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In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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29 Nov 2017, 22:58
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Re: In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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30 Nov 2017, 01:16
Bunuel wrote: In the figure above, if A, B, C are the areas of the respective circles, what is the value of (A + B)/C? (A) 1 (B) 3/2 (C) π/2 (D) 2 (E) 3π/2 Attachment: The attachment 20171130_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 \(\frac{16 * \pi}{2}\) \(\frac{\frac{9 * \pi}{2} + \frac{16 * \pi}{2}}{\frac{25 * \pi}{2}} = 1\) The answer is A. If there are any mistakes please let me know
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Re: In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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30 Nov 2017, 12:05
Bunuel wrote: In the figure above, if A, B, C are the areas of the respective circles, what is the value of (A + B)/C? (A) 1 (B) 3/2 (C) π/2 (D) 2 (E) 3π/2 Attachment: 20171130_0950_001.png Let the three areas be : \(A = π*\frac{B^2}{2}\) \(B = π*\frac{P^2}{2}\) \(C = π*\frac{H^2}{2}\) Where B, P and H are three sides of the triangle and H is the Hypotenuse. Thus, \(\frac{(A+B)}{C} = \frac{(π*B^2 + π*P^2)}{(π*H^2)}\) Using Pythagoras Theorem, we can write \(B^2 + P^2 = H^2\) Thus \(\frac{(A+B)}{C} = \frac{(B^2 + P^2)}{H^2} = \frac{H^2}{H^2} = 1\) The correct answer is Option A.
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In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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30 Nov 2017, 16:37
EgmatQuantExpert wrote: Bunuel wrote: In the figure above, if A, B, C are the areas of the respective circles, what is the value of (A + B)/C? (A) 1 (B) 3/2 (C) π/2 (D) 2 (E) 3π/2 Attachment: 20171130_0950_001.png Let the three areas be : \(A = π*\frac{B^2}{2}\) \(B = π*\frac{P^2}{2}\) \(C = π*\frac{H^2}{2}\) Where B, P and H are three sides of the triangle and H is the Hypotenuse. Thus, \(\frac{(A+B)}{C} = \frac{(π*B^2 + π*P^2)}{(π*H^2)}\) Using Pythagoras Theorem, we can write \(B^2 + P^2 = H^2\) Thus \(\frac{(A+B)}{C} = \frac{(B^2 + P^2)}{H^2} = \frac{H^2}{H^2} = 1\) The correct answer is Option A. 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. 345, or 51213? Now that I see your answer, I would think so  just checking.
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Re: In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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30 Nov 2017, 18:53
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



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Re: In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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30 Nov 2017, 20:09
Bunuel wrote: In the figure above, if A, B, C are the areas of the respective circles, what is the value of (A + B)/C? (A) 1 (B) 3/2 (C) π/2 (D) 2 (E) 3π/2 Attachment: 20171130_0950_001.png 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



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Re: In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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02 Dec 2017, 15:46
(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
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Re: In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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03 Dec 2017, 10:15
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



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Re: In the figure above, if A, B, C are the areas of the respective circle [#permalink]
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03 Dec 2017, 18:50
Bunuel wrote: In the figure above, if A, B, C are the areas of the respective circles, what is the value of (A + B)/C? (A) 1 (B) 3/2 (C) π/2 (D) 2 (E) 3π/2 Attachment: 20171130_0950_001.png 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
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Re: In the figure above, if A, B, C are the areas of the respective circle
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