In the diagram above, <PQR is a right angle, and QS is : Problem Solving (PS)

mehdi2012

Joined: 06 May 2012

Last visit: 18 Mar 2013

Posts: 1

Own Kudos:

10

[10]

Given Kudos: 4

Posts: 1

Kudos: 10

[10]

06 May 2012, 02:16

9

Kudos

1

Bookmarks

is it true?

25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **

from * and ** we have:

pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145

Joy111

Joined: 18 Oct 2010

Last visit: 25 Dec 2013

Posts: 57

Own Kudos:

593

Given Kudos: 26

Posts: 57

Kudos: 593

06 May 2012, 15:58

Kudos

Bookmarks

Bunuel

BhaskarPaul

Attachment:

Geometry.jpg

In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined

Perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular QS divides right triangle PRQ into two similar triangles PQS and QRS (which are also similar to the big triangle PRQ). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).

So, PS/QS=QS/SR --> QS^2=PS*SR=25*4=100 --> QS=10 --> area of PQR equals to 1/2*PR*QS=1/2*(PS+SR)*QS=1/2*29*10=145.

Answer: B.

For more on check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.

ya that helps , great

Joy111

Joined: 18 Oct 2010

Last visit: 25 Dec 2013

Posts: 57

Own Kudos:

593

[1]

Given Kudos: 26

Posts: 57

Kudos: 593

[1]

14 May 2012, 04:23

1

Kudos

Bookmarks

mehdi2012

is it true?

25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **

from * and ** we have:

pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145

This is also a good way to solve this problem if we find it difficult to apply similarity of triangles, though we have to find the squares of some numbers .

rahulmathur9818

Joined: 19 Feb 2013

Last visit: 16 Apr 2023

Posts: 3

Given Kudos: 2

Posts: 3

Kudos: 0

31 Jul 2013, 07:00

Kudos

Bookmarks

mehdi2012

is it true?

25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **

from * and ** we have:

pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145

How do we figure out which side is proportional to which using angles? :S

avinashrao9

Joined: 02 Feb 2012

Last visit: 26 May 2014

Posts: 18

Own Kudos:

23

Given Kudos: 35

GPA: 4

Posts: 18

Kudos: 23

01 Aug 2013, 20:37

Kudos

Bookmarks

Bunuel

BhaskarPaul

Attachment:

Geometry.jpg

In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined

Perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular QS divides right triangle PRQ into two similar triangles PQS and QRS (which are also similar to the big triangle PRQ). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).

So, PS/QS=QS/SR --> QS^2=PS*SR=25*4=100 --> QS=10 --> area of PQR equals to 1/2*PR*QS=1/2*(PS+SR)*QS=1/2*29*10=145.

Answer: B.

For more on check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.

Corresponding angles are QPS and QRS right.
In that case, it must be PS/QS=SR/QS..

Please let me know where I am wrong...

healthjunkie

Joined: 14 Oct 2013

Last visit: 01 Jun 2016

Posts: 37

Own Kudos:

26

Given Kudos: 120

Schools: Duke '18 (A) Ross '18 (A) Stanford '18 (D)

Posts: 37

Kudos: 26

24 May 2015, 20:27

Kudos

Bookmarks

Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.

healthjunkie

Joined: 14 Oct 2013

Last visit: 01 Jun 2016

Posts: 37

Own Kudos:

26

Given Kudos: 120

Schools: Duke '18 (A) Ross '18 (A) Stanford '18 (D)

Posts: 37

Kudos: 26

25 May 2015, 09:35

Kudos

Bookmarks

Ah that makes sense, I was assuming the smaller triangles were similar to eachother as opposed them both being similar because they're similar to the larger triangle. Makes sense now!

sytabish

Joined: 02 May 2014

Last visit: 04 Oct 2017

Posts: 70

Own Kudos:

148

Given Kudos: 475

Schools: ESADE '16 HKU'16 SMU '16

GMAT 1: 620 Q46 V30 Verified score

Schools: ESADE '16 HKU'16 SMU '16

GMAT 1: 620 Q46 V30 Verified score

Posts: 70

Kudos: 148

02 Jun 2015, 20:32

Kudos

Bookmarks

healthjunkie

Ah that makes sense, I was assuming the smaller triangles were similar to eachother as opposed them both being similar because they're similar to the larger triangle. Makes sense now!

Hi healthjunkie ,

All 3 triangles are similar. PQR ~ PSQ ~ QSR . U can try to prove it using AAA. Lemme know if u have any doubts on that.

Thanks!

sananoor

Joined: 24 Jun 2012

Last visit: 11 Apr 2022

Posts: 301

Own Kudos:

470

Given Kudos: 331

Location: Pakistan

Concentration: Strategy, International Business

Schools: Booth '19 HBS '19 Ross '19 Mason '21 (A$) Babson '21 (A$) GWU '21 (A$) AU Kogod"21 (A$)

GPA: 3.76

Products:

Schools: Booth '19 HBS '19 Ross '19 Mason '21 (A$) Babson '21 (A$) GWU '21 (A$) AU Kogod"21 (A$)

Posts: 301

Kudos: 470

02 Jun 2015, 22:02

Kudos

Bookmarks

when we draw perpendicular then the sides of the two triangle are in same ratio
small side of large triangle/ large side of large triangle = small side of small triangle/large side of small triangle
we know PQS is large triangle and QSR is smaller triangle
PQS and QSR share the same height QS and angle QSR = QSP which confirms that the sides must be in same ratio
thus we know PS = 25 and RS = 4
we can write
25/QS = QS/4
100 = QS^2
10 = QS
now we know height = 10 and PR = 25+4 = 29
area of the triangle = 1/2(10)(29) = 145
Its B

mvictor

Board of Directors

Joined: 17 Jul 2014

Last visit: 14 Jul 2021

Posts: 2,129

Own Kudos:

1,247

Given Kudos: 236

Location: United States (IL)

Concentration: Finance, Economics

Schools: Stanford '20 (WD)

GMAT 1: 650 Q49 V30 Verified score

GPA: 3.92

WE:General Management (Transportation)

Products:

Schools: Stanford '20 (WD)

GMAT 1: 650 Q49 V30 Verified score

Posts: 2,129

Kudos: 1,247

11 Apr 2016, 19:09

Kudos

Bookmarks

Joy111

Attachment:

Geometry.jpg

In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined

right from the start I knew there's something with the similar triangles...
P is the same, and 90 degree angle is the same. it means that Q1 angle is equal to R angle.
and Q2 angle is equal to P angle.

now we have 2 right triangles..similar to each other.
leg 25 and height x for example
leg 4 and height x.

so:
25 corresponds to the x side of the smaller triangle, scale factor thus must be 25/x
x corresponds to the side equal to 4. so scale factor is x/4
of course the scale factor is the same...so we can set these two equal: 25/x = x/4 = cross multiply -> x^2 = 100, x=10.
base=29, height =10. 29x10/2 = 145.

LeonidK

Joined: 28 Jul 2016

Last visit: 29 Nov 2018

Posts: 122

Own Kudos:

40

[1]

Given Kudos: 42

Posts: 122

Kudos: 40

[1]

18 Mar 2017, 14:26

1

Kudos

Bookmarks

mehdi2012

is it true?

25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **

from * and ** we have:

pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145

correct
I fell into a trap with area calculation

it's important not to forget to divide 29*10 by 2...

sometimes when I'm happy to find solution I make stupid very trivial mistakes

Akash777

Joined: 27 Apr 2016

Last visit: 29 May 2019

Posts: 10

Own Kudos:

1

Given Kudos: 17

Location: Japan

Schools: Darden '21

Posts: 10

Kudos: 1

10 Apr 2019, 22:27

Kudos

Bookmarks

I agree that these are similar triangle( AAA theorem). I am struggling to validate the correct corresponding sides.
For instance in two smaller triangle,
(1) PS/QS=QS/SR
(2) PS/QS=SR/QS

All mentioned sides are bases of a right angled triangle. We are not even given the angles that could have indicated the correct corresponding sides. Thus the could be either

anmolgmat14

Joined: 12 Feb 2019

Last visit: 04 Jan 2020

Posts: 10

Own Kudos:

2

Given Kudos: 17

Posts: 10

Kudos: 2

25 Jun 2019, 18:07

Kudos

Bookmarks

VeritasKarishma in the naming part for 2nd triangle, the common angle is R for both the triangles (PQR and QSR) shouldn't it be the first letter in the naming of triangles ?

just as you did it for the above triangles PQR and PSQ ?

VeritasKarishma

healthjunkie

Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.

What made you think that QS/SR=QS/PS is the correct relation?
I assume you found that the two small triangles are similar to each other. The point is how are they similar to each other? They are similar because they are both similar to the big triangle PQR.
Angle PQR = Angle QSP = angle QSR = 90 degrees
Ange P is common to PQR and PSQ so by AA, triangle PQR is similar to triangle PSQ - note the naming of the triangles. The angles which are equal are placed in corresponding positions. Angle P is common so it is the first vertex of each triangle. Then angle Q = angle S so we have Q and S as second vertices and the leftover as third vertices to get PQR and PSQ.

Similarly, angle R is common to triangle PQR and triangle QSR so by AA, triangle PQR is similar to triangle QSR - the naming of the triangles should be in order to ensure that you get the corresponding sides correctly.

So triangle PQR is similar to triangles PSQ and QSR. Now you know the corresponding sides:

PS/QS (Sides made by first two underlined letters)= SQ/SR (sides made by next two letters) = PQ/QR (sides made by first and third letters)

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 22 Jun 2025

Posts: 16,059

Own Kudos:

73,842

[1]

Given Kudos: 472

Location: Pune, India

Expert

Expert reply

Posts: 16,059

Kudos: 73,842

[1]

25 Jun 2019, 22:14

1

Kudos

Bookmarks

anmolgmat14

VeritasKarishma in the naming part for 2nd triangle, the common angle is R for both the triangles (PQR and QSR) shouldn't it be the first letter in the naming of triangles ?

just as you did it for the above triangles PQR and PSQ ?

VeritasKarishma

healthjunkie

Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.

What made you think that QS/SR=QS/PS is the correct relation?
I assume you found that the two small triangles are similar to each other. The point is how are they similar to each other? They are similar because they are both similar to the big triangle PQR.
Angle PQR = Angle QSP = angle QSR = 90 degrees
Ange P is common to PQR and PSQ so by AA, triangle PQR is similar to triangle PSQ - note the naming of the triangles. The angles which are equal are placed in corresponding positions. Angle P is common so it is the first vertex of each triangle. Then angle Q = angle S so we have Q and S as second vertices and the leftover as third vertices to get PQR and PSQ.

Similarly, angle R is common to triangle PQR and triangle QSR so by AA, triangle PQR is similar to triangle QSR - the naming of the triangles should be in order to ensure that you get the corresponding sides correctly.

So triangle PQR is similar to triangles PSQ and QSR. Now you know the corresponding sides:

PS/QS (Sides made by first two underlined letters)= SQ/SR (sides made by next two letters) = PQ/QR (sides made by first and third letters)

The triangles can be named in any way. They can even be PQR and SRQ are similar. But, for our ease of finding corresponding sides, we name them as per corresponding angles
So PQR => angle P, then 90 degrees angle, then angle R
and QSR => angle Q, then 90 degrees angle, then angle R

You switch to RQP and RSQ if you wish but it makes no difference.

sharanjaymalkani

Joined: 19 Dec 2018

Last visit: 18 Jul 2021

Posts: 3

Own Kudos:

1

Given Kudos: 24

Posts: 3

Kudos: 1

02 Jul 2019, 06:53

Kudos

Bookmarks

For the triangle PQR would'nt the height be PQ base be QR and hypotenuse be PR. Bunuel could you explain this

Kinshook

Major Poster

Joined: 03 Jun 2019

Last visit: 22 Jun 2025

Posts: 5,615

Own Kudos:

5,111

Given Kudos: 161

Location: India

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

WE:Engineering (Transportation)

Products:

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

Posts: 5,615

Kudos: 5,111

21 Sep 2019, 09:52

Kudos

Bookmarks

Joy111

Attachment:

Geometry.jpg

In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined

PS/QS = QS/SR
25*4 = QS^2 = 100
QS = 10

Area PQR = 1/2 * PR * QS = 1/2 * 29*10 = 145

IMO B

BrentGMATPrepNow

Major Poster

Joined: 12 Sep 2015

Last visit: 13 May 2024

Posts: 6,757

Own Kudos:

33,893

[1]

Given Kudos: 799

Location: Canada

Expert

Expert reply

Posts: 6,757

Kudos: 33,893

[1]

13 Sep 2020, 10:42

1

Kudos

Bookmarks

Joy111

Attachment:

Geometry.jpg

In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined

Let's label the three angles in ∆PQR as follows

This means we can label the angles in the red and blue triangles as follows

This means, the remaining angles in the red and blue triangles are as follows:

Since the red and blue triangles have the SAME three angles, they are SIMILAR triangles.
This means the ratios of their corresponding sides must be equal.

In the blue triangle, the side BETWEEN the circle and the square has length 25
In the red triangle, the side BETWEEN the circle and the square has length h
So one ratio is: 25/h

In the blue triangle, the side BETWEEN the star and the square has length h
In the red triangle, the side BETWEEN the star and the square has length 4
So another ratio is: h/4

Since the ratios of corresponding sides must be equal, we can write: 25/h = h/4
Cross multiply to get: h² = 100

Solve: h = 10 or h = -10
Since the height must be POSITIVE, we can be certain that h = 10

So the area of ∆PQR = (base)(height)/2 = (29)(10)/2 = 145

Answer: B

Cheers,
Brent

In the diagram above, <PQR is a right angle, and QS is

Prep Toolkit

My Rewards

GMAT Problem Solving (PS) Questions

In the diagram above, <PQR is a right angle, and QS is

Prep Toolkit

My Rewards