2007, 08:07
Display posts from previous: Sort by