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The above figure shows a circle with center O. what is the area of the circle?
1) The perimeter of the shaded region is (pi+2.828) -
Central angle/360 = perimeter of segment/perimeter of circle
1/4 = pi+2.828/2 * pi * r
2*pi*r = 4pi+2*2.828.
r can be found with calculation. So area of the circle can also found.
Sufficient.

2) Area of the shaded region is (pi-2)
Not sufficient.

Ans A.
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In the given figure, we can see a square AOEC is inscribed in a quarter of a circle with center O. If we can find the radius of the circle that would be sufficient to find the area of the circle= \(pi*r^2\). Also, since angle BOD is 90 degrees, it forms a quarter of the circle.

1) Perimeter of the shaded region would mean the circumference of the circle from B to D which constitutes for a quarter of the total circumference. \(1/4*2*pi*r=pi+2.828\). Since we know pi is approx 3.14 or 22/7, we can find the radius of the circle. Sufficient.

(2) Area of the shaded region= Area of the quarter of the circle- Area of the square AOEC. We can draw a diagonal (d) OC for the square AOEC joining O and C. Notice that OC is also the radius of the circle. Let the side of square= a. We know that diagonal of a square (d)= \sqrt{2}*a. So \(a^2= r^2/2\). Area of a square is \(side^2= a^2= r^2/2\). Area of shaded region= \(pi-2= 1/4* pi* r^2- r^2/2\). Since we we know that pi is approx 3.14 or 22/7, we can find the radius of the circle. Sufficient.

Answer: D.
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Answer is D

Statement 1:
Angle BOD is 90, which is 360/4 --> the perimeter of BCD represents half of the perimeter of the circle --> The perimeter of the circle is 4 times pi+2.828 --> 4pi + 4*2.828 = 2pi*Radius --> we can find the Radius and thus the area of the circle (Area = Radius * pie^2)

Statement 2:
AOEC is a square, thus triangle OCE is a right isosceles triangle. If OE=EC=X, then 2X^2 = R^2 (see screenshot.)
Then the area of square AOCE is x^2=R^2/2 --> The area of CMNK is 4 * R^2/2 = 2R^2.
The area of the shaded yellow region is 4*(pie-2), which is 4 times the original shaded region.
The total area of the circle is 2R^2 (square MNKC) plus 4*(pie-2) (yellow shaded region.) This can be equal to pie*Radius^2. Based on this formula we can find the Radius and thus the area of the circle.
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The above figure shows a circle with center O. what is the area of the circle?
1) The perimeter of the shaded region is (pi+2.828)
2) Area of the shaded region is (pi-2)
let side of square (OE) = s and radius of OD = r

#1
The perimeter of the shaded region is (pi+2.828)
the said perimeter can be re written as pi + 3
which would be for circle side 1/4
1/4 * 2pi*r - ( perimeter of square AOCE) = pi+3
pi*r-12/8 = s ---(1)
digonal of square = r√2=s
determine r and area can be determined
sufficient
#2
Area of the shaded region is (pi-2)
1/4 * pi *r^2 - (s^2) = pi-2
use r√2=s
sufficient
determine r then area can be determined
sufficient
OPTION D
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Given: The above figure shows a circle with center O.

Asked: what is the area of the circle?

Area of the circle = \(\pi r^2\)

1) The perimeter of the shaded region is \((\pi+2.828)\)
Perimeter of shaded region = \(2r + \pi r/2 = \pi+2.828\)
Radius of the circle r is known
Area of the circle is known
SUFFICIENT

2) Area of the shaded region is \(\pi-2\)
Let us assume side of the square = x
Radius of the circle \(r = x\sqrt{2}\)
Area of shaded region = \(\pi r^2 /4 - x^2 = (\pi/4 - 1/2)r^2 = \pi - 2\)
r^2 = 4; r = 2
Area of the circle = \(\pi r^2 = 4 \pi \)
SUFFICIENT

IMO D


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So the question is to find the area of a circle which is pi*R^2. So the hidden question is to find the radius of the circle.
Statement 1:
1) The perimeter of the shaded region is (pi+2.828)

let Side of square is 'a'. Then by Pythagorean theorem, we get
2a^2=R^2
therefore a = R/root2

The perimeter of the shaded region = (1/4 * 2 * pi * R)+ 2 *a + 2(R-a)
since a= R/root2
(pi+2.828) = (1/4 * 2 * pi * R)+ 2 * R/root2 + 2(R-R/root2)

This is equation with one variable. No need to solve further.
Hence Statment 1 is sufficient.

2) Area of the shaded region is (pi-2)

Area of the shaded region = 1/4*pi*R^2 -(side of the square)^2
let Side of square is 'a'. Then by Pythagorean theorem, we get
2a^2=R^2
therefore a = R/root2
substituting this value in the above equation we get
(pi-2) =(1/4* pi* R^2) - (R/ root2)

This is an equation with one variable hence we can find the value of R. No need to solve further since we can find R.

Thus statement 2 is sufficient.

This answer is D.
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The answer to this question is D

Statement 1: perimeter of shaded region is pi + 2.828. Perimeter is made up of two sides of square + quarter of circumference + two (radius - side of square) extra lengths.

Let the side of the square be x, then radius is x*2^(1/2), extra length is x*2^(1/2)-x
x + x + x*2^(1/2)-x + x*2^(1/2)-x + (2 *pi* x*2^(1/2))/4 = pi + 2.828. Solving this will give us the value of the x, then radius and area

Statement 2: Area of the shaded region = pi-2
Shaded region = Quarter circle area - square
Shaded region = (pi * r^2)/4 - x^2, radius = x*2^(1/2) = pi - 2, solve for x we get side of square, radius and area of complete circle
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Answer : D Each Alone sufficient. From both we can both Radius

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The above figure shows a circle with center O. ACEO is a Square. what is the area of the circle?
1) The perimeter of the shaded region is (pi+2.828)
2) Area of the shaded region is (pi-2)

Let side of square = x
and radius of circle = r

we need to find r to find area of circle.

r is also the diagonal of square so r = x sq root 2

Statement 1-
x+x+r-x+r-x+(2.pi.r)/4 = pi+2.828
we can find r from this as we know relation between x and r , hence 1 sufficient

Statement -2
area of shaded region is pi-2

that is: area of quarter circle - area of square = pi-2
pi.r^2/4 - x^2 = pi-2
we can again find r here as we know the relation between x and r . Hence 2 is also sufficient.

Hence D is answer.
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The above figure shows a circle with center O. ACEO is a Square. what is the area of the circle?

Let's say that the side of a square and the radius of a circle are x and R, respectively.
--> Radius = ?
(Statement1): The perimeter of the shaded region is (π+2.828)
--> \((R-x)+ x+x +(R-x) + \frac{2πR}{4}= π+2.828\)
\(2R (1+π/4)= π+ 2.828\)
\(R= \frac{(π+ 2.828)}{(1+π/4)}\)
Sufficient

(Statement2): Area of the shaded region is (π-2)
\(2x^{2}= R^{2}\)
--> \(x^{2}= \frac{R^{2}}{2}\)

\(\frac{πR^{2}}{4} - x^{2}= π-2\)
\(\frac{πR^{2}}{4} - \frac{R^{2}}{2} =π-2\)
\(\frac{R^{2} (π-2)}{4} = π-2\)
--> \(R=2\)
Sufficient

Answer (D)
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i think answer is "D" as we can make two equation from both of the.

Just need to use that here side of square is r/2^1/2 (here r is the radius of the circle)
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