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555-605 Level|   Geometry|      
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Bunuel
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In the figure above, ABCD is a square inscribed in the circle with 09 Oct 2015, 02:33
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77% (01:14) correct 23% (01:19) wrong based on 179 sessions

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In the figure above, ABCD is a square inscribed in the circle with center O. If the length of line segment AB equals 12, what is the area of the circle ?

(A) 12π
(B) 36π
(C) 72π
(D) 144π
(E) 288π


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C
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In the figure above, ABCD is a square inscribed in the circle with 09 Oct 2015, 07:21
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Since square ABCD is inscribed in the circle, and line segment AB is 12 , shouldn't the question be what is the area of circle , considering that the answers are in form pi.


If that is the case, Ans is 72 pi.

the diameter (diagonal BD or AC) will be the hypotenuse of the triangle formed in the semi circle.
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In the figure above, ABCD is a square inscribed in the circle with 09 Oct 2015, 07:54
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I agree with the person above. If we are talking about the area of the square....then it is simply 12^2 = 144.

If we are talking about the circle, then we need to find the radius, which is (6^2+6^2)^0.5 = 6 root(2)

So the area of the circle will be pi*r^2 = 72 pi
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In the figure above, ABCD is a square inscribed in the circle with 07 Jul 2019, 22:27
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stevkang8
I agree with the person above. If we are talking about the area of the square....then it is simply 12^2 = 144.

If we are talking about the circle, then we need to find the radius, which is (6^2+6^2)^0.5 = 6 root(2)

So the area of the circle will be pi*r^2 = 72 pi


Hello Stev,

In the above figure = (AD)^2 = AB^2 + BD^2
AD = root288

AD = diameter of the circle . Therefore radium = root288/2

Area = 1/2 *pi* * ((root288)/2)^2

So shouldnt the answer be 36pi.

Can you please explain what is wrong?
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In the figure above, ABCD is a square inscribed in the circle with 11 Jul 2019, 11:34
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stevkang8
I agree with the person above. If we are talking about the area of the square....then it is simply 12^2 = 144.

If we are talking about the circle, then we need to find the radius, which is (6^2+6^2)^0.5 = 6 root(2)

So the area of the circle will be pi*r^2 = 72 pi


Hello Stev,

In the above figure = (AD)^2 = AB^2 + BD^2
AD = root288

AD = diameter of the circle . Therefore radium = root288/2

Area = 1/2 *pi* * ((root288)/2)^2

So shouldnt the answer be 36pi.

Can you please explain what is wrong?

i think u have used the additional 1/2 in the area formula of circle....
kunalbean :)
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In the figure above, ABCD is a square inscribed in the circle with 12 Jul 2019, 02:10
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Bunuel

In the figure above, ABCD is a square inscribed in the circle with center O. If the length of line segment AB equals 12, what is the area of the circle ?

(A) 12π
(B) 36π
(C) 72π
(D) 144π
(E) 288π


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Bunuel , Can you please explain the solution in detail ?

Why am I getting 36 Pi as an answer.
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In the figure above, ABCD is a square inscribed in the circle with 16 Jul 2019, 12:31
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By Formula, The diagonal of a square is s\(\sqrt{2}\).

Hence the diagonal of the given square is 12\(\sqrt{2}\). the diagonal of the square is also, the diameter of the circle. Hence the Radius of a circle is \(\frac{12Sqrt2}{2}\).

which is \(6\sqrt{2}\)

The Area of the Circle = is π x R x R = 72 π

The Answer is the option C. Hope that helps!
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In the figure above, ABCD is a square inscribed in the circle with 16 Jul 2019, 14:04
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Diagonals of a square intersect in a right angle.(they are perpendicular).

OA=OB=R
(AOB is a right Triangle)
R^2+R^2=144
2R^2=144
R^2=72
The area of the circle is
pi*R^2= 72pi
The answer choice is C.

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In the figure above, ABCD is a square inscribed in the circle with 09 Oct 2024, 12:39
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