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

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13 Apr 2007, 09:12

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

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?

see attached JPEG

3/2

7/4

15/8

16/9

2

I donot know the best way;

b = 15
l = 20
h = 25.

solving for the areas of the triangles, the ratio = 16/9.

[quote="fresinha12"]OK..you need to memorize the famous triangles..right angle triangles, usually come in the size 3:4:5, 6:8:10 5:12:13..check it for yourself..

so we know the perimeter: is 60

3X+4x+5x=60; x= 5...

OK..so we know that PQ=20, PR=25 and QR=15.

Ok...lets see the second triangle PQS is again 3:4:5, cause we know that QR=15, the other line is 12, so we know the other line is 9.

alright..now area of triangle = 0.5 base*height.. well get rid of the 0.5 cause they will cancel out in the ratio.

so we have...PQS=base=12 heigth=16; i.e 25-9. so 12*16
QRS base=9 height=12...

now PQS/QRS=16/9

Hope this helps...

Hi Fresinha, could you please explain what was your criteria to select the right triangle 3:4:5? (..3x+4x+5x=60) ?

The greatest common divisor of the three numbers 3, 4, and 5 is 1.

Pythagorean triples with this property are called primitive.

From primitive Pythagorean triples, you can get other, imprimitive ones, by multiplying each of a, b, and c by any positive whole number d > 1. This is because
a^2 + b^2 = c^2

if and only if (da)^2 + (db)^2 = (dc)^2.

Thus (a,b,c) is a Pythagorean triple if and only if (da,db,dc) is. For example, (6,8,10) and (9,12,15) are imprimitive Pythagorean triples.

Guess we should know few of the primitive triplets

- If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other

- In a right triangle, the length of the altitude from the right angle to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse

- In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. Each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg

- The areas of two similar triangles are proportional to the squares of any two homologous sides.

However, I still can't solve the question. I just run around in circles. Maybe these can help jar something loose in someone else...