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Re: In the figure above, inscribed polygon PQRST is equilateral. If the [#permalink]

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14 Nov 2017, 23:37

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Circumference of PQRST= P*D=10P Given that polygon is equilateral we understand that circumference of PQR would be 2/5 of PQRST, hence 2/5*10P=4P Answer B

Re: In the figure above, inscribed polygon PQRST is equilateral. If the [#permalink]

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15 Nov 2017, 02:06

Alexey1989x wrote:

Circumference of PQRST= P*D=10P Given that polygon is equilateral we understand that circumference of PQR would be 2/5 of PQRST, hence 2/5*10P=4P Answer B

Re: In the figure above, inscribed polygon PQRST is equilateral. If the [#permalink]

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15 Nov 2017, 02:35

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rocko911 wrote:

Alexey1989x wrote:

Circumference of PQRST= P*D=10P Given that polygon is equilateral we understand that circumference of PQR would be 2/5 of PQRST, hence 2/5*10P=4P Answer B

Can u plz explain this?

I'll try. The answer to the question boils down to finding the circumference of the circle, which can be found by using the formula: 2*P (3,14)*r and finding the proportion of arc PQR to the circle PQRST.

1) Let's calculate the circumference 2*P (3,14)*r =10P since we are given the value of the diameter. 2) Let's find the relationship between arc PQR to circumference of the PQRST Let's assume that circle has point O as its centre. Then we cut polygon into 5 equal triangles with one shared vertice O: PQO; QOR;ROS;SOT;TPO. Each trangle is 1/5 of the polygon and its length equals to 1/5 of the circumference of the circle. So, we see that arc PQR covers two triangles and hence we can deduce that PQR would constitute 2/5 of the circumference of the circle.

2/5*10P=4P

Hope it's clear. If not, you should go to the basics of geometry.

rocko911 - in fairness, if I recall correctly (and I may not), many properties of pentagons inscribed in circles involve trig.

Not true here, but you do have to know (or derive): An equilateral \(n\)-gon inscribed in a circle divides the circle's circumference into \(n\) pieces.

The distance between the vertices is the same. The vertices lie on the circle. The chords that intercept the arcs are equal. If two chords are equal in measure, then their corresponding minor arcs (PQ and QR) are equal in measure.

Here, an inscribed equilateral pentagon's vertices divide the circumference into 5 arcs.

We are asked for the arc length of arc PQR. That length is a fraction of the circumference's length.

Count arc segments: from P to Q, then Q to R, there are 2 arc segments out of 5 total. Arc PQR hence is \(\frac{2}{5}\) of the circumference of the circle.

Circumference of circle? Diameter = 10 = 2r. So r = 5 Circumference = \(2\pi*r=10\pi\)

Length of arc PQR? \((\frac{2}{5} * 10\pi) = 4\pi\)

Since polygon PQRST is equilateral, it breaks the circle into 5 arcs of equal length. Since PQR represents 2 arcs, that is equivalent to 2/5 of the circumference. Since the diameter of the circle is 10, the circumference is 10π and arc PQR is:

10π x 2/5 = 4π

Answer: B
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In the figure above, inscribed polygon PQRST is equilateral. If the [#permalink]

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27 Nov 2017, 21:18

Made a mistake edited the same

The sum of all the angles of a regular(equal sides) polygon=(n-2)*180 n=total number of sides here n=5 The sum of angles=(5-2)*180=540 Each angle=540/5=108 Let us create a triangle POQ with OP and OQ being its sides and they split the angle P and Q to half each ie 54. Now the sum of the triangle is 180 hence the angle O will be 72. Arc PQR has 2 such angles=72+72=144 Length of arc=144/360*(2pi5)-Here 5=radius of the circle which is given-half of diameter(10/2)

=4Pi

Last edited by gps5441 on 28 Nov 2017, 01:36, edited 1 time in total.

In the figure above, inscribed polygon PQRST is equilateral. If the [#permalink]

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28 Nov 2017, 00:34

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gps5441 wrote:

The sum of all the angles of a regular(equal sides) polygon=(n-2)*180 n=total number of sides here n=5 The sum of angles=(5-2)*180=360 Each angle=360/5=72 Arc PQR has 2 such angles=72+72=144(if o is the center of the circle than angle opq +angle orq)

Length of arc=144/360*(2pi5)-Here 5=radius of the circle which is given-half of diameter(10/2) =4Pi

If n is 5, then (n-2) * 180 = (5-2) * 180 = 3 * 180 = 540. Each Angle of the polygon would have to be 540/5 = 108.

Still, going by the idea we create an isosceles triangle POQ, we split the 108's in two = leaving two angles of 54 and a remainder of 72 stemming from the origin of the circle. Double the 72 to acquire the central angle that encompasses the arc PQR, 144. 144 / 360 = (12 * 12) / (12 * 30) = (12 / 30) = (6 / 15) = (2 / 5)

Multiply that by the circumference 10pi to get 4pi

genxer123's tidbit about n-sided, equilateral polygons inscribed in a circle though would've saved a lot more time; thanks for sharing that.

Re: In the figure above, inscribed polygon PQRST is equilateral. If the [#permalink]

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28 Nov 2017, 01:36

Silver65x wrote:

gps5441 wrote:

The sum of all the angles of a regular(equal sides) polygon=(n-2)*180 n=total number of sides here n=5 The sum of angles=(5-2)*180=360 Each angle=360/5=72 Arc PQR has 2 such angles=72+72=144(if o is the center of the circle than angle opq +angle orq)

Length of arc=144/360*(2pi5)-Here 5=radius of the circle which is given-half of diameter(10/2) =4Pi

If n is 5, then (n-2) * 180 = (5-2) * 180 = 3 * 180 = 540. Each Angle of the polygon would have to be 540/5 = 108.

Still, going by the idea we create an isosceles triangle POQ, we split the 108's in two = leaving two angles of 54 and a remainder of 72 stemming from the origin of the circle. Double the 72 to acquire the central angle that encompasses the arc PQR, 144. 144 / 360 = (12 * 12) / (12 * 30) = (12 / 30) = (6 / 15) = (2 / 5)

Multiply that by the circumference 10pi to get 4pi

genxer123's tidbit about n-sided, equilateral polygons inscribed in a circle though would've saved a lot more time; thanks for sharing that.

Thank you for pointing out the mistake in my mistake