Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
18 Feb 2015, 05:00
Kudos
Bookmarks
| FROM Veritas Prep Blog: Determining the Area of Similar Triangles on the GMAT |
 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. Answer (C) The question could take a long time if we do not remember the Pythagorean triplets and the area of similar triangles property. Takeaways:
|
This Blog post was imported into the forum automatically. We hope you found it helpful. Please use the Kudos button if you did, or please PM/DM me if you found it disruptive and I will take care of it.
-BB
18 Feb 2015, 05:02
Kudos
Bookmarks
25 Nov 2018, 09:15
Kudos
Bookmarks
Bunuel
Can you pls explain the highlighted part? I understood the concept of the ratio of similar triangles but not able to apply the concept to this question.
