MGMAT - triangles

[jayhawk2001]

OA after a few reply. I'm not sure how this can be solved in under 2 min.

[bww]

Is the answer 16/9?

I got this by taking into account that many right triangles have lengths of 3, 4, 5 or multiples of those numbers. In that case, the smaller triangle has lengths of 9, 12, and 15 while the larger triangle has lengths of 12, 16, and 20.

The perimeter then checks out: 20+15+25=60

Area of larger triangle to smaller triangle is 96:54 or 16:9.

[jayhawk2001]

Is the answer 16/9?

I got this by taking into account that many right triangles have lengths of 3, 4, 5 or multiples of those numbers. In that case, the smaller triangle has lengths of 9, 12, and 15 while the larger triangle has lengths of 12, 16, and 20.

The perimeter then checks out: 20+15+25=60

Area of larger triangle to smaller triangle is 96:54 or 16:9.

You have the correct answer but I'm not sure how you converged on
multiples of 3,4,5 for right-angled triangles.

A 45-45-90 degree triangle will have its sides in the ratio of 1:1:root-2
and a 30-60-90 will have its sides in the ratio 1:root-3:2. These are
just 2 samples from the entire set.

So, there are many possibilities for the sides of a rt-angled triangle.

Can you please explain how you converged on the sides being 3, 4 and
5.

MGMAT has given an explanation that involves 14 steps of calculations,
which I thought was an overkill for solving a problem under 2 minutes.
So, just trying to understand if there's a quicker way to converge on
the answer.

[abbyyip]

MGMAT has given an explanation that involves 14 steps of calculations

Is there any way you can post this explaination?

[jayhawk2001]

Is there any way you can post this explaination?

Here you go


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.

[bww]

Unfortunately, my only explanation is drawing from what I read during GMAT prep a few months ago. Apparently the Pythagorean triangle that appears most frequently on the exam are those with sides of lengths 3:4:5 or multiples of such. I spent probably 15+ minutes trying to figure out how to even approach this problem before I recalled that piece of info. Double checking took about 10 seconds as I plugged in the 3:4:5 proportions and then calculated the perimeter. Luckily, it worked out. Like I said, I was really stumped, but the 10 seconds I spent chasing an idea proved useful. Sorry I don't have a better explanation!

[abkhan]

Unfortunately, my only explanation is drawing from what I read during GMAT prep a few months ago. Apparently the Pythagorean triangle that appears most frequently on the exam are those with sides of lengths 3:4:5 or multiples of such. I spent probably 15+ minutes trying to figure out how to even approach this problem before I recalled that piece of info. Double checking took about 10 seconds as I plugged in the 3:4:5 proportions and then calculated the perimeter. Luckily, it worked out. Like I said, I was really stumped, but the 10 seconds I spent chasing an idea proved useful. Sorry I don't have a better explanation!

What bww has used is called Pythagorean Triplets. These are sets of numbers numbers with a relation for pythagorean equation and quiet usefull. other sets are (5:12:13) (3:4:5) etc.