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In the diagram to the right, triangle PQR has a right angle [#permalink]
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

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.

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 !!!!!!!!!!!!!!!

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

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.