In the figure above, if A, B, C are the areas of the respective circle

EgmatQuantExpert , thanks for the algebraic approach.

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 diameter circle A =a,(this also the side of the triangle) , diameter of circle B=b(side of Triangle) , diameter of circle C= c (side of Triangle)

Radius of Circle A =a/2, radius circle B=b/2, radius of circle C=c/2

Triangle is right triangle, so a^2+b^2=c^2

(A+B)/C=(pi a^2/4 +pi b^2/4)/pi c^2/4
={pi/4(a^2+b^2)}/pi/4(c^2)
=(a^2+ b^2)/c^2
= 1

Sent from my Redmi Note 3 using GMAT Club Forum mobile app

In a right angle tringle we have
A sqr + B sqr = C sqr
A sqr/4 + B sqr/4 = C sqr /4 (divide both sides by 4)
multiply both sides by Pi sqr. we get

area of Cirles A + B = C
so, (A+B)/c = 1

(Area A1 + Area A2) / Area A3=


(Area A + Area B) / Area C=

(πa^2/4 + πb^2/4) / (πc^2/4)=

a^2 + b^2 /c^2 = 1

Reason, Per Pythagoras,
c^2= a^2+b^2

Hence, Ans A

Posted from my mobile device

diagram is misleading and i spent some time searching for A, B, C.

we can see that the there is a right angled triangle and its sides are chords of the 3 circles.

we know that R1^2 + R2^2 = R3^2 ---- equation 1
so area of the 3 circles will be pi*R1^2,pi*R2^2 and pi*R3^2

so (A+B)/C =(pi*R1^2 + pi*R2^2)/pi*R3^2
= (R1^2 + R2^2)/ R3^2
=R3^2/R3^2 -- using equation 1
= 1

answer A

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.