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# In the figure above, if semicircles A and B each have area 4pi

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In the figure above, if semicircles A and B each have area 4pi [#permalink]

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15 Mar 2018, 23:57
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In the figure above, if semicircles A and B each have area $$4\pi$$, what is the area of semicircle C?

(A) $$4\pi$$

(B) $$4\pi \sqrt{2}$$

(C) $$6\pi$$

(D) $$8\pi$$

(E) 16

[Reveal] Spoiler:
Attachment:

2018-03-16_1011_003.png [ 9.14 KiB | Viewed 426 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, if semicircles A and B each have area 4pi [#permalink]

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15 Mar 2018, 23:57
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Bunuel wrote:

In the figure above, if semicircles A and B each have area $$4\pi$$, what is the area of semicircle C?

(A) $$4\pi$$

(B) $$4\pi \sqrt{2}$$

(C) $$6\pi$$

(D) $$8\pi$$

(E) 16

[Reveal] Spoiler:
Attachment:
2018-03-16_1011_003.png

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In the figure above, if semicircles A and B each have area 4pi [#permalink]

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16 Mar 2018, 00:59
Bunuel wrote:

In the figure above, if semicircles A and B each have area $$4\pi$$, what is the area of semicircle C?

(A) $$4\pi$$

(B) $$4\pi \sqrt{2}$$

(C) $$6\pi$$

(D) $$8\pi$$

(E) 16

[Reveal] Spoiler:
Attachment:
2018-03-16_1011_003.png

Since both semi-circles A and B have the same area $$4*\pi$$
they will have similar radii(r), which can be calculated as follows:

$$\frac{1}{2}*\pi*r^2 = 4*\pi$$ -> $$r^2 = 8$$ ->$$r = 2\sqrt{2}$$

The diameter of the two semicircle $$2r = 4\sqrt{2}$$ are the equal sides of an isosceles right-angled triangle.

The sides of these type of triangle are in the ratio $$1:1:\sqrt{2}$$.

The hypotenuse of this right-angled triangle is the diameter of the semicircle C, which is $$\sqrt{2} * 4\sqrt{2} = 8$$

Therefore, the area of the semicircle C is $$\frac{1}{2}*\pi*r^2 = \frac{1}{2}*\pi*4^2 = 8*\pi$$(Option D)
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Re: In the figure above, if semicircles A and B each have area 4pi [#permalink]

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16 Mar 2018, 01:06
Bunuel wrote:

In the figure above, if semicircles A and B each have area $$4\pi$$, what is the area of semicircle C?

(A) $$4\pi$$

(B) $$4\pi \sqrt{2}$$

(C) $$6\pi$$

(D) $$8\pi$$

(E) 16

[Reveal] Spoiler:
Attachment:
2018-03-16_1011_003.png

Area of semicircle A and B = 4π = (1/2)πr^2
'i.e.Radius of semicircles A and B = 2√2
i.e. Diameter of semicircles A and B = 4√2

i.e. Hypotenuse of right triangle = Diameter of Semicircle C = √2 *4√2 (Using 45º-45º-90º triangle ratio)

i.e. Diameter of Circle C = 8
i.e. RAdius of semicircle C = 4

Area of Semicircle = (1/2)π4^2 = 8π

Answer: option D
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Re: In the figure above, if semicircles A and B each have area 4pi [#permalink]

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19 Mar 2018, 16:15
Bunuel wrote:

In the figure above, if semicircles A and B each have area $$4\pi$$, what is the area of semicircle C?

(A) $$4\pi$$

(B) $$4\pi \sqrt{2}$$

(C) $$6\pi$$

(D) $$8\pi$$

(E) 16

[Reveal] Spoiler:
Attachment:
2018-03-16_1011_003.png

Since area = πr^2, we see that a semicircle area is (1/2)(πr^2). Thus, for either semicircle, we can solve for the radius:

(1/2)(πr^2) = 4π

r^2 = 8

r = 2√2

The radius of each semicircle is 2√2, so the diameter of each of the semicircles A and B is 4√2.

The diameter of semicircle A is the length of one of the legs of the triangle, and the diameter of semicircle B is the length of the other leg. We can thus use the Pythagorean theorem to determine the length of the hypotenuse of the triangle, which is also the diameter of semicircle C.

(4√2)^2 + (4√2)^2 = c^2

32 + 32 = c^2

64 = c^2

8 = c

So the radius of semicircle C is 4, and hence the area of semicircle C is:

(1/2)(π x 4^2) = (1/2)(16π) = 8π

Answer: D
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Re: In the figure above, if semicircles A and B each have area 4pi   [#permalink] 19 Mar 2018, 16:15
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# In the figure above, if semicircles A and B each have area 4pi

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