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All the arcs inscribed in the square
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20 Mar 2024, 07:58
20 Mar 2024, 07:58
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All the arcs inscribed in the square have equal radii. Each of the arcs touches its neighbor at exactly one point.
GRe arcs.jpg [ 31.5 KiB | Viewed 794 times ]
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Attachment:
GRe arcs.jpg [ 31.5 KiB | Viewed 794 times ]
Quantity A |
Quantity B |
The area of the shaded region |
The area of the UNshaded region |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: All the arcs inscribed in the square
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27 May 2024, 15:45
27 May 2024, 15:45
1
using algebra here could get ugly, so I'll choose numbers. Let say our square has a side length of 12, this means that the radius of the circles is 3. We have 4 semi circles and 4 quarter of circles. Now that we have all our information we can solve the problem.
The area of the circle is=144
The area of the semi of 4 semi circles= 4(9pi/2)=2(9pi)
The area of the 4 quarter of circles is= 4(9pi/4)= 9pi
We add those like (2x+x=3x) it would be 3(9pi)
Now subtract 144-3(9pi)= 84.78 (area of the unshaded region)
The are of the shaded region is 144-84.78=59.22
Hence the correct answer is B
The area of the circle is=144
The area of the semi of 4 semi circles= 4(9pi/2)=2(9pi)
The area of the 4 quarter of circles is= 4(9pi/4)= 9pi
We add those like (2x+x=3x) it would be 3(9pi)
Now subtract 144-3(9pi)= 84.78 (area of the unshaded region)
The are of the shaded region is 144-84.78=59.22
Hence the correct answer is B
Re: All the arcs inscribed in the square
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25 Sep 2024, 18:08
25 Sep 2024, 18:08
1
With a simple trick, the algebraic solution isn't so bad. Let r be the radius of each semi-circle and quarter-circle.
One side of the square = 4r, so the area of the square = 16(r^2).
For the unshaded area, notice that the 4 semi-circles make two full circles,
and the 4-quarter circles make 1 full circle for a total of 3 circles worth of area.
The area of one circle is (π)(r^2), so the unshaded area = 3π(r^2).
To find the shaded area we subtract the unshaded area from the total area of the square.
This will equal 16(r^2) - 3π(r^2) and we want to compare this to 3π(r^2).
We can divide by r^2 and add 3π to each expression leaving us with 16 (shaded region) and 6π (unshaded region). 6π ≈ 6(3.14) > 16.
So the unshaded region is larger than the shaded region and the answer is B.
You may also notice that (1/2)(16) = 8 and 3π > 8,
meaning the unshaded region's area takes up more than half of the total area of the square and is therefore larger than the shaded region.
One side of the square = 4r, so the area of the square = 16(r^2).
For the unshaded area, notice that the 4 semi-circles make two full circles,
and the 4-quarter circles make 1 full circle for a total of 3 circles worth of area.
The area of one circle is (π)(r^2), so the unshaded area = 3π(r^2).
To find the shaded area we subtract the unshaded area from the total area of the square.
This will equal 16(r^2) - 3π(r^2) and we want to compare this to 3π(r^2).
We can divide by r^2 and add 3π to each expression leaving us with 16 (shaded region) and 6π (unshaded region). 6π ≈ 6(3.14) > 16.
So the unshaded region is larger than the shaded region and the answer is B.
You may also notice that (1/2)(16) = 8 and 3π > 8,
meaning the unshaded region's area takes up more than half of the total area of the square and is therefore larger than the shaded region.
