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

Question

A circle is circumscribed around a square and inscribed in a larger square. A point inside the larger square is chosen at random. Which of the following best approximates the probability that this point is inside the circle but outside the square inscribed in the circle?

(A) 28%
(B) 31%
(C) 35%
(D) 39%
(E) 43%

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Official Answer

A

(A) 28% 
(B) 31% 
(C) 35% 
(D) 39% 
(E) 43%

Are You Up For the Challenge: 700 Level Questions

Lipun · Answer

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

ArunSharma12 · Answer

img1.png let the side of larger side be 2a. radius of the circle is a. area of larger square = 4a^2 area of the sector circle outside inner square = \frac{\pi* a^2*90}{360}-\frac{a^2sin(90)}{2} = \frac{\pi* a^2}{4}-\frac{a^2}{2} area of all four sectors = \pi* a^2- 2a^2 required probability = area of all four sectors / area of larger square required probability = \frac{\pi* a^2- 2a^2 }{ 4a^2} = \frac{\pi- 2 }{ 4} = \frac{1.14 *100}{ 4} = 28.5 %

vipulshahi · Answer

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
=   / 2*a^2
=   / 2
= 11-7 / 14
= 4/14
= 2/7
= .28

= 28%

IMO(A)

NitishJain · Answer

IMO  A

Let the side of outer square = x
Area = x^2 
 
Side of outer square = diameter of inscribed circle = x
Radius of circle =  \frac{x}{2} 
Area of circle =  \frac{22}{7}  ( \frac{x}{2} )^2
 =  \frac{11}{14}   x^2

Diameter of circle = Diagonal of inner square =x
Side of inner square = x/ \sqrt{2} 
Area of inner square =  x^2 /2

Required Probability = ( \frac{11}{14}   x^2  -  x^2 /2 )/  x^2  =  \frac{2}{7}  ~ 28%

SandeepS030489 · Answer

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

yashikaaggarwal · Answer

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

Posted from my mobile device

fauji · Answer

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

lacktutor · Answer

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

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