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

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The attachment 2017-11-30_0950_001.png is no longer available

Let's assume that dimensions of right triangle are 6, 8 and 10 and their radii are 3, 4 and 5 respectively. The C area is \(\frac{\pi * r^2}{2} = \frac{25 * \pi}{2}\) The A area is \(\frac{9 * \pi}{2}\) The B is \(\frac{16 * \pi}{2}\)

I used Vorovski 's method, but I felt as if I were guessing until I saw your solution.

Question: given that the solution depends on Pythagorean theorem, may we choose any values for the legs and hypotenuse of the right triangle that we know satisfy \(a^2 + b^2 = c ^2\)? E.g. 3-4-5, or 5-12-13?

Now that I see your answer, I would think so -- just checking.

Let the radius of the semicircles A1, A2 and A3 be a, b and c, respectively. Since the triangle is a right triangle, we have by the Pythagorean Theorem:

(2a)^2 + (2b)^2 = (2c)^2

4a^2 + 4b^2 = 4c^2

a^2 + b^2 = c^2

Note that the area of semicircle A1 is ½ x π x a^2; area of semicircle A2 is ½ x π x b^2 and area of semicircle ½ x π x c^2. Thus, (A1 + A2)/A3 = (½ x π x a^2 + ½ x π x b^2)/(½ x π x c^2) = (a^2 + b^2)/c^2 = 1.

We can verify this if we let the diameter of A1 = 6, diameter of A2 = 8, and diameter of A3 = 10, and thus the radius of each is 3, 4, and 5, respectively.

Finally, the area of A1 is ½ x π x 3^2 = 4.5π, the area of A2 is ½ x π x 5^2 = 8π, and the area of A3 is ½ x π x 5^2 = 12.5π.

Thus, (A1 + A2)/A3 = (4.5π + 8π)/12.5π = 1.

Answer: A
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions