Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 29 Dec 2013
Posts: 9

In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
Updated on: 29 Mar 2014, 03:25
16
This post was BOOKMARKED
Question Stats:
83% (01:54) correct 17% (01:57) wrong based on 185 sessions
HideShow timer Statistics
Attachment: File comment: Image of the triangle
Capture.PNG [ 14.11 KiB  Viewed 22418 times ]
In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR? A. 72 B. 96 C. 108 D. 150 E. 200
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by sudeeptasahu29 on 29 Mar 2014, 00:17.
Last edited by Bunuel on 29 Mar 2014, 03:25, edited 1 time in total.
Renamed the topic and edited the question.



MBA Section Director
Status: Back to work...
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 5102
Location: India
City: Pune
GPA: 3.4
WE: Business Development (Manufacturing)

Re: In the figure to the right, triangle PQR has a right angle [#permalink]
Show Tags
29 Mar 2014, 01:54
sudeeptasahu29 wrote: Triangle PQR has a right angle at Q and the line segment QS is perpendicular to PR. If line segment PS = 16 and SR has a length of 9, what is the area of triangle PQR?
a. 72 b. 96 c. 108 d. 150 e. 200 In right triangle PQR right angled at Q, QS^2 = PS * SR > qs^2 = 16*9 > qs = 4*3 = 12 Area = \(\frac{1}{2} QS*PR\) > \(\frac{1}{2} 12*25\) > 150
_________________
Chances of Getting Admitted After an Interview [Data Crunch]
Must Read Forum Topics Before You Kick Off Your MBA Application
New GMAT Club Decision Tracker  Real Time Decision Updates



Intern
Joined: 29 Dec 2013
Posts: 9

Re: In the figure to the right, triangle PQR has a right angle [#permalink]
Show Tags
29 Mar 2014, 02:00
Dear Narenn,
Thank you for the reply. Please explain how you got this relation QS^2 = PS * SR??



MBA Section Director
Status: Back to work...
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 5102
Location: India
City: Pune
GPA: 3.4
WE: Business Development (Manufacturing)

Re: In the figure to the right, triangle PQR has a right angle [#permalink]
Show Tags
29 Mar 2014, 02:39
1
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
sudeeptasahu29 wrote: Dear Narenn,
Thank you for the reply. Please explain how you got this relation QS^2 = PS * SR?? It's a standard relationship based on similar triangle theory. We should note that triangle PQR, triangle PSQ, and triangle QSR are similar In similar triangles, ratios of corresponding sides are equal. We know that triangle PSQ is similar to triangle QSR so, \(\frac{base of PSQ}{height of PSQ} = \frac{base of QSR}{height of QSR}\) \(\frac{QS}{PS} = \frac{RS}{QS}\) > \(QS^2\) = PS*RS Similarly you can also prove that \(PQ^2\)= PS*PR and \(QR^2\) = SR*PR
_________________
Chances of Getting Admitted After an Interview [Data Crunch]
Must Read Forum Topics Before You Kick Off Your MBA Application
New GMAT Club Decision Tracker  Real Time Decision Updates



Math Expert
Joined: 02 Sep 2009
Posts: 44599

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
29 Mar 2014, 03:39
1
This post received KUDOS
Expert's post
9
This post was BOOKMARKED
sudeeptasahu29 wrote: In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR? A. 72 B. 96 C. 108 D. 150 E. 200 Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: ifarcpqraboveisasemicirclewhatisthelengthof144057.html#p1154669According to the above, triangles PQR, PSQ and QSR must be similar. Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR > QS^2 = PS*SR = 16*9 > QS = 4*3 = 12. The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150. Answer: D. Hope it's clear.
_________________
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



Intern
Joined: 17 Oct 2013
Posts: 45
Location: India
Concentration: Strategy, Statistics
WE: Analyst (Computer Software)

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
16 Jun 2014, 00:35
Bunuel wrote: sudeeptasahu29 wrote: In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR? A. 72 B. 96 C. 108 D. 150 E. 200 Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: ifarcpqraboveisasemicirclewhatisthelengthof144057.html#p1154669According to the above, triangles PQR, PSQ and QSR must be similar. Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR > QS^2 = PS*SR = 16*9 > QS = 4*3 = 12. The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150. Answer: D. Hope it's clear. Hello Bunnel, how can we ascertain that PSQ and QSR but not PSQ and RSQ that are similar?



Math Expert
Joined: 02 Sep 2009
Posts: 44599

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
16 Jun 2014, 00:38
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
Kconfused wrote: Bunuel wrote: sudeeptasahu29 wrote: In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR? A. 72 B. 96 C. 108 D. 150 E. 200 Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: ifarcpqraboveisasemicirclewhatisthelengthof144057.html#p1154669According to the above, triangles PQR, PSQ and QSR must be similar. Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR > QS^2 = PS*SR = 16*9 > QS = 4*3 = 12. The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150. Answer: D. Hope it's clear. Hello Bunnel, how can we ascertain that PSQ and QSR but not PSQ and RSQ that are similar? Triangle QSR is the same as RSQ.
_________________
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



Intern
Joined: 17 Oct 2013
Posts: 45
Location: India
Concentration: Strategy, Statistics
WE: Analyst (Computer Software)

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
16 Jun 2014, 05:24
1
This post received KUDOS
Thanks Bunnel! Wouldn't that mix up the corresponding sides? Obviously QSR and RSQ when matched with PSQ would result in different set of corresponding sides My question is, how can we find the right set of corresponding sides to match up in similar triangles? Thank you in advance! I've always had this problem with similar triangles.



Math Expert
Joined: 02 Sep 2009
Posts: 44599

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
18 Jun 2014, 06:03



Intern
Joined: 15 Jul 2012
Posts: 35

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
06 Jul 2014, 10:30
Kconfused wrote: Thanks Bunnel! Wouldn't that mix up the corresponding sides? Obviously QSR and RSQ when matched with PSQ would result in different set of corresponding sides My question is, how can we find the right set of corresponding sides to match up in similar triangles? Thank you in advance! I've always had this problem with similar triangles. i had the same issue as in finding corresponding sides in similiar triangle. i suggest you watch some videos in khan academy relating to similiar triangles. there are few videos pertaining to it .watch them they are really helpful.



Intern
Joined: 19 Jan 2016
Posts: 49

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
09 Dec 2016, 02:53
Hi Bunnel,
I tried solving this problem by using the 454590 triangle approach for triangle QSR !! Why is this approach incorrect ?



Math Expert
Joined: 02 Sep 2009
Posts: 44599

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
09 Dec 2016, 02:59



Manager
Status: IF YOU CAN DREAM IT, YOU CAN DO IT
Joined: 03 Jul 2017
Posts: 198
Location: India
Concentration: Finance, International Business

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
02 Aug 2017, 21:33
Can we solve this by using the following approach?? the area of triangle PQR = area of triangle PQS+ area of QRS But the side QS must be equal to what ?? IS that equal to PS or SR Kindly help



Manager
Joined: 19 Aug 2016
Posts: 71

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
28 Oct 2017, 00:19
Bunuel wrote: sudeeptasahu29 wrote: In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR? A. 72 B. 96 C. 108 D. 150 E. 200 Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: http://gmatclub.com/forum/ifarcpqrab ... l#p1154669According to the above, triangles PQR, PSQ and QSR must be similar. Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR > QS^2 = PS*SR = 16*9 > QS = 4*3 = 12. The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150. Answer: D. Hope it's clear. But 16+9 is the hypotenuse of the triangle PQR Why are we using it as the height?



Math Expert
Joined: 02 Sep 2009
Posts: 44599

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
28 Oct 2017, 01:07
zanaik89 wrote: Bunuel wrote: sudeeptasahu29 wrote: In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR? A. 72 B. 96 C. 108 D. 150 E. 200 Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: http://gmatclub.com/forum/ifarcpqrab ... l#p1154669According to the above, triangles PQR, PSQ and QSR must be similar. Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR > QS^2 = PS*SR = 16*9 > QS = 4*3 = 12. The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150. Answer: D. Hope it's clear. But 16+9 is the hypotenuse of the triangle PQR Why are we using it as the height? We are considering PR = 25 as the base of triangle PQR and QS as the height: The area of triangle PQR = 1/2*base*height = 1/2*PR*QS = 1/2*(16+9)*12 = 150.
_________________
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



Manager
Joined: 06 Aug 2017
Posts: 79
GMAT 1: 570 Q50 V18 GMAT 2: 610 Q49 V24

In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
28 Oct 2017, 12:04
2
This post received KUDOS
Here is the basic mathematical way of solving this question. Attachment: File comment: Image related to solution of this problem
Traingle.jpg [ 9.23 KiB  Viewed 6771 times ]
Let QR = x, PQ = y and QS = z Given PS = 16 and SR = 9 Triangle PQR is right angle triangle \(x^2 + y^2 = (16+9)^2 = 25^2\) Equation 1 Triangle QSR is right angle triangle \(z^2 + 9^2 = x^2\) Equation 2 Triangle QPS is right angle traingle \(z^2 + 16^2 = y^2\) Equation 3 Putting the values of x^2 and y^2 from equation 2 and 3 into equation 1, we get \(z^2 + 9^2 + z^2 + 16^2 = 25^2\) \(2z^2 + 9^2 + 16^2 = 25^2\) Equation 4 Solving equation 4 for z, we get z=12 Now Area of the triangle is (\(\frac{1}{2}\))*base*hieght So, area of triangle PQR is (\(\frac{1}{2}\))*PR*QS = (\(\frac{1}{2}\))*(16+9)*12 = (\(\frac{1}{2}\))*25*12 = 150 Hence D
_________________
 Kudos are the only way to tell whether my post is useful.
GMATPREP1: Q47 V36  680 Veritas Test 1: Q43 V34  630 Veritas Test 2: Q46 V30  620 Veritas Test 3: Q45 V29  610



Manager
Joined: 30 Dec 2016
Posts: 159

In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
20 Dec 2017, 01:12
Hi. Can we solve this by using the property : The median from right angle in a triangle is half of the hypotenuse? Implying QS = 1/2* PR and then using Pythagorean Theorem ? Is this method valid as well ?
_________________
Regards SandySilva
____________ Hit kudos if my post helped (:



Math Expert
Joined: 02 Sep 2009
Posts: 44599

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
20 Dec 2017, 01:44



Intern
Joined: 04 Jan 2017
Posts: 7

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]
Show Tags
27 Mar 2018, 10:49
sudeeptasahu29 wrote: Attachment: Capture.PNG In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR? A. 72 B. 96 C. 108 D. 150 E. 200 Standard type of right triangles are with sides 3 4 5 , 5 12 13, 8 15 17, and 7 24 25. 1) In this question we have hypotenuse as 25 (a multiple of 5) so we can choose 3 4 5 and 5 12 13. 2) Now hypotenuse is the largest side of a right triangle , this leaves only 3 4 5. In this case multiply each side of 3 4 5 with 5 (since hypotenuse is a multiple of 5) we get 15 20 25. Now calculate the area! (15*20)/2=150 hit kudos if you like or tell me where I am wrong




Re: In the diagram to the right, triangle PQR has a right angle
[#permalink]
27 Mar 2018, 10:49






