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# In the figure above, AC is tangent to circle B at point A. What is the

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Math Expert
Joined: 02 Sep 2009
Posts: 61302
In the figure above, AC is tangent to circle B at point A. What is the  [#permalink]

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26 Nov 2019, 00:00
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35% (medium)

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In the figure above, AC is tangent to circle B at point A. What is the area of the shaded region?

(1) The area of the circle is 49 in^2.
(2) AB:BC = 1:2

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CircleShadegee.jpg [ 12.11 KiB | Viewed 226 times ]

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VP
Joined: 19 Oct 2018
Posts: 1301
Location: India
Re: In the figure above, AC is tangent to circle B at point A. What is the  [#permalink]

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26 Nov 2019, 02:37
Area of shaded region = Area(ABC)- sector of the circle
hence, we need radius and angle B or C

Statement 1- it gives us radius, We don't know angle B or C
Insufficient'

Statement 2- it gives us angle B and C. we have no clue about radius.
Insufficient

Combining both equations, we will have radius and angle B or C

Sufficient

Bunuel wrote:

In the figure above, AC is tangent to circle B at point A. What is the area of the shaded region?

(1) The area of the circle is 49 in^2.
(2) AB:BC = 1:2

Are You Up For the Challenge: 700 Level Questions

Attachment:
VP
Joined: 24 Nov 2016
Posts: 1208
Location: United States
In the figure above, AC is tangent to circle B at point A. What is the  [#permalink]

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26 Nov 2019, 03:30
Bunuel wrote:

In the figure above, AC is tangent to circle B at point A. What is the area of the shaded region?

(1) The area of the circle is 49 in^2.
(2) AB:BC = 1:2

AC is tang to Circle B, then BAC is a right triangle;
Shaded region = Sector area - Triangle area.

(1) The area of the circle is 49 in^2. insufic

$$πr^2=49…r^2=49/π…r=\sqrt{49/π}$$

(2) AB:BC = 1:2 insufic

AB:BC=1:2
AC: $$4=1-x^2…x=\sqrt{3}$$
AB:AC:BC=1x:x√3:2x=30:60:90

(1&2) sufic

$$AB=r$$
$$T_{area}=r(r√3)/2$$
$$Sector_{area}=49*60/360$$
$$Shaded_{area}=T_{area}-Sector_{area}$$

Ans (C)
In the figure above, AC is tangent to circle B at point A. What is the   [#permalink] 26 Nov 2019, 03:30
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