sanju09 wrote:A thump machine is used to make a circular puncture of diameter 2 units from a square sheet of side 2 units, as shown in the attachment. The puncture is made in such a way that it touches one corner, P, of the square sheet, and the diameter of the puncture originating at P coincides with a diagonal of the square. What fraction of sheet area is safe from the puncture?
(A) (7 - π)/3
(B) (14 - 3Ï€)/6
(C) (6 - π)/8
(D) (Ï€ - 2)/4
(E) (4 - π)/4

The area that is punctured = Triangle AB + Semi Circle AB (APB makes an angle of 90 degree on the arc)
= 1/2*sqr(2)*sqrt(2) +1/2 * π *1*1 = 1+1/2π

Area that is safe from puncture = 2*2 - (1+1/2Ï€ ) = 3- 1/2Ï€ = 6-Ï€/2

ratio = (6-Ï€/2)/4 = 6-Ï€/8

I'm sort of confused because I dont know what the answer choices mean but I'll solve it anyway.

C=center of the circle
A=the point at which the circle intersects the square on the left side
B=the point at which the circle intersects the square on the bottom
As=Area of the square
Aapb=Area of triangle of APB
Ac=Area of circle "above" line segment AB

The problem can be reduced to

AREA = As - Aapb - Ac

Looking at angle APB which is 90 and using the central angle theorem we see that the angle ACB is 180 (or a line). Now the problem reduces. We can see that Ac is half of the circle and that Aapb is a 45-45-90 triangle with hypotenuse 2. Given that the area of the square is 2*2...

AREA = (2) * (2) - (1/2) * (root(2)) * (root(2)) - (1/2) * pi * (1)^2

Since the radius is 1 and the legs of the 45-45-90 triangle are root 2

AREA NOT PUNCTURED = 4 - 1 - 1/2 * pi = 3 - 1/2 * pi

Helpful?

AHHH that wasnt an n it was a pi! I understand the answers now. The rationale still stands and ajith is correct.