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

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Math Expert
Joined: 02 Sep 2009
Posts: 46051
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
00:00

Difficulty:

25% (medium)

Question Stats:

80% (01:12) correct 20% (01:28) wrong based on 43 sessions

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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:

2017-11-30_0950_001.png [ 42.74 KiB | Viewed 1011 times ]

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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 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 $$\frac{16 * \pi}{2}$$

$$\frac{\frac{9 * \pi}{2} + \frac{16 * \pi}{2}}{\frac{25 * \pi}{2}} = 1$$

The answer is A. If there are any mistakes please let me know
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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 1480
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
1
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:
2017-11-30_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:
2017-11-30_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. 3-4-5, or 5-12-13?

Now that I see your answer, I would think so -- just checking.
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Joined: 25 Nov 2017
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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)

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

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Joined: 04 Dec 2016
Posts: 115
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:
2017-11-30_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
Manager
Joined: 01 Feb 2017
Posts: 136
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

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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:
2017-11-30_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.

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Re: In the figure above, if A, B, C are the areas of the respective circle   [#permalink] 03 Dec 2017, 18:50
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