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by Frankenstein » Thu Aug 04, 2011 11:37 pm
Hi,
Triangles PQR, PSQ, and QSR are all similar because the internal angles of all these triangles are equal. So, if a,b,c are the lengths of sides of PQR, then the sides of PSQ will be ka,kb,kc and those of QSR will be la,lb,lc where k and l are constants.
Perimeter of PQR is 60
So, a+b+c = 60
Sum of perimeters of PQS and QSR = (k+l)(a+b+c) = 60+2*QS = 60+24 = 84
So, (k+l)(60) = 84 => k+l = 1.4 --eqn(1)
Area of triangle PQR = (1/2)a*b
Area of triangle PSR = (1/2)(ka)*(kb) = (1/2)abk^2
Area of triangle QSR = (1/2)(la)*(lb) = (1/2)abl^2
Area of PQR = area of PSR + area of QSR
So, ab = (k^2+l^2)ab
So, k^2+ l^2 = 1 -- eqn(2)
From (1) and (2): k = 0.8, l = 0.6 because PQ>QR
So, ratio of areas is k^2/l^2 = (0.8/0.6)^2 = 16/9

Hence, D

There might be simpler solution that this. But, at the moment I don't know. I will post it if I find any.
Cheers!

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by Anurag@Gurome » Thu Aug 04, 2011 11:58 pm
jainrahul1985 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

OA D

Image: https://www.postimage.org/image.php?v=aVzed6i
Image

Let us assume that PQ = a, QR = b, and PR = c
By Pythagoras Theorem, in triangle PQR, c² = a² + b²
Area of triangle PQR = (1/2) * a * b = (1/2) * c * 12, which implies ab/2 = 12c/2 or ab = 12c
It is given that the perimeter of PQR = 60, which means a + b + c = 60
PQ > QR implies a > b
We know ab in terms of c, a + b = 60 - c and we also know the value of a² + b² in terms of c². So let us apply the formula of (a + b)² = a² + b² + 2ab so that we get an equation in terms of variable c.
After substituting the values in terms of c, we get,
(60 - c)² = c² + 2 * 12c
3600 + c² - 120c = c² + 24c
3600 = 144c
c = 25
ab = 12c implies ab = 12 * 25 = 300
a + b = 60 - c implies a + b = 60 - 25 = 35 implies b = 35 - a
Solving above 2 equations we get, a(35 - a) = 300
a² - 35a + 300 = 0
a² - 15a - 20a + 300 = 0
a(a - 15) - 20(a - 15) = 0
a = 15, 20 implies b = 20, 15
It is given that a > b, so the only possible value of a = 20 and b = 15.

If an altitude is drawn from the vertex with the right angle to the hypotenuse then the triangle is divided into two smaller triangles which are both similar to the original and therefore similar to each other. So, the areas of these triangles will be in the ratio of the square of respective sides.
Since, a = 20, b = 15, so a : b = 20 : 25 = 4 : 3
To have the ratio of areas, we should square the above ratio, so (4 : 3)² = 16 : 9

The correct answer is D.
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by gmatboost » Fri Aug 05, 2011 10:11 am
Anurag@Gurome wrote: If an altitude is drawn from the vertex with the right angle to the hypotenuse then the triangle is divided into two smaller triangles which are both similar to the original and therefore similar to each other.
I definitely recommend you remember the above fact stated by Anurag.

If you are able to recognize that all 3 triangles are similar, you can often take a huge shortcut (if you are short on time or otherwise stumped) and assume that the triangles belong to one of the families of common Pythagorean Triples. This isn't really the best way to go, but it is a way to go if you're not sure what else to do.

Since we have three different triangles, and two of them have a side length of 12, AND we know the perimeter is a nice round number like 60, it's not a huge stretch to assume that they might all be 3-4-5 triangles.

If we make this assumption, we would conclude that for the biggest one, 3x + 4x + 5x = 60, or 12x = 60. That means x=5. That means the triangle is a 15-20-25 triangle.

This means the hypotenuses (sp?) of the other two 3-4-5 triangles are 15 and 20. This means they are 9-12-15 and 12-16-20 triangles respectively. This works well, since we already know that they must share a side of 12 (QS).

So, the ratio of the areas is [spoiler](12*16)/(12*9) = 16/9.[/spoiler]

Again, I want to emphasize that this is not a great go-to method, but if you see right triangles and have no clue what to do, consider that there is a very good chance they are one of the following:
1. 3-4-5- family
2. 5-12-13 family
3. 30-60-90
4. 45-45-90
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by GMATGuruNY » Fri Aug 05, 2011 11:54 am
jainrahul1985 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

OA D

Image: https://www.postimage.org/image.php?v=aVzed6i
We can reason our way to the correct answer here.

The GMAT tends to use triangles that are very pretty.
Since the perimeter of ∆PQR is an integer, all the sides are likely to be integers.
Thus, ∆PQR is probably a pythagorean triple: 3-4-5, 5-12-13, etc.

To yield a perimeter of 60, the sum of the sides of the correct pythagorean triple would need to be a factor of 60.
Since 3+4+5 = 12, and 60/12 = 5, it is likely that ∆PQR = 5 * (3-4-5) = 15-20-25.
With these dimensions, P = 15+20+25 = 60.
Thus, ∆PQR likely looks like this:

Image

To confirm that we have the correct dimensions for ∆PQR:
PQ>QR.
If we call QR the base and PQ the height, bh = 20*15 = 300.
If we call PR the base and QS the height, bh = 25*12 = 300.
Since the product of the base and the height is the same in each case, we have determined the correct dimensions of ∆PQR.

As Anurag noted, when a height is drawn through the right angle of a right triangle, 3 similar triangles are formed.
Thus, ∆PQS is similar to ∆RQS.
The hypotenuse of ∆PQS = 20.
The hypotenuse of ∆RQS = 15.
Thus, corresponding sides in PQS and RQS yield a ratio of 20:15 = 4:3.
Given similar triangles with corresponding sides in a ratio of x:y, the ratio of the areas = x² : y².
Thus, the ratio of the areas = 4² : 3² = 16:9.

The correct answer is D.
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by winniethepooh » Fri Aug 05, 2011 7:41 pm
I don't think any of these solutions would complete in maximum 2 minutes!
It will surely take more.
So what I want to ask is : is this a Gmat question?
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by gmatboost » Sun Aug 07, 2011 7:13 pm
I agree that this is at the harder end of GMAT questions. For this reason, I recommend you consider the shortcuts mentioned by me and by Mitch as well as the algebraic solution posted by Anurag.

But remember, it is not the case that you need to complete each question in under two minutes. It is especially not true that you should complete the hardest and most complex questions in under 2 minutes.

I blogged about this recently here: https://blog.gmatboost.com/2011/08/02/my ... 2-minutes/

Finally, it's fair to assume you will get flustered by at least 1 question during the test. Part of the skill of test taking is learning to identify those questions, and then moving on to save time for others.

If you were really rushed and needed to answer this question in 1 minute, you should try to find a reasonable triangle that has a perimeter of 60. There are two choices here: 10-24-26 or 15-20-25. Then, if you recognize that the two triangles share the height of 12, the ratio of their areas will equal the ratio of their bases, PS and SR.

So, you should pick an answer that allows PR to be either 25 or 26. In this case PS:SR would need to be either 3:2 or 16:9 for PR to be 25 or 26.
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by Anaira Mitch » Fri Dec 30, 2016 1:31 am
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) a2 + b2 = c2 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 = (a2 + b2) + 2ab Expansion of (a + b)2
(6) (a + b)2 = c2 + 24c Substitute (1) and (2) into right side of (5)
(7) (60 - c)2 = c2 + 24c Substitute (a + b) = 60 - c from (3)
(8) 3600 - 120c + c2 = c2 + 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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