A punching machine is used to create a circular hole of diameter 2 uni

Question

https://gmatclub.com/forum/download/file.php?id=59998 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 Are You Up For the Challenge: 700 Level Questions 1.jpg Untitled.png

anjanita · Answer

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

TestPrepUnlimited · Answer

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

Fdambro294 · Answer

Since the circle’s diameter is in line with the square’s diagonal, we can break the overlap part of the circle that needs to be removed into 4 pieces:

(1st) draw a diagonal from P to the diagonally opposite vertex of the Square

Label the intersection of the diagonal and the opposite end of the circle point X. Line segment PX will be the diameter of the circle (2 units) and will fall on top of the diagonal of the square.

The center of the circle (point O) will be a little closer to vertex P then the center of the Square

(2nd) since diameter PX lies on the square’s diagonal, it will bisect the 90 degree angle at vertex P.

Draw a radius from center O to the intersection point of the square on the square’s left side. Call this point Y.

Angle YPX is an inscribed angle of the circle. It is 45 degrees.

The central angle subtended by the same arc is at YOX and will be equal to = (2) (45) = 90 degrees

There are now two parts inside the upper part of the circle that shares area with the square:

Triangle YOP - which is a 45-45-90 degree triangle with Leg = radius = 1

And sector area of the circle Defined by central angle YOX, which is 90 degrees

The area inside the circle shared with inside the square BELOW diameter PX will be mirror images or these two parts.

Thus, the shaded area of the circle that is inside the square can be found by:

(2) * (Area Triangle YOP) + (2) * (Sector Area of the Circle defined by central angle YOX)

Since the answer choices have 7 as the DEN, most likely they are using (22/7) to approximate (pi)

This area then is equal to:

(2)   + (2)   =

1 + 11/7 =

18/7

Area remaining in Square with side of 2:

(2)^2 - (18/7) =

(28/7) - (18/7) =

10/7 square units

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A punching machine is used to create a circular hole of diameter 2 uni

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