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In the figure above, AC = BC = 8, angle C = 90°, and the circular arc

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In the figure above, AC = BC = 8, angle C = 90°, and the circular arc  [#permalink]

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New post 10 Mar 2015, 05:07
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In the figure above, AC = BC = 8, angle C = 90°, and the circular arc has its center at point C. Find the area of the shaded region.

A. \(8\pi-32\)
B. \(16\pi-32\)
C. \(16\pi-64\)
D. \(32\pi-32\)
E. \(32\pi-64\)


Kudos for a correct solution.

Attachment:
gm-tuaaof_img1.png
gm-tuaaof_img1.png [ 5.79 KiB | Viewed 2134 times ]

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Re: In the figure above, AC = BC = 8, angle C = 90°, and the circular arc  [#permalink]

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New post 10 Mar 2015, 05:33
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Hi

Area of sector = (90/360)*pi(8^2) = 16pi

Area of triangle = (1/2)*8*8 = 32

Area of shaded region = area of sector - area of triangle = 16pi-32
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Re: In the figure above, AC = BC = 8, angle C = 90°, and the circular arc  [#permalink]

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New post 10 Mar 2015, 18:58
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Area of shaded region = (area of circle/4) - area of triangle = (pi* 8^2)/4 - (8*8/4) = 16 pi - 32
Therefore answer is B

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Re: In the figure above, AC = BC = 8, angle C = 90°, and the circular arc  [#permalink]

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New post 15 Mar 2015, 20:11
Bunuel wrote:
Image
In the figure above, AC = BC = 8, angle C = 90°, and the circular arc has its center at point C. Find the area of the shaded region.

A. \(8\pi-32\)
B. \(16\pi-32\)
C. \(16\pi-64\)
D. \(32\pi-32\)
E. \(32\pi-64\)


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

There’s not a single formula we can use to get the answer, but by combining a few formulas, we can calculate this.

First, think about the circle. The circle has radius r = 8, so its total area would be \(A=\pi{r^2}=64\pi\).
gm-tuaaof_img5

This entire figure is a quarter of the circle, so that area would be \(quarter \ circle=16\pi\).

Now, the shaded area (technically known as a circular segment), would have an area of
(circular segment) = (quarter circle) – (triangle ABC)

Well, we already have the area of the quarter circle. The triangle would have an area of (1/2)bh = 32. Therefore, the area of the segment is \(area=16\pi-32\).

Answer = (B)
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the figure above, AC = BC = 8, angle C = 90°, and the circular arc  [#permalink]

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New post 27 Feb 2018, 09:24
Top Contributor
Bunuel wrote:
Attachment:
gm-tuaaof_img1.png
In the figure above, AC = BC = 8, angle C = 90°, and the circular arc has its center at point C. Find the area of the shaded region.

A. \(8\pi-32\)
B. \(16\pi-32\)
C. \(16\pi-64\)
D. \(32\pi-32\)
E. \(32\pi-64\)


Kudos for a correct solution.


Area of shaded region = (area of sector) - (area of triangle)

Area of circle = π(radius)²
Area of triangle = (base)(height)/2


The sector ABC is 1/4 of a circle of radius 8
So, area of sector = (1/4)(π)(8²)
= 16π


area of triangle = (8)(8)/2
= 32


So, area of shaded region = (16π) - (32)

Answer: B

Cheers,
Brent
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Re: In the figure above, AC = BC = 8, angle C = 90°, and the circular arc &nbs [#permalink] 27 Feb 2018, 09:24
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In the figure above, AC = BC = 8, angle C = 90°, and the circular arc

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