A circle is circumscribed around a square and inscribed in a larger sq

My way of answering this question:

Use of smart numbers:

Let the side of larger square be 14 units. Therefore, the area of larger square = 14*14 = 196 sq units

Since the circle is inscribed in larger square, diameter of the circle will be equal to side of the larger square i.e. 14 units
Area of circle = (22/7)*7*7 = 154 sq units (radius will be half of diameter)

Now since the smaller square is circumscribed by the circle, diagonal of the square will be equal to diameter of the circle i.e. 14 units
diagonal of square = side*SRQT 2. Therefore, side = diagonal/SRQT 2 = 14/SRQT 2
Area of smaller square = (14*14)/(SRQT 2*SRQT 2) = 196/2 = 98 sq units

Area required in the question = area of circle - area of smaller square = 154-98 = 56 sq unit
probability = area in question/total area of larger square = 56/196 = ~28%

Correct Answer = A

Let radius of circle be 'a'. So, side length of outer square = 2*a, and side length of inscribed square = \(\sqrt{2}*a\).

Area of circle = \(3.14*a^2\)
Area of outer square = \(4*a^2\)
Area of inner square = \(2*a^2\)

P(inside circle w.r.t outer square) = P1 = \(\frac{3.14*a^2}{4*a^2}\) = \(\frac{3.14}{4}\)
P(inside inscribed square w.r.t. circle) = P2 = \(\frac{2*a^2}{3.14*a^2}\) = \(\frac{2}{3.14}\)

P = P1*(1-P2) = \(\frac{3.14}{4}\)*(\(1-\frac{2}{3.14}\)) = \(\frac{1.14}{4}\)

We know, 1/4 = 25%, so 1.14/4 will be slightly > 25% (~28%).

Ans: A

Let side of inner square = a
radius of circle = a/sqrt(2)
Side of larger square = a*sqrt(2)
Probability(point inside the circle but outside the inner square)
= (Area of circle - Area of inner square) / Area of outer square
= [ 3.14*a^2/ 2 - a^2 ] / 2*a^2
= [11/7-1] / 2
= 11-7 / 14
= 4/14
= 2/7
= .28

= 28%

IMO(A)

Let the side of outer square = 7
Area of outer square = S^2 = 7^2 = 49 ----------(1)

Side of Outer Square = Diameter of Circle Inscribed in the Square
Radius = Diameter/2
Area of Circle inscribed in the outer square = πr^2
=> 22/7*7/2*7/2 = 77/2 = 38.5 -------(2)

Diameter of Circle inscribed in the outer square = Diagonal Of square inscribed in the circle.
Diagonal = 2(Side) ^2
7^2 = 2(S)^2
49/2 = S^2
S = 7/√2
Area of inner square = S^2 = 7/√2 => 49/2 = 24.5 --------(3)

The probability that this point is inside the circle but outside the square inscribed in the circle = Area of circle - Area of inner square/area of outer square.
Putting values from equation 1,2&3
=> (38.5-24.5/49)*100
=> (14/49)*100
=> (2/7)*100
=> 28.57 or 28(approximately)

Answer is A

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Approach

Assume radius of circle: r

Side of outer square : 2r and Area : \(4r^2\)

Side of inner square: \(\sqrt{2}r\) and Area : \(2r^2\)

Now, The probability that this point is inside the circle but outside the square inscribed in the circle:

\(\frac{(Area of circle - Area of inner Square)}{Area of outer Square}\) => \(\frac{(r^2(π-2))}{4r^2}\)

On solving we get ~28%

Option A

Let's say that the side of larger square = 10 (Area = \(10^{2}\)= 100)
---> the area of the circle should be \(πR^{2}= π5^{2}= 25π\)
---> the side of the smaller square = 5√2 --> Area = 50

The point is in the area of (25π-50). (Look at the picture)
---> The probability = \(\frac{(25π-50)}{ 100} = \frac{(25*\frac{22}{7} - 50)}{100}= \frac{\frac{200}{7}}{100}= \frac{2}{7} ≈ 28.5\) %

Answer (A).

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