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In the diagram to the right, triangle PQR has a right angle

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In the diagram to the right, triangle PQR has a right angle [#permalink] New post 23 Aug 2007, 12:55
In the diagram to the right, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?
3/2
7/4
15/8
16/9
2
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 [#permalink] New post 23 Aug 2007, 13:24
16/9. But it took me close to 8 mins
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 [#permalink] New post 23 Aug 2007, 18:38
can you explain your method dacrap?

I have not been able to solve it..

I keep coming with 1 as the answer :shock:
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 [#permalink] New post 23 Aug 2007, 19:48
I'm not able to get it either =(
Could smb explain the answer to this Q?
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 [#permalink] New post 23 Aug 2007, 20:11
Seriously ive been sittin here for like 15min and my mind is blank...
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 [#permalink] New post 23 Aug 2007, 22:23
ok i solved it to 16/9 but it took me 10 minutes to do this :evil:


Here's how :



area of PQS/ area of RQS = (1/2) PS*QS/ (1/2)SR*QS = PS/SR ---eqn 1


PQ + QR + PR = 60
PQ + QR = 60 - PR ------eqn 2

1/2 * PQ *QR = 1/2 * PR * QS

solving and substituting the value of QS we get

PQ * QR = 12 . PR ....eqn 3



now,

(PQ + QR )^2 = PQ^2 + QR^2 + 2. PQ.PR -----eqn 4

in eqn 4 substitute the value of PQ + QR from eqn 2 and PQ*QR from eqn 3 and you will get

(60 - pr)^2 = PR^2 + 24 PR

solve the above equation to find

PR = 25

substitute the value of PR in eqn 2 and eqn 3

eqn 2 becomes :


PQ + QR = 35

eqn 3 becomes :


PQ * QR = 300

solving the above two equations :

PQ.(35 - PQ) = 300

PQ^2 -35PQ +300 = 0

PQ = 20,15

so, QR = 20 or 15

but given that side PQ > QR

so, PQ = 20
QR = 15


NOW YOU HAVE ALL THE 3 SIDES OF TRIANGLE as :

PR = 25
PQ = 20
QR = 15


PS = SQRT (PQ^2 - QS^2) = 16

SR = SQRT (QR^2 - QS^2) = 9

PS/SR = 16/9
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 [#permalink] New post 24 Aug 2007, 06:35
empty_spaces, good job!
OA D - 16/9
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 [#permalink] New post 24 Aug 2007, 08:20
Piter wrote:
empty_spaces, good job!
OA D - 16/9


I feel we can do this in under 2 mins using similarity of triangles. Is there an OE for this ?
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 [#permalink] New post 24 Aug 2007, 11:46
first time I've been able to help w/a question, so stuyding seems to be paying off. Properties of 3-4-5 right triangles helped here:

says perimeter is 60
right triangle with sides 15-20-25 fits [ 5*triangle with sides 3-4-5]

Therefore;
area of PQS = 1/2(12)(16)=96
area of RQS = 1/2(12)(9) = 54

96/54=16/9
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 [#permalink] New post 24 Aug 2007, 13:37
OE
Quote:
We are given a right triangle PQR with perimeter 60 and a height to the hypotenuse QS of length 12. We're asked to find the ratio of the area of the larger internal triangle PQS to the area of the smaller internal triangle RQS.

First let's find the side lengths of the original triangle. Let c equal the length of the hypotenuse PR, and let a and b equal the lengths of the sides PQ and QR respectively. First of all we know that:

(1) a^2 + b^2 = c^2 Pythagorean Theorem for right triangle PQR
(2) ab/2 = 12c/2 Triangle PQR's area computed using the standard formula (1/2*b*h) but using a different base-height combination:
- We can use base = leg a and height = leg b to get Area of PQR = ab/2
- We can also use base = hypotenuse c and height = 12 (given) to get Area of PQR = 12c/2
- The area of PQR is the same in both cases, so I can set the two equal to each other: ab/2 = 12c/2.

(3) a + b + c = 60 The problem states that triangle PQR's perimeter is 60

(4) a > b PQ > QR is given

(5) (a + b)^2 = (a^2 + b^2) + 2ab Expansion of (a + b)^2
(6) (a + b)^2 = c^2 + 24c Substitute (1) and (2) into right side of (5)
(7) (60 – c)^2 = c^2 + 24c Substitute (a + b) = 60 – c from (3)
(8) 3600 – 120c + c^2 = c^2 + 24c
(9) 3600 = 144c
(10) 25 = c

Substituting c = 25 into equations (2) and (3) gives us:

(11) ab = 300
(12) a + b = 35

which can be combined into a quadratic equation and solved to yield a = 20 and b = 15. The other possible solution of the quadratic is a = 15 and b = 20, which does not fit the requirement that a > b.

Remembering that a height to the hypotenuse always divides a right triangle into two smaller triangles that are similar to the original one (since they all have a right angle and they share another of the included angles), therefore all three triangles are similar to each other. Therefore their areas will be in the ratio of the square of their respective side lengths. The larger internal triangle has a hypotenuse of 20 (= a) and the smaller has a hypotenuse of 15 (= b), so the side lengths are in the ratio of 20/15 = 4/3. You must square this to get the ratio of their areas, which is (4/3)^2 = 16/9.

The correct answer is D.
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 [#permalink] New post 24 Aug 2007, 13:39
gnr646 wrote:
first time I've been able to help w/a question, so stuyding seems to be paying off. Properties of 3-4-5 right triangles helped here:

says perimeter is 60
right triangle with sides 15-20-25 fits [ 5*triangle with sides 3-4-5]

Therefore;
area of PQS = 1/2(12)(16)=96
area of RQS = 1/2(12)(9) = 54

96/54=16/9


I think this is inherently dangerous. 5, 12, 13 also forms a right triangle and we can have 10, 24, 26 as the sides. in this case the answer would be totally different.

Himalayan / KillerSquirrel, Give us a faster method !!!!!!!!!!!!!!!
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 [#permalink] New post 24 Aug 2007, 14:31
A 10, 24, 26 (5-12-13) triangle would not work with the given information because it would result in right triangle with a hypotenuse equal to 10 and one leg equal to 12.

Say:
PQ=10
QR=24
PR=26

The problem tells us QS is equal to 12.

Given this, right triangle PQS would have hypotenuse PQ=10 and side QS=12 which is impossible.

The only right triangle ratio that can work with this problem is 3-4-5
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 [#permalink] New post 24 Aug 2007, 16:28
Piter wrote:
In the diagram to the right, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?


3/2
7/4
15/8
16/9
2



The ratio of sides can not be qr: pq:pr (10:24:26). if so, then qs (12) would be longer than qr. this is not possible cuz in qrs triangle, h is qr and p is qs. p can not be > h.

in that case the ratio would not be 16/9 too.



so the correct ratio of the sides should be 15:20:25.

then the area pqs/ area rqs = (1/2 x 16 x 12) / (1/2 x 9 x 12) = 16/9


here the question comes, how to find the sides of the trianagle. i will keep on working and will post later.
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 [#permalink] New post 24 Aug 2007, 22:03
This is the best for me. killer might have better idea.

empty_spaces wrote:
ok i solved it to 16/9 but it took me 10 minutes to do this :evil:

Here's how :

area of PQS/ area of RQS = (1/2) PS*QS/ (1/2)SR*QS = PS/SR ---eqn 1

PQ + QR + PR = 60
PQ + QR = 60 - PR ------eqn 2

1/2 * PQ *QR = 1/2 * PR * QS

solving and substituting the value of QS we get

PQ * QR = 12 . PR ....eqn 3

now, (PQ + QR )^2 = PQ^2 + QR^2 + 2. PQ.PR -----eqn 4

in eqn 4 substitute the value of PQ + QR from eqn 2 and PQ*QR from eqn 3 and you will get

(60 - pr)^2 = PR^2 + 24 PR

solve the above equation to find

PR = 25

substitute the value of PR in eqn 2 and eqn 3

eqn 2 becomes :


PQ + QR = 35

eqn 3 becomes :


PQ * QR = 300

solving the above two equations :

PQ.(35 - PQ) = 300

PQ^2 -35PQ +300 = 0

PQ = 20,15

so, QR = 20 or 15

but given that side PQ > QR

so, PQ = 20
QR = 15

NOW YOU HAVE ALL THE 3 SIDES OF TRIANGLE as :

PR = 25
PQ = 20
QR = 15

PS = SQRT (PQ^2 - QS^2) = 16
SR = SQRT (QR^2 - QS^2) = 9

PS/SR = 16/9
  [#permalink] 24 Aug 2007, 22:03
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