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Let S be a point on a circle whose center is R. If PQ is a chord that [#permalink]
Any other way ?
Also, why is this not a square. Diagnols bisect at 90 in a square as well


edited: answer to myself rhombus also has bisectors at 90 degrees

Posted from my mobile device

Originally posted by ShankSouljaBoi on 31 Oct 2018, 20:13.
Last edited by ShankSouljaBoi on 25 Nov 2019, 04:31, edited 1 time in total.
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Re: Let S be a point on a circle whose center is R. If PQ is a chord that [#permalink]
Bunnel, could you please explain? Thanks so much!
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Re: Let S be a point on a circle whose center is R. If PQ is a chord that [#permalink]
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Please have a reference of the above figure posted by GMATinsight for the solution i am posting right now.

We know that radius of circle is OB and OC both.
Pick triangles OCS and SCB , they are congruent by SAS (CS=CS Angle = 90 and OS = SB given)
---> OC= CB
or OC = OB = CB ---> Equilateral triangle . Hence angle COA is 60 + 60 = 120 (Tirangle OAB is also equilateral by same reeasoning as above).

Now (120/360*2pi*OC)/2piOC = required answer = 1/3.

Please reply, if you did not get this LT2018
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Re: Let S be a point on a circle whose center is R. If PQ is a chord that [#permalink]
Since the chord PQ is the purpendicular bisector of SR (which is the radius r), it divides it into r/2 & r/2.

This forms 2 triangles with hypotenuse r and height r/2. Using properties of triangles (& trigonometry), CosX = Adj/hyp

=> CosX = 1/2 => x= 60 degrees. => total angle is 120 degrees.

=> the arc length is 120/360 times the total circumference = 1/3

Ans B
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Re: Let S be a point on a circle whose center is R. If PQ is a chord that [#permalink]
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Bunuel 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. 1/π

B. 1/3

C √3/(π+2)

D. 1/(2√2)

E. 2√3/(3π)

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
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Let S be a point on a circle whose center is R. If PQ is a chord that [#permalink]
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*midpoint* is the keyword as it indirectly leads us to the ratio of the sides of isocèles triangles.in the figure attached we can deduce that sides are in the ratio:radius:radius/2, or in the ratio x:2x ( this is a 30-60-90) isocèles triangle), and the angles formed at the circumference of the circle is 30(this 30 is formed on both ends of the arc so the inscribed angle is 30+30 . This is the inscribed angle, so according to central angle theorem, angle formed at the centre is twice the inscribed angle so 60*2=120, this is 120/360 of the the circumference of the entire circle.
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Let S be a point on a circle whose center is R. If PQ is a chord that [#permalink]
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