Last visit was: 20 Nov 2025, 01:26 It is currently 20 Nov 2025, 01:26
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
605-655 Level|   Geometry|      
User avatar
mailnavin1
User avatar
Joined: 24 Jul 2009
Last visit: 04 Aug 2012
Posts: 119
Own Kudos:
356
 [46]
Given Kudos: 30
Status:swimming against the current
Location: Chennai, India
Concentration: Finance
Posts: 119
Kudos: 356
 [46]
Triangle PQR is right angled at Q. QT is perpendicular to PR Updated on: 24 May 2014, 10:09
Originally posted by mailnavin1 on 12 Dec 2010, 06:28.
Last edited by Bunuel on 24 May 2014, 10:09, edited 2 times in total.
Renamed the topic and edited the question.
7
Kudos
Add Kudos
39
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

68% (02:19) correct 32% (02:54) wrong based on 554 sessions

History

Date
Time
Result
Not Attempted Yet
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
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
778,429
 [15]
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,429
 [15]
Triangle PQR is right angled at Q. QT is perpendicular to PR 12 Dec 2010, 07:23
11
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
mailnavin1
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
Show more

Look at the diagram:
Attachment:
untitled.PNG
untitled.PNG [ 5.73 KiB | Viewed 71594 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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
778,429
 [5]
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,429
 [5]
Triangle PQR is right angled at Q. QT is perpendicular to PR 24 May 2014, 10:23
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
russ9
Bunuel
mailnavin1
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?
Show more

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
General Discussion
User avatar
LM
User avatar
Joined: 03 Sep 2006
Last visit: 04 Apr 2015
Posts: 444
Own Kudos:
7,673
 [1]
Given Kudos: 33
Posts: 444
Kudos: 7,673
 [1]
Triangle PQR is right angled at Q. QT is perpendicular to PR 13 Dec 2010, 05:51
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
ajit257
User avatar
Joined: 28 Aug 2010
Last visit: 08 Apr 2011
Posts: 121
Own Kudos:
4,058
Given Kudos: 11
Posts: 121
Kudos: 4,058
Triangle PQR is right angled at Q. QT is perpendicular to PR 24 Jan 2011, 13:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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 ?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
778,429
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,429
Triangle PQR is right angled at Q. QT is perpendicular to PR 24 Jan 2011, 13:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ajit257
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 ?
Show more

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.
avatar
girlinslc
avatar
Joined: 14 Feb 2011
Last visit: 06 Dec 2012
Posts: 3
Own Kudos:
5
Given Kudos: 1
Posts: 3
Kudos: 5
Triangle PQR is right angled at Q. QT is perpendicular to PR 22 Feb 2011, 16:27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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!
User avatar
vivesomnium
User avatar
Joined: 09 Feb 2011
Last visit: 18 Mar 2018
Posts: 174
Own Kudos:
494
Given Kudos: 13
Concentration: General Management, Social Entrepreneurship
Schools: HBS '14 (A)
GMAT 1: 770 Q50 V47
Schools: HBS '14 (A)
GMAT 1: 770 Q50 V47
Posts: 174
Kudos: 494
Triangle PQR is right angled at Q. QT is perpendicular to PR 04 May 2011, 02:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ajit257
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 ?
Show more

You might want to refer to this, i found it very comprehensive, like a high school chapter :)https://www.nos.org/Secmathcour/eng/ch-17.pdf
User avatar
subhashghosh
User avatar
Retired Moderator
Joined: 16 Nov 2010
Last visit: 25 Jun 2024
Posts: 896
Own Kudos:
1,279
Given Kudos: 43
Location: United States (IN)
Concentration: Strategy, Technology
Products:
My Reviews
Posts: 896
Kudos: 1,279
Triangle PQR is right angled at Q. QT is perpendicular to PR 04 May 2011, 07:59
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1/2 * QT * PR = 1/2 * PQ * QR

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

Answer - C
User avatar
russ9
User avatar
Joined: 15 Aug 2013
Last visit: 20 Apr 2015
Posts: 174
Own Kudos:
400
Given Kudos: 23
Posts: 174
Kudos: 400
Triangle PQR is right angled at Q. QT is perpendicular to PR 24 May 2014, 09:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
mailnavin1
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.
Show more

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?
User avatar
russ9
User avatar
Joined: 15 Aug 2013
Last visit: 20 Apr 2015
Posts: 174
Own Kudos:
400
Given Kudos: 23
Posts: 174
Kudos: 400
Triangle PQR is right angled at Q. QT is perpendicular to PR 24 May 2014, 12:10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
russ9


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
Show more

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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
778,429
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,429
Triangle PQR is right angled at Q. QT is perpendicular to PR 24 May 2014, 14:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
russ9
Bunuel
russ9


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.
Show more

Yes, you can equate the ratios of corresponding sides in all three triangles.
User avatar
SchruteDwight
User avatar
Joined: 03 Sep 2018
Last visit: 30 Mar 2023
Posts: 167
Own Kudos:
113
Given Kudos: 923
Location: Netherlands
Schools: HEC MiM "22 (S) LBS MiM "21 (A)
GPA: 4
Products:
My Reviews
Schools: HEC MiM "22 (S) LBS MiM "21 (A)
Posts: 167
Kudos: 113
Triangle PQR is right angled at Q. QT is perpendicular to PR 21 Oct 2018, 14:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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)?
User avatar
Kritisood
User avatar
Joined: 21 Feb 2017
Last visit: 19 Jul 2023
Posts: 492
Own Kudos:
1,272
Given Kudos: 1,090
Location: India
GMAT 1: 700 Q47 V39
Products:
My Reviews
GMAT 1: 700 Q47 V39
Posts: 492
Kudos: 1,272
Triangle PQR is right angled at Q. QT is perpendicular to PR 16 Jul 2020, 02:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel theres a property that median from the right angle is half of the hypotenuse. would there be a way for that to be used here?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,408
Own Kudos:
778,429
 [1]
Given Kudos: 99,987
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,429
 [1]
Triangle PQR is right angled at Q. QT is perpendicular to PR 16 Jul 2020, 02:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kritisood
Bunuel theres a property that median from the right angle is half of the hypotenuse. would there be a way for that to be used here?
Show more

QT is perpendicular to the hypotenuse, not the median. A perpendicular to the hypotenuse coincides with the median to the hypotenuse only if a right triangle is also isosceles. This is not so in the question.
User avatar
SnorLax_7
User avatar
Joined: 19 Nov 2022
Last visit: 22 Sep 2025
Posts: 87
Own Kudos:
29
Given Kudos: 1,925
Posts: 87
Kudos: 29
Triangle PQR is right angled at Q. QT is perpendicular to PR 08 Sep 2023, 13:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Corresponding sides are opposite the corresponding (equal) angles. Corresponding (equal) angles are marked in the diagram below:
Hi Bunuel,
How you arrived at the conclusion that these are the corresponding sides on the basis of corresponding angles (equal) ?
I am facing difficulty in understanding this. can you please elaborate ?
Moderators:
Bunuel
Math Expert
105408 posts
Krunaal
Tuck School Moderator
805 posts