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In the xy-plane, two circular regions are centere at (-6, 5)
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Updated on: 24 Oct 2020, 02:19
Updated on: 24 Oct 2020, 02:19
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83% (01:05) correct
16% (01:33) wrong based on 18 sessions
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In the xy-plane, two circular regions are centered at (-6, 5). One of the regions is tangent to the x-axis and the other is tangent to the y-axis.
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.
Quantity A |
Quantity B |
The area of the larger region minus the area of the smaller region |
\(9\pi\) |
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.
Originally posted by bunterocks on 23 Oct 2020, 20:57.
Last edited by Carcass on 24 Oct 2020, 02:19, edited 1 time in total.
Last edited by Carcass on 24 Oct 2020, 02:19, edited 1 time in total.
Formatted
Re: In the xy-plane, two circular regions are centere at (-6, 5)
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24 Oct 2020, 06:57
24 Oct 2020, 06:57
3
The larger circle has radius equal 6. The smaller circle has radius equal to 5.
The area of the larger circle minus the smaller circle is \(36\pi -25\pi=11\pi\)
Answer is A.
The area of the larger circle minus the smaller circle is \(36\pi -25\pi=11\pi\)
Answer is A.
Re: In the xy-plane, two circular regions are centere at (-6, 5)
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26 Oct 2020, 02:29
26 Oct 2020, 02:29
1
Drawing an x-y axis with the point helps visualise this problem better. The point (-6,5) lies in the second quadrant. We are told this is the centre of two circles, one that touches the x-axis and another that touches the y-axis. Naturally, the circle touching the x-axis would be smaller since the centre point (-6,5) is closer to the x axis. Similarly, the larger circle would touch the y-axis. Since we have the point we can determine the radii. The smaller circle will have a radius of 5 whereas the larger one will have a radius of 6.
Now we can find the difference in area,
\(36\pi -25\pi=11\pi\)
Now we can find the difference in area,
\(36\pi -25\pi=11\pi\)
