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amitdgr
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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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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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fresinha12
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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GMAT TIGER
fresinha12
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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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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spiridon
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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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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scthakur..
how did u figure out that PQS and QSR are similar triangles ??
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scthakur
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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GMAT TIGER
scthakur
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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mbajingle
GMAT TIGER
scthakur
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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GMAT TIGER
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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