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Let S be a point on a circle whose center is R. If PQ is a chord that

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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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New post 31 Oct 2018, 02:21
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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. 1/π

B. 1/3

C √3/(π+2)

D. 1/(2√2)

E. 2√3/(3π)
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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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New post 12 Nov 2018, 08:59
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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)



Ok Let us solve with property of triangles 30-60-90..
look at the attached figure..
Take triangle RPO....
It is a right angled triangle with sides RO:PO:RP in the ratio of r/2:PO:r or 1:PO:2.
This is nothing but a 30-60-90 triangle with ratio of sides as 1:\(\sqrt{3}\):2
so Angle PRO is 60 and similarly Angle QRO is also 60

Thus the central angle is 120 and the arc made by it will be \(\frac{120}{360}=\frac{1}{3}\)

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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New post 31 Oct 2018, 09:41
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)


The figure looks like attached figure.

Here we see that the quadrilaterla is a rhombus which in itself is equilavelt to two equilateral triangles.

i.e. Angle at the centre = 60+60 = 120º

i.e. Length of Arc = (120/360)* Circumference = (1/3)*Circumference

i.e. Answer: Option 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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New post Updated on: 25 Nov 2019, 04:31
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]

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New post 12 Nov 2018, 08:14
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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New post 12 Nov 2018, 08:35
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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]

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New post 21 Nov 2019, 23:14
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

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Re: Let S be a point on a circle whose center is R. If PQ is a chord that   [#permalink] 21 Nov 2019, 23:14
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