Answer Questions 35 and 36 on the basis of the information given below:

35.A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminum of width 2 units, as shown below. The hole is punched such that the circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square.

The proportion of the sheet area that remains after punching is:
A.(Π + 2)/8
B.(6 - Π)/8
C.(4 - Π)/4
D.(Π - 2)/4
E.(14 - 3Π)/6

36
Find the area of the part of the circle (round punch) falling outside the square sheet.
A. Π/4
B.(Π - 1)/2
C.(Π - 1)/4
D.(Π - 2)/2
E.(Π - 2)/4

1) If the circle cuts the square on point A and B then we are joining PA and PB
PC will be the diameter of the circle extending which we will get the diagonal of the big square.
So now in PACB we get that the diagonals PC and BA cut each other at point 0
So, PO=OC=BO=OA=1
So triangle area of triangle PAB= 1/2*PO*AB = 1/2*PO*(AO+OB) = 1/2*1*2 =1
and area of semicircle = pi*r^2/2 =pi/2
So, the proportion of the sheet area remains after punching is
= 2^2-area of ( triangle PAB + Semicircle) = 4-(1+pi/2) = (6-pi)/8

2) Now the area outside the big square = area of circle- area of (semicircle+triangle PAB)
= pi-(pi/2+1) = (pi-2)/2
6 years agoHelpfull: Yes(1) No(0)
36.D 1/2*area of circle-area of triangle formed by semicircle
35.B area of squire-(area of circle-area of the part of the circle falling outside)/area of circle
9 years agoHelpfull: Yes(0) No(0)
well this question is tough to explain without graphical represntation but answer will be
! B
!! D
8 years agoHelpfull: Yes(0) No(0)
dimensions of aluminum sheet is not given...how can we solve without it?
8 years agoHelpfull: Yes(0) No(2)

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