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16 Nov 2022, 11:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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A metalworker has two sheets of metal. The first sheet is in the shape of an equilateral parallelogram with two opposite angles of measure 60 degrees each. The second sheet is in the shape of a square. If the metalworker cuts out the largest possible circle from each sheet, then the areas of the two circles will be equal. What is the ratio of the area of the first sheet to the area of the second sheet?
A. 1 to √3
B. 2 to √3
C. 4 to √3
D. √3 to 1
E. √3 to 2
A. 1 to √3
B. 2 to √3
C. 4 to √3
D. √3 to 1
E. √3 to 2
17 Nov 2022, 01:34
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Let's first find the area of sheet 1 and the area of the largest circle in it
Let's say the side of each parallelogram is P and two opposite angles of 60-degree measures are given
In parallelogram consecutive angles are complementary, based on the other two angles will be 120 degrees each.
Based on the figure attached of a parallelogram, we can find its height of it based on the Pythagoras theorem.
Triangle of 30:60:90 = 1:\(\sqrt{3}\):2, the height of the parallelogram would be \(\frac{\sqrt{3}P}{2}\)
So area of the parallelogram sheet 1 = base * height = P * \(\frac{\sqrt{3}P}{2}\) = \(\frac{\sqrt{3}P^2}{2}\)
Now, the Height of the parallelogram = Diameter of the largest inscribed circle = \(\frac{\sqrt{3}P}{2}\)
so area of that circle 1 = \(\frac{\pi*Diameter^2}{4}\) = \((\frac{\pi}{4})*(\frac{\sqrt{3}P}{2})^2\)
Now, Let's count the area of sheet 2 and the area of the circle inscribed in it
Let's length of the square sheet S
So area of the square sheet 2 = \(S^2\)
Height of the square = diameter of the largest circle inscribed in it = S
So the area of circle 2 = \(\frac{\pi*Diameter^2}{4}\) = \(\frac{\pi*S^2}{4}\)
Now the area of both circles is the same,
\((\frac{\pi}{4})*(\frac{\sqrt{3}P}{2})^2\) = \(\frac{\pi*S^2}{4}\)
\(\frac{3*P^2 }{4 }\) = \(S^2\)
\(\frac{P^2 }{ S^2 }\) = \(\frac{4}{3}\)
Now, the ratio of the area of both sheets = \(\frac{Area of sheet 1 }{ Area of sheet 2}\)
= \(\frac{(\sqrt{3}P^2)}{2*(S^2)}\)
= \(\frac{\sqrt{3}}{2}\)*\(\frac{4}{3}\)
= \(\frac{2}{\sqrt{3}}\)
That the answer (B)
Let's say the side of each parallelogram is P and two opposite angles of 60-degree measures are given
In parallelogram consecutive angles are complementary, based on the other two angles will be 120 degrees each.
Based on the figure attached of a parallelogram, we can find its height of it based on the Pythagoras theorem.
Triangle of 30:60:90 = 1:\(\sqrt{3}\):2, the height of the parallelogram would be \(\frac{\sqrt{3}P}{2}\)
So area of the parallelogram sheet 1 = base * height = P * \(\frac{\sqrt{3}P}{2}\) = \(\frac{\sqrt{3}P^2}{2}\)
Now, the Height of the parallelogram = Diameter of the largest inscribed circle = \(\frac{\sqrt{3}P}{2}\)
so area of that circle 1 = \(\frac{\pi*Diameter^2}{4}\) = \((\frac{\pi}{4})*(\frac{\sqrt{3}P}{2})^2\)
Now, Let's count the area of sheet 2 and the area of the circle inscribed in it
Let's length of the square sheet S
So area of the square sheet 2 = \(S^2\)
Height of the square = diameter of the largest circle inscribed in it = S
So the area of circle 2 = \(\frac{\pi*Diameter^2}{4}\) = \(\frac{\pi*S^2}{4}\)
Now the area of both circles is the same,
\((\frac{\pi}{4})*(\frac{\sqrt{3}P}{2})^2\) = \(\frac{\pi*S^2}{4}\)
\(\frac{3*P^2 }{4 }\) = \(S^2\)
\(\frac{P^2 }{ S^2 }\) = \(\frac{4}{3}\)
Now, the ratio of the area of both sheets = \(\frac{Area of sheet 1 }{ Area of sheet 2}\)
= \(\frac{(\sqrt{3}P^2)}{2*(S^2)}\)
= \(\frac{\sqrt{3}}{2}\)*\(\frac{4}{3}\)
= \(\frac{2}{\sqrt{3}}\)
That the answer (B)
Attachments
parallelogram.png [ 4.77 KiB | Viewed 4715 times ]
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17 Nov 2022, 09:53
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Given that the circle area is equal for both circles cut out of square and equilateral parallelogram.
let the diameter of this circle be d.
Then the side of the square = d
Based on the first statement, the height of the parallelogram is also = d
That makes the side of the parallelogram = 2d/\sqrt{3}
A1/A2 = 2/\sqrt{3}
Answer is B
let the diameter of this circle be d.
Then the side of the square = d
Based on the first statement, the height of the parallelogram is also = d
That makes the side of the parallelogram = 2d/\sqrt{3}
A1/A2 = 2/\sqrt{3}
Answer is B
Attachments
IMG-2007.jpg [ 1.12 MiB | Viewed 4636 times ]
