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06 Nov 2022, 03:24
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Question Stats:
67% (02:20) correct
33%
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A circular piece of metal of maximum size is cut out of a larger square piece. Then a square piece of maximum size is cut out of that circular piece. How much metal is wasted in total?
(A) 1/2 the area of the original square
(B) 1/4 the area of the original square
(C) 1/2 the area of the circular piece
(D) 1/4 the area of the circular piece
(E) none of these
(A) 1/2 the area of the original square
(B) 1/4 the area of the original square
(C) 1/2 the area of the circular piece
(D) 1/4 the area of the circular piece
(E) none of these
06 Nov 2022, 05:10
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Let the side of the square be a. the area of the original square is a^2
the circle cut from the square will have a radius of 1/2a since the diameter runs at center of the square.
The square cut from this circle will have a side of 2sqrt {(1/2)a}^2 = since the radius of the circle intersects at 90 degrees at the center forming a right angle. so the side of the square will have a/ sqrt of 2. the area then of the new square will be (a^2)/2. so the lost area is a^2 - (a^2)/2= (a^2)/2. A
the circle cut from the square will have a radius of 1/2a since the diameter runs at center of the square.
The square cut from this circle will have a side of 2sqrt {(1/2)a}^2 = since the radius of the circle intersects at 90 degrees at the center forming a right angle. so the side of the square will have a/ sqrt of 2. the area then of the new square will be (a^2)/2. so the lost area is a^2 - (a^2)/2= (a^2)/2. A
06 Nov 2022, 06:19
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Let's assume the length of the side of the outer square is 4 units as shown below.
Its given that circular piece of metal of maximum size is cut out of this square. Therefore, the length of the diameter of the circle is equal to the side of the square.
Diameter of the circle is = 2 units
Now, a square piece of maximum size is cut out of that circular piece. Hence, the diagonal of the inner square is equal to the diameter of the circle.
Let's assume the side of the inner square (PQRS) = x
Diagonal of PQRS = \(\sqrt{2}x\)
\(\sqrt{2}x\) = 4
x = \(\frac{4}{\sqrt{2}}\)
Area of inner square (PQRS)= \(x^2\) = \(\frac{16}{2}\) = 8 units
Area of outer square (ABCD) = \(4^2\) = 16 units
Metal wasted= 16 - 8 = 8 units
Let's refer to the given options -
(A) 1/2 the area of the original square - True
Option A
Its given that circular piece of metal of maximum size is cut out of this square. Therefore, the length of the diameter of the circle is equal to the side of the square.
Diameter of the circle is = 2 units
Now, a square piece of maximum size is cut out of that circular piece. Hence, the diagonal of the inner square is equal to the diameter of the circle.
Let's assume the side of the inner square (PQRS) = x
Diagonal of PQRS = \(\sqrt{2}x\)
\(\sqrt{2}x\) = 4
x = \(\frac{4}{\sqrt{2}}\)
Area of inner square (PQRS)= \(x^2\) = \(\frac{16}{2}\) = 8 units
Area of outer square (ABCD) = \(4^2\) = 16 units
Metal wasted= 16 - 8 = 8 units
Let's refer to the given options -
(A) 1/2 the area of the original square - True
Option A
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