It is currently 20 Oct 2017, 13:25

### 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

# PS : Triangle

Author Message
VP
Joined: 30 Jun 2008
Posts: 1034

Kudos [?]: 707 [0], given: 1

### Show Tags

29 Oct 2008, 10:17
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

ps
Attachments

mgmat-triangle.jpg [ 24.07 KiB | Viewed 895 times ]

_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 707 [0], given: 1

Current Student
Joined: 28 Dec 2004
Posts: 3350

Kudos [?]: 319 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

### Show Tags

29 Oct 2008, 10:32
the triangle is a 3:4:5 where the x factor is 5..

so height is 20 and base=15, which leaves area to b e 1/2*15*20 or 150

Kudos [?]: 319 [0], given: 2

VP
Joined: 30 Jun 2008
Posts: 1034

Kudos [?]: 707 [0], given: 1

### Show Tags

29 Oct 2008, 10:46
fresinha12 wrote:
the triangle is a 3:4:5 where the x factor is 5..

so height is 20 and base=15, which leaves area to b e 1/2*15*20 or 150

That was fast !! But I did not understand

1) how do we know triangle is 3:4:5
2) x factor is 5 ( what is x factor how did you get it as 5 ?)

isn't the base supposed to be 16+9 = 25 ?
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 707 [0], given: 1

Current Student
Joined: 28 Dec 2004
Posts: 3350

Kudos [?]: 319 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

### Show Tags

29 Oct 2008, 10:49
you should memorize the famous right triangles they are 3:4:5, 5:12:13 etc

in this case it was obvious that hyp is 25 which is a multiple of 5..so i knew the x factor is 5..

3:4:5
15:20:25

Kudos [?]: 319 [0], given: 2

Senior Manager
Joined: 21 Apr 2008
Posts: 269

Kudos [?]: 165 [0], given: 0

Location: Motortown

### Show Tags

29 Oct 2008, 11:02
D - 150

I went by the traditional way of 3 traingles(PQS, QSR, PQR), 3 variables(PQ, QS, QR), 3 equations

Area = 1/2 * 25 * QS
After solving the 3 equations you will get QS = 12, So area = 150

But, I would rather solve it with faster approach

Kudos [?]: 165 [0], given: 0

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 843 [1], given: 19

### Show Tags

29 Oct 2008, 11:15
1
KUDOS
fresinha12 wrote:
the triangle is a 3:4:5 where the x factor is 5..

so height is 20 and base=15, which leaves area to b e 1/2*15*20 or 150

But where in the question it is mentioned that the sides of the traingle PQR are in the ratio of 3:4:5?

I also got 150 but differently:
QS = x
PQ = a
QR = b

a^2 = 16^2 + x^2
b^2 = 9^2 + x^2

25^2 = a^2 + b^2
25^2 = 16^2 + x^2 + 9^2 + x^2
625 = 2x^2 + 337
x = 12

So a = 20 and b = 15

area = 1/2 (20x15) = 150
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 843 [1], given: 19

VP
Joined: 30 Jun 2008
Posts: 1034

Kudos [?]: 707 [0], given: 1

### Show Tags

29 Oct 2008, 11:23
GMAT TIGER wrote:
fresinha12 wrote:
the triangle is a 3:4:5 where the x factor is 5..

so height is 20 and base=15, which leaves area to b e 1/2*15*20 or 150

But where in the question it is mentioned that the sides of the traingle PQR are in the ratio of 3:4:5?

I also got 150 but differently:
QS = x
PQ = a
QR = b

a^2 = 16^2 + x^2
b^2 = 9^2 + x^2

25^2 = a^2 + b^2
25^2 = 16^2 + x^2 + 9^2 + x^2
625 = 2x^2 + 337
x = 12

So a = 20 and b = 15

area = 1/2 (20x15) = 150

Ahh!! Good ol' Pythagoras ..... Thanks GMAT TIGER... +1!
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 707 [0], given: 1

Senior Manager
Joined: 04 Aug 2008
Posts: 372

Kudos [?]: 37 [0], given: 1

### Show Tags

29 Oct 2008, 11:34
there are several shorcuts to calculate sides for right triangles as fresinha said and one on them is
3-4-5

others are
5-12-13
6-8-10
8-15-17
7-24-25
etc

basically all that satisfy pythagorean theorem

here we need to recognize (visually) that hypothenuse 16+9=25 stands approximatelly in 3-4-5 ratio
and therefore the explanation follows
hypo = 25
therefore height is (15*15)-(9*9)=sqrt144=12

plug in the triangle formula and calculate 300/2=150
_________________

The one who flies is worthy. The one who is worthy flies. The one who doesn't fly isn't worthy

Kudos [?]: 37 [0], given: 1

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 843 [0], given: 19

### Show Tags

29 Oct 2008, 20:55
spiridon wrote:
there are several shorcuts to calculate sides for right triangles as fresinha said and one on them is
3-4-5

others are
5-12-13
6-8-10
8-15-17
7-24-25
etc

basically all that satisfy pythagorean theorem

here we need to recognize (visually) that hypothenuse 16+9=25 stands approximatelly in 3-4-5 ratio
and therefore the explanation follows
hypo = 25
therefore height is (15*15)-(9*9)=sqrt144=12

plug in the triangle formula and calculate 300/2=150

How do you use those shortcuts such as "3-4-5"? Can you be more elabroative?

I could not do so.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 843 [0], given: 19

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

### Show Tags

29 Oct 2008, 22:48
I will solve this using the similar triangles. PQS and QSR are similar triangles such that
PS/SQ = SQ/SR
or SQ = sqrt(PS*SR) = 4*3 = 12

Hence, area = 1/2*12*25 = 150.

Kudos [?]: 279 [0], given: 0

Intern
Joined: 16 Jun 2008
Posts: 6

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

### Show Tags

30 Oct 2008, 08:05
scthakur..
how did u figure out that PQS and QSR are similar triangles ??

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

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 843 [0], given: 19

### Show Tags

30 Oct 2008, 09:28
scthakur wrote:
I will solve this using the similar triangles. PQS and QSR are similar triangles such that
PS/SQ = SQ/SR
or SQ = sqrt(PS*SR) = 4*3 = 12

Hence, area = 1/2*12*25 = 150.

Intresting but on what basis you say that PS/SQ = SQ/SR is true?
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 843 [0], given: 19

Intern
Joined: 30 Oct 2008
Posts: 30

Kudos [?]: 2 [0], given: 0

### Show Tags

30 Oct 2008, 21:40
GMAT TIGER wrote:
scthakur wrote:
I will solve this using the similar triangles. PQS and QSR are similar triangles such that
PS/SQ = SQ/SR
or SQ = sqrt(PS*SR) = 4*3 = 12

Hence, area = 1/2*12*25 = 150.

Intresting but on what basis you say that PS/SQ = SQ/SR is true?

Similar triangles. Its a theorem which says if the angles of two triangle are same then the corresponding sides of the triangle will be in proportion.

Kudos [?]: 2 [0], given: 0

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 843 [0], given: 19

### Show Tags

30 Oct 2008, 22:16
mbajingle wrote:
GMAT TIGER wrote:
scthakur wrote:
I will solve this using the similar triangles. PQS and QSR are similar triangles such that
PS/SQ = SQ/SR
or SQ = sqrt(PS*SR) = 4*3 = 12

Hence, area = 1/2*12*25 = 150.

Intresting but on what basis you say that PS/SQ = SQ/SR is true?

Similar triangles. Its a theorem which says if the angles of two triangle are same then the corresponding sides of the triangle will be in proportion.

No dispute on these basics. But how do you show that the corresponding angles are same?
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 843 [0], given: 19

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

### Show Tags

30 Oct 2008, 22:29
GMAT TIGER wrote:
No dispute on these basics. But how do you show that the corresponding angles are same?

Ok, here we go.

Between triangles PQR and PQS
angle PQR = angle PSQ = 90 degree.
angle QPS is common.
Hence, the third angle QRP equals angle PQS and this also makes triangles PQS and QSR similar.

Hope, this clarifies.

Kudos [?]: 279 [0], given: 0

Re: PS : Triangle   [#permalink] 30 Oct 2008, 22:29
Display posts from previous: Sort by