jimjohn 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

i get 16/9 but after spending too much time trying to figure it out.

i'll try my best to explain as best as i can (see my diagram below)

stem only tells you 1) these three triangles are all right triangles (so you can easily apply pythagorean theorem), 2) the perimeter of the largest triangle, 3) the length of the bisector, and 4) PQ > QR. From this you have to think about the possible lengths to find the sides.

I thought about the common right triangle sides: 3-4-5, 6-8-10, etc and saw that 3-4-5 = 12 and 60 is a multiple of 12 (5x). So the possible sides are [3-4-5]*5 = 15-20-25. But first have to test it out. We know that PQ > QR and the largest side is the hypotenuse. So PQ = 20, QR = 15, and PR = 25. From there I used side QS = 12 to figure out the splits between PS and RS. (Lucky thing it worked out!!)

PS^2 + QS^2 = PQ^2

PS^2 = PQ^2 - QS^2

400 - 144 = 256

sqrt256 = 16

QS^2 + RS^2 = QR^2

RS^2 = QR^2 - QS^2

225 - 144 = 81

sqrt 81 = 9

so the ratio of the area of triangle PQS to the area of triangle RQS is

PQS = 1/2bh

1/2 * 12 * 16 = 96

RQS = 1/2bh

1/2 * 12 * 9 = 54

96:54 = 16:9

Attachments

geometry.JPG [ 9.88 KiB | Viewed 2967 times ]