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09 Jan 2021, 08:04
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
55% (03:03) correct
45%
(03:29)
wrong
based on 51
sessions
History
Date
Time
Result
Not Attempted Yet
A punching machine is used to create a circular hole of diameter 2 unit from a square sheet of aluminum of width 2 unit, as shown in the figure. 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. What is the area of the sheet that remains after punching ?
A. 5/7 square unit
B. 9/7 square unit
C. 10/7 square unit
D. 11/7 square unit
E. 22/7 square unit
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09 Jan 2021, 08:34
Kudos
Bookmarks
I think the answer is C
The area of the square = 2^2 = 4
The shaded region consists half of the circle (radius = 1 unit) and one right angle isosceles triangle with hypotenuse equals to the diameter of the circle i.e. 2 unit
Area of the half circle with diameter equals to 2 unit (radius = 1 unit) = (π*(1^2))/2
Let us assume the length of the each of the two equal sides of the right angle isosceles triangle = a
Therefore, 2*(a^2) = 4 = a^2 = 2
Therefore the area of the right angle isosceles triangle = (1/2)*a*a = 1
therefore, the Area of the shaded region = (π*(1^2))/2 + (1/2)*2
=22/14 + 1
= 11/7 +1
= 18/7
Therefore, the area of the sheet that remains after punching = 4 - 18/7 = 10/7
The area of the square = 2^2 = 4
The shaded region consists half of the circle (radius = 1 unit) and one right angle isosceles triangle with hypotenuse equals to the diameter of the circle i.e. 2 unit
Area of the half circle with diameter equals to 2 unit (radius = 1 unit) = (π*(1^2))/2
Let us assume the length of the each of the two equal sides of the right angle isosceles triangle = a
Therefore, 2*(a^2) = 4 = a^2 = 2
Therefore the area of the right angle isosceles triangle = (1/2)*a*a = 1
therefore, the Area of the shaded region = (π*(1^2))/2 + (1/2)*2
=22/14 + 1
= 11/7 +1
= 18/7
Therefore, the area of the sheet that remains after punching = 4 - 18/7 = 10/7
15 Jan 2021, 15:01
Kudos
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Bunuel
See the attachment to break down the graph, the shaded area can be broken down into 1 half of the circle + one half of the smaller square.
One half of the smaller square has an area of 2*1/2 = 1, one half of the circle has an area of \(\pi /2\). Taking a peek at the answers, we may replace pi with the famous approximation \(\frac{22}{7}\) so the area of the shaded region is \(1 + \frac{11}{7}\).
Finally subtract that from 4 which is the area of the bigger square, we get \(3 - \frac{11}{7} = \frac{10}{7}\).
Ans: C
Attachments
HalfSqHalfCircle.png [ 25.04 KiB | Viewed 3830 times ]
