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PS : Triangle

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amitdgr
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PS : Triangle [#permalink] New post   29 Oct 2008, 10:17
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E

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Re: PS : Triangle [#permalink] New post   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
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Re: PS : Triangle [#permalink] New post   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 ?
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Re: PS : Triangle [#permalink] New post   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
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Re: PS : Triangle [#permalink] New post   29 Oct 2008, 11:02
D - 150

fresinha12: Can you please explain your method :roll:

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 :wink:
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Re: PS : Triangle [#permalink] New post   29 Oct 2008, 11:15
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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
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Re: PS : Triangle [#permalink] New post   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!
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Re: PS : Triangle [#permalink] New post   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
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Re: PS : Triangle [#permalink] New post   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.
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Re: PS : Triangle [#permalink] New post   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.
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Re: PS : Triangle [#permalink] New post   30 Oct 2008, 08:05
scthakur..
how did u figure out that PQS and QSR are similar triangles ??
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Re: PS : Triangle [#permalink] New post   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 :o but on what basis you say that PS/SQ = SQ/SR is true? :roll:
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Re: PS : Triangle [#permalink] New post   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 :o but on what basis you say that PS/SQ = SQ/SR is true? :roll:


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.
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Re: PS : Triangle [#permalink] New post   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 :o but on what basis you say that PS/SQ = SQ/SR is true? :roll:


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?
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Re: PS : Triangle [#permalink] New post   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.



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Re: PS : Triangle [#permalink]
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