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805+ (Hard)|   Geometry|      
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Bunuel
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The radius of the circle with center A is 10 cm, BC = 10*2^(1/2)cm and 19 Mar 2021, 03:00
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27% (03:20) correct 73% (03:07) wrong based on 26 sessions

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The radius of the circle with center A is 10 cm, \(BC = 10\sqrt{2}\) cm and the smaller circle with center O is the largest circle that can be inscribed inside the segment BGC. What is the approximate area of the shaded region in cm^2?


A. \(20\pi - 50\)

B. \(22.5\pi - 50\)

C. \(25\pi - 50\)

D. \(27.5\pi - 50\)

E. \(30\pi - 50\)

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2021-03-08_20-56-37.png
2021-03-08_20-56-37.png [ 40.83 KiB | Viewed 1909 times ]
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B
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The radius of the circle with center A is 10 cm, BC = 10*2^(1/2)cm and 19 Mar 2021, 03:25
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Shaded area = Area of the quarter- Area of triangle ABC- Area of inscribed circle

25PI-50+25Pi(1.5-sqrt2)~ 22.5Pi-50

Option B
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The radius of the circle with center A is 10 cm, BC = 10*2^(1/2)cm and 19 Mar 2021, 07:50
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Bunuel

The radius of the circle with center A is 10 cm, \(BC = 10\sqrt{2}\) cm and the smaller circle with center O is the largest circle that can be inscribed inside the segment BGC. What is the approximate area of the shaded region in cm^2?


A. \(20\pi - 50\)

B. \(22.5\pi - 50\)

C. \(25\pi - 50\)

D. \(27.5\pi - 50\)

E. \(30\pi - 50\)

Show Spoiler
Attachment:
2021-03-08_20-56-37.png
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The solution of this question is explained in the link below (YouTube channel video).


The required area = \(91\pi/4 - 50\) = \(22.75\pi - 50\) ~ \(22.5\pi - 50\)

Answer B

(You may subscribe to my YouTube channel for more videos)
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The radius of the circle with center A is 10 cm, BC = 10*2^(1/2)cm and 19 Mar 2021, 20:20
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sujoykrdatta
Bunuel

The radius of the circle with center A is 10 cm, \(BC = 10\sqrt{2}\) cm and the smaller circle with center O is the largest circle that can be inscribed inside the segment BGC. What is the approximate area of the shaded region in cm^2?


A. \(20\pi - 50\)

B. \(22.5\pi - 50\)

C. \(25\pi - 50\)

D. \(27.5\pi - 50\)

E. \(30\pi - 50\)

Show Spoiler
Attachment:
2021-03-08_20-56-37.png


The solution of this question is explained in the link below (YouTube channel video).


The required area = \(91\pi/4 - 50\) = \(22.75\pi - 50\) ~ \(22.5\pi - 50\)

Answer B

(You may subscribe to my YouTube channel for more videos)
Show more



Hello,

could you reexplain the calculation of the inner circle area? why BC and AP and how did you get the length of BC?
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CEdward
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The radius of the circle with center A is 10 cm, BC = 10*2^(1/2)cm and 26 Mar 2021, 15:15
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Just wondering what I did wrong here? Or why my calculation doesn't simplify neatly.

1. First, split the triangle into two. Using Pythagorean theorem, we get the height 5√2 so the area of the triangle is 50.
2. Determine the area of the smaller circle. Since it's the largest possible circle, we know that we can figure out its radius as follows: (10 - 5√2)/2 ...subtract the height of the triangle from the radius gives you the diameter. Hence, the radius of the smaller circle is (10 - 5√2)/2. The area of the smaller circle is then (150 - 100√2)/4
3. The area of the whole region encompassed by the two radii is 100π/4 = 25π

Finally, 25π - (150 - 100√2)/4 - 50 should give you the shaded region.

I don't know where I messed up.
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