Author 
Message 
TAGS:

Hide Tags

Manager
Status: swimming against the current
Joined: 24 Jul 2009
Posts: 226
Location: Chennai, India

Triangle PQR is right angled at Q. QT is perpendicular to PR [#permalink]
Show Tags
12 Dec 2010, 06:28
4
This post received KUDOS
12
This post was BOOKMARKED
Question Stats:
61% (01:52) correct 39% (02:30) wrong based on 361 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Gonna make it this time
Last edited by Bunuel on 24 May 2014, 10:09, edited 2 times in total.
Renamed the topic and edited the question.



Math Expert
Joined: 02 Sep 2009
Posts: 44422

Re: Triangles [#permalink]
Show Tags
12 Dec 2010, 07:23
6
This post received 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 34777 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 03 Sep 2006
Posts: 850

Re: Triangles [#permalink]
Show Tags
13 Dec 2010, 05:51
1
This post received KUDOS
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
Joined: 28 Aug 2010
Posts: 255

Re: Triangles [#permalink]
Show Tags
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 ?
_________________
Verbal:newtotheverbalforumpleasereadthisfirst77546.html Math: newtothemathforumpleasereadthisfirst77764.html Gmat: everythingyouneedtoprepareforthegmatrevised77983.html  Ajit



Math Expert
Joined: 02 Sep 2009
Posts: 44422

Re: Triangles [#permalink]
Show Tags
24 Jan 2011, 13:39



Intern
Joined: 14 Feb 2011
Posts: 6

Re: Triangles [#permalink]
Show Tags
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!



Senior Manager
Joined: 09 Feb 2011
Posts: 265
Concentration: General Management, Social Entrepreneurship

Re: Triangles [#permalink]
Show Tags
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/ch17.pdf



TOEFL Forum Moderator
Joined: 16 Nov 2010
Posts: 1505
Location: United States (IN)
Concentration: Strategy, Technology

Re: Triangles [#permalink]
Show Tags
04 May 2011, 07:59
1/2 * QT * PR = 1/2 * PQ * QR QT = 18/root(45) = 6/root(5) Answer  C
_________________
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: 286

Re: Triangles [#permalink]
Show Tags
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: \(PR=3\sqrt{5}\) > \(QT=\frac{6}{\sqrt{5}}\). 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?



Math Expert
Joined: 02 Sep 2009
Posts: 44422

Re: Triangles [#permalink]
Show Tags
24 May 2014, 10:23
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: inthediagramtotherighttrianglepqrhasarightangle169506.htmlinthefigureabovethecirclesarecenteredato00and162390.htmlinthefigureabovead4ab3andcd168366.htmlinthefigureabovesegmentsrsandturepresenttwopositi167496.htmlintrianglepqrtheangleq9pq6cmqr8cmxis161440.htmlintriangleabctotherightifbc3andac4then88061.htmlifarcpqraboveisasemicirclewhatisthelengthof144057.htmlintriangleabcifbc3andac4thenwhatisthe126937.htmlwhatarethelengthsofsidesnoandopintrianglenop130657.htmlinthediagramwhatisthelengthofab127098.htmllengthofac119652.htmlintheabovecircleab4bc6ac5andad6what106009.html
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 15 Aug 2013
Posts: 286

Re: Triangles [#permalink]
Show Tags
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: inthediagramtotherighttrianglepqrhasarightangle169506.htmlinthefigureabovethecirclesarecenteredato00and162390.htmlinthefigureabovead4ab3andcd168366.htmlinthefigureabovesegmentsrsandturepresenttwopositi167496.htmlintrianglepqrtheangleq9pq6cmqr8cmxis161440.htmlintriangleabctotherightifbc3andac4then88061.htmlifarcpqraboveisasemicirclewhatisthelengthof144057.htmlintriangleabcifbc3andac4thenwhatisthe126937.htmlwhatarethelengthsofsidesnoandopintrianglenop130657.htmlinthediagramwhatisthelengthofab127098.htmllengthofac119652.htmlintheabovecircleab4bc6ac5andad6what106009.htmlI 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.



Math Expert
Joined: 02 Sep 2009
Posts: 44422

Re: Triangles [#permalink]
Show Tags
24 May 2014, 14:35
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: inthediagramtotherighttrianglepqrhasarightangle169506.htmlinthefigureabovethecirclesarecenteredato00and162390.htmlinthefigureabovead4ab3andcd168366.htmlinthefigureabovesegmentsrsandturepresenttwopositi167496.htmlintrianglepqrtheangleq9pq6cmqr8cmxis161440.htmlintriangleabctotherightifbc3andac4then88061.htmlifarcpqraboveisasemicirclewhatisthelengthof144057.htmlintriangleabcifbc3andac4thenwhatisthe126937.htmlwhatarethelengthsofsidesnoandopintrianglenop130657.htmlinthediagramwhatisthelengthofab127098.htmllengthofac119652.htmlintheabovecircleab4bc6ac5andad6what106009.htmlI 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



NonHuman User
Joined: 09 Sep 2013
Posts: 6553

Re: Triangle PQR is right angled at Q. QT is perpendicular to PR [#permalink]
Show Tags
02 May 2017, 07:51
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: Triangle PQR is right angled at Q. QT is perpendicular to PR
[#permalink]
02 May 2017, 07:51






