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Let S be a point on a circle whose center is R. If PQ is a c
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24 Feb 2020, 10:01
24 Feb 2020, 10:01
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Let S be a point on a circle whose center is R. If PQ is a chord that passes perpendicularly through the midpoint of RS, then the length of arc PSQ is what fraction of the circle’s circumference?
A. \(\frac{1}{\pi}\)
B. \(\frac{1}{3}\)
C. \(\frac{3}{\pi+2}\)
D. \(\frac{1}{2 \sqrt{2}}\)
E. \(\frac{2 \sqrt{3} }{3 \pi}\)
Kudos for the right answer and explanation[/m]
A. \(\frac{1}{\pi}\)
B. \(\frac{1}{3}\)
C. \(\frac{3}{\pi+2}\)
D. \(\frac{1}{2 \sqrt{2}}\)
E. \(\frac{2 \sqrt{3} }{3 \pi}\)
Kudos for the right answer and explanation[/m]
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Re: Let S be a point on a circle whose center is R. If PQ is a c
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22 Oct 2021, 06:38
22 Oct 2021, 06:38
2
Carcass wrote:
Let S be a point on a circle whose center is R. If PQ is a chord that passes perpendicularly through the midpoint of RS, then the length of arc PSQ is what fraction of the circle’s circumference?
A. \(\frac{1}{\pi}\)
B. \(\frac{1}{3}\)
C. \(\frac{3}{\pi+2}\)
D. \(\frac{1}{2 \sqrt{2}}\)
E. \(\frac{2 \sqrt{3} }{3 \pi}\)
A. \(\frac{1}{\pi}\)
B. \(\frac{1}{3}\)
C. \(\frac{3}{\pi+2}\)
D. \(\frac{1}{2 \sqrt{2}}\)
E. \(\frac{2 \sqrt{3} }{3 \pi}\)
Here's what the diagram looks like:
From here let's add lines from the center (R) to P and Q.
At the same time, let's say the radius of the circle is 2, which means we get the following measurements:
We now have two small right triangles in our diagram.
Since we know the length of two of the three sides, we can apply the Pythagorean theorem to find the length of the third sides:
Notice that the two right triangles have the lengths 1, 2, and √3, which are the lengths of the base 30-60-90 right triangle.
This means we add the following angles to our diagram:
We can now see that angle PRQ = 120°
The entire circumference of the circle encompasses an angle of 360°
So the fraction of the circumference occupied by arc PSQ = 120°/360° = 1/3
Answer: B
General Discussion
Re: Let S be a point on a circle whose center is R. If PQ is a c
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13 Mar 2020, 14:05
13 Mar 2020, 14:05
1
Let the midpoint of \(RS\) be \(M\). Consider the right angled triangle \(PMR\). Observe that \(PR\) is the radius of the circle and hence it is double the length of \(RM\), which itself is half of the radius \(RS\). Let angle \(PRM\) be \(x\). Then, \(cos(x)=RM/PR =RM/(2*RM) =1/2\). Therefore, \(x=60\) degrees. A similar triangle can be constructed using point \(Q\) instead of \(P\) from which we get the total angle displaced by the arc \(PSQ = 2*60 = 120\) degrees
Required ratio = \((120/360*circumference) / circumference = 1/3\)
Required ratio = \((120/360*circumference) / circumference = 1/3\)
