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Geometry: Square and circle problem

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Geometry: Square and circle problem

by AndyB » Sun Oct 09, 2011 3:30 am
Hi Everyone,

I am having difficulty in solving a simple geometry problem.Need help in understanding the problem and also the solution.

Please help!!!

1) A punching machine is used to create a circular diameter 2 unit from a square sheet of aluminium 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 the diagonal of the square.The area of the sheet that remains after pucnching is...???

a)10/7 square unit
b)22/7 square unit
c)5/7 square unit
d)9/7 square unit
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by GMATGuruNY » Sun Oct 09, 2011 5:43 am
AndyB wrote:Hi Everyone,

I am having difficulty in solving a simple geometry problem.Need help in understanding the problem and also the solution.

Please help!!!

1) A punching machine is used to create a circular diameter 2 unit from a square sheet of aluminium 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 the diagonal of the square.The area of the sheet that remains after pucnching is...???

a)10/7 square unit
b)22/7 square unit
c)5/7 square unit
d)9/7 square unit
Image

An inscribed angle is formed by two chords. Angle APC is an inscribed angle formed by chords PA and PC.
An inscribed angle of 90 degrees intersects the diameter.
Thus, AC is the diameter of the circle.
It is given that PB is a diameter of the circle.
Since diameters intersect at the center, O is the center of the circle.

Thus:
OA, OP and OC are all radii of length 1.
Triangles APO and POC are 45-45-90 triangles.
Since the sides of a 45-45-90 triangle are proportioned s : s : s√2, PA = √2 and PC = √2.

Thus:
The area of triangle APC = (1/2)bh = (1/2)(√2)(√2) = 1.
The area of semicircle ABC = (1/2)πr² = (1/2)π(1²) = π/2.
The area of the square = s² = 2² = 4.

Thus:
The remaining area after punching = square - triangle APC - semicircle ABC = 4 - 1 - π/2 ≈ 1.4.

The correct answer is A.
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by parul9 » Sun Oct 09, 2011 5:49 am
GMATGuruNY wrote:
AndyB wrote:Hi Everyone,

I am having difficulty in solving a simple geometry problem.Need help in understanding the problem and also the solution.

Please help!!!

1) A punching machine is used to create a circular diameter 2 unit from a square sheet of aluminium 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 the diagonal of the square.The area of the sheet that remains after pucnching is...???

a)10/7 square unit
b)22/7 square unit
c)5/7 square unit
d)9/7 square unit
Image

An inscribed angle is formed by two chords. Angle APC is an inscribed angle formed by chords PA and PC.
An inscribed angle of 90 degrees intersects the diameter.
Thus, AC is the diameter of the circle.
It is given that PB is a diameter of the circle.
Since diameters intersect at the center, O is the center of the circle.

Thus:
OA, OP and OC are all radii of length 1.
Triangles APO and POC are 45-45-90 triangles.
Since the sides of a 45-45-90 triangle are proportioned s : s : s√2, PA = √2 and PC = √2.

Thus:
The area of triangle APC = (1/2)bh = (1/2)(√2)(√2) = 1.
The area of semicircle ABC = (1/2)πr² = (1/2)π(1²) = π/2.
The area of the square = s² = 2² = 4.

Thus:
The remaining area after punching = square - triangle APC - semicircle ABC = 4 - 1 - π/2 ≈ 1.4.

The correct answer is A.
Superb explanation!
Thanks!
Is there any material available for working on geometry fundamentals?

Parul
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by ritzzzr » Mon Oct 10, 2011 5:04 am
We join the points where the circle intersects the sides of the square be A and B,thus forming a right angled triangle PAB so angle subtended at center will be 180(property of circle) hence the points of intersection form the diameter ,

Now requuired area:

area of square - (area of triangle PAB + Area of semi circle AB)

Area of square= 2^2=4

Area of Triangle PAB = 1/2*AB(which is diameter)*OP(that is radius and O is Center)
=1/2*2*1
=1
Area of semi circle =1/2*22/7*1*1
=11/7
So required Area = 4-(1+11/7)
=4-(18/7)
=(28-18)/7
=10/7
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by AndyB » Mon Oct 17, 2011 1:36 am
Thanks everyone..For your time@#$!@#
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