Recall the important property that we discussed about the relation between the areas of the two similar triangles last week – if the ratio of their sides is ‘k’, the ratio of their areas will be k^2. As mentioned last week, it’s an important property and helps you easily solve otherwise difficult questions. The question I have in mind today also brings in focus the Pythagorean triplets.

There are some triplets that you should know out cold: (3, 4, 5), (5, 12, 13) and (8, 15, 17). Usually you will find one of these three or their multiples on GMAT. Given a right triangle and two sides, say the two legs, of length 20 and 48, we need to immediately bring them down to the lowest form 20:48 = 5:12. So we know that we are talking about the 5, 12, 13 triplet and the hypotenuse will be 13*4 = 52. These little things help us save a lot of time. Why is it that some people get done with the Quant section in less than an hour while others fall short on time? It is these little things that an adept test taker has mastered which make all the difference.

Anyway, let us go on to the question we have in mind.

Question: In the figure given below, the length of PQ is 12 and the length of PR is 15. The area of right triangle STU is equal to the area of the shaded region. If the ratio of the length of ST to the length of TU is equal to the ratio of the length of PQ to the length of QR, what is the length of TU?

(A) (9?2)/4

(B) 9/2

(C) (9?2)/2

(D) 6?2

(E) 12

Solution: The information given in the question seems to overwhelm us but let’s take it a bit at a time.

“length of PQ is 12 and the length of PR is 15”

PQR is a right triangle such that PQ = 12 and PR = 15. So PQ:PR = 4:5. Recall the 3-4-5 triplet. A multiple triplet of 3-4-5 is 9-12-15. This means QR = 9.

“ratio of the length of ST to the length of TU is equal to the ratio of the length of PQ to the length of QR”
ST/TU = PQ/QR

The ratio of two sides of PQR is equal to the ratio of two sides of STU and the included angle between the sides is same ( = 90). Using SAS, triangles PQR and STU are similar.

“The area of right triangle STU is equal to the area of the shaded region”

Area of triangle PQR = Area of triangle STU + Area of Shaded Region

Since area of triangle STU = area of shaded region, (area of triangle PQR) = 2*(area of triangle STU)

In similar triangles, if the sides are in the ratio k, the areas of the triangles are in the ratio k^2. If the ratio of the areas is given as 2 (i.e. k^2 is 2), the sides must be in the ratio ?2 (i.e. k must be ?2).

Since QR = 9, TU must be 9/?2. But there is no 9/?2 in the options – in the options the denominators are rationalized. TU = 9/?2 = (9*?2)/(?2*?2) = (9*?2)/2.

The question could take a long time if we do not remember the Pythagorean triplets and the area of similar triangles property.

Takeaways:

1. Pythagorean triplets you should know: (3, 4, 5), (5, 12, 13) and (8, 15, 17) and their multiples.
2. In similar triangles, if the sides are in the ratio k, the areas of the triangles are in the ratio k^2.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

### 2 Responses

1. Ben says:

Karishma,

I am having trouble understanding why an alternative method yields a wrong answer:

Since we know qr is 9 and pq is 12, we know the area of the entire triangle is (9*12)/2 = 54. We also know that the smaller triangle is 1/2 the area of the entire triangle (which you state yourself).

Knowing this, i tried to solve for side tu given the equations tu * st = 54 (since the area after dividing by 2 will be 27, half of the full area), and tu/st = 3/4, knowing that the sides are similar. Solving for tu i seem to arrive at 6 * sqrt 2, which is answer choice D. where am i going wrong?

• Karishma says:

You have done everything right except the last step:
tu * st = 54
tu/st = 3/4
Gives 3x*4x = 54
x^2 = 9/2
x = 3/sqrt2
x = 3*(sqrt2)/2

tu = 3x = 3*3*(sqrt2)/2 = 9*(sqrt2)/2