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Square FGHJ has sides of length x. Region FGJ is a sector of
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15 Nov 2019, 17:14
15 Nov 2019, 17:14
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#greprepclub Square FGHJ has sides .jpg [ 11.3 KiB | Viewed 8563 times ]
Square \(FGHJ\) has sides of length x. Region \(FGJ\) is a sector of the circle with center \(F\).
Quantity A |
Quantity B |
The area of the shaded region |
\(\frac{x^2}{4}\) |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Kudos for the right answer and solution.
Re: Square FGHJ has sides of length x. Region FGJ is a sector of
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15 Nov 2019, 23:12
15 Nov 2019, 23:12
1
B .
Area of square is x^2
Area of circle is Pi * x^2.
Area of sector is Pi *x^2/4
Area of shaded region is x^2 - Pi * x^2/4 = (4 -Pi) x^2 /4
As Pi >3 , (4- Pi) <1 , So area is less than x^2/4
Area of square is x^2
Area of circle is Pi * x^2.
Area of sector is Pi *x^2/4
Area of shaded region is x^2 - Pi * x^2/4 = (4 -Pi) x^2 /4
As Pi >3 , (4- Pi) <1 , So area is less than x^2/4
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Square FGHJ has sides of length x. Region FGJ is a sector of
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08 Feb 2021, 06:40
08 Feb 2021, 06:40
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Carcass wrote:
Attachment:
#greprepclub Square FGHJ has sides .jpg
Square \(FGHJ\) has sides of length x. Region \(FGJ\) is a sector of the circle with center \(F\).
Quantity A |
Quantity B |
The area of the shaded region |
\(\frac{x^2}{4}\) |
Area of shaded region = (area of square FGHJ) - (area of sector FGJ)
(area of square FGHJ
Since each side of the square has length x, the area = \(x^2\)
area of sector FGJ
Area of sector = \((\frac{n}{360})(\pi)(radius^2)\), where \(n =\) the central angle
Since FGHJ is a square, we know that the central angle (angle GFJ) \(= 90°\)
We also know that the radius \(= x\)
So, the area of sector FGJ \(= (\frac{90}{360})(\pi)(x^2)= (\frac{1}{4})(\pi)(x^2)=\) \(\frac{\pi x^2}{4}\)
Area of shaded region = \(x^2\) \(-\) \(\frac{\pi x^2}{4}\)
= \(\frac{4x^2}{4}\) \(-\) \(\frac{\pi x^2}{4}\)
= \(\frac{4x^2 - \pi x^2}{4}\)
We get:
Quantity A: \(\frac{4x^2 - \pi x^2}{4}\)
Quantity B: \(\frac{x^2}{4}\)
Multiply both quantities by \(4\) to get:
Quantity A: \(4x^2 - \pi x^2\)
Quantity B: \(x^2\)
Divide book quantities by \(x^2\) to get:
Quantity A: \(4 - \pi \)
Quantity B: \(1\)
In other words we have:
Quantity A: \(4 - 3.14 = 0.86 \)
Quantity B: \(1\)
Answer: B
