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what is the radius (in cm) of the biggest possible circle

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what is the radius (in cm) of the biggest possible circle [#permalink]

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06 Jan 2018, 04:17
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what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$

A)10($$\sqrt{3}$$ -1)
B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$
C) 10(2$$\sqrt{3}$$-3)
D) 5
E) 10$$\sqrt{3}$$
[Reveal] Spoiler: OA
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Joined: 23 Oct 2017
Posts: 64
Re: what is the radius (in cm) of the biggest possible circle [#permalink]

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06 Jan 2018, 05:15
Rocky1304 wrote:
what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$

A)10($$\sqrt{3}$$ -1)
B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$
C) 10(2$$\sqrt{3}$$-3)
D) 5
E) 10$$\sqrt{3}$$

---------------
Draw a circle with a sector subtending 120 degrees at the center. Then draw an inscribed circle. Let its raius be r
Now the r will be tangent to the radius of the bigger circle leading to 30-60-90 triangle.
sin (60) = r/10-r
since sin 60 = sqrt(3)/2

On solving r =10(2*sqrt(3)- 3))
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8030
Location: Pune, India
Re: what is the radius (in cm) of the biggest possible circle [#permalink]

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06 Jan 2018, 05:45
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Rocky1304 wrote:
what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$

A)10($$\sqrt{3}$$ -1)
B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$
C) 10(2$$\sqrt{3}$$-3)
D) 5
E) 10$$\sqrt{3}$$

Since I would like to see how various sides are related, I need to first draw the diagram.
The moment I look at the options and see $$\sqrt{3}$$ everywhere and a play of 2 on 10 cm (the radius), I start thinking about 30-60-90 triangle.

Attachment:

IMG_8275.jpg [ 1.75 MiB | Viewed 715 times ]

So this is what I have. Since the circle is inscribed in the sector, it will touch the radii at one point. So the radius of the inscribed (smaller) circle will be tangential to radius of the large circle (the one to which the sector belongs). SO I find the 30-60-90 triangle I was looking for.

$$\sqrt{3}x + 2x = 10$$
$$x = 10*(2 - \sqrt{3})$$

The radius of the small circle is $$\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})$$

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Aug 2009 Posts: 5776 Re: what is the radius (in cm) of the biggest possible circle [#permalink] Show Tags 06 Jan 2018, 06:20 Rocky1304 wrote: what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$ A)10($$\sqrt{3}$$ -1) B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$ C) 10(2$$\sqrt{3}$$-3) D) 5 E) 10$$\sqrt{3}$$ Hi.. as VeritasPrepKarishma has pointed out it is a 30-60-90 triangle.. Another way to know that - the center of the inscribed angle has to be on the bisector of this sector , so half of 120 = 60 degree .. Also the two radius containing this sector will be tangential to the circle so 90 degree.. If I am short of time, i would use the choices for my benefit.. the center of the inscribed angle has to be on the bisector of this sector, which itself is equal to the radius .. radius is 10, and the centre of the inscribed circle will be closer to the arc, so the radius will be LESS than 10/2 or 5 ONLY C is less than 5, therefore our answer Note- the choices need to be in some order - ascending or descending _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html GMAT online Tutor Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8030 Location: Pune, India Re: what is the radius (in cm) of the biggest possible circle [#permalink] Show Tags 08 Jan 2018, 05:14 VeritasPrepKarishma wrote: Rocky1304 wrote: what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$ A)10($$\sqrt{3}$$ -1) B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$ C) 10(2$$\sqrt{3}$$-3) D) 5 E) 10$$\sqrt{3}$$ Since I would like to see how various sides are related, I need to first draw the diagram. The moment I look at the options and see $$\sqrt{3}$$ everywhere and a play of 2 on 10 cm (the radius), I start thinking about 30-60-90 triangle. So this is what I have. Since the circle is inscribed in the sector, it will touch the radii at one point. So the radius of the inscribed (smaller) circle will be tangential to radius of the large circle (the one to which the sector belongs). SO I find the 30-60-90 triangle I was looking for. $$\sqrt{3}x + 2x = 10$$ $$x = 10*(2 - \sqrt{3})$$ The radius of the small circle is $$\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})$$ Answer (C) Quote: Hi Karishma I understood that 10 is the radius of the sector so that is why it is 3‾√x+2x=10 But I did not understand how did you form this equation. x=10∗(2−3‾√) We have rationalised the denominator to get this: $$\sqrt{3}x + 2x = 10$$ $$x = \frac{10}{(\sqrt{3} + 2)}$$ $$x = \frac{10*(\sqrt{3} - 2)}{(\sqrt{3} + 2)*(\sqrt{3} - 2)}$$ $$x = \frac{10*(\sqrt{3} - 2)}{\sqrt{3}^2 - 2^2}$$ (because (a+b)(a-b) = a^2 - b^2) $$x = \frac{10*(\sqrt{3} - 2)}{3 - 4}$$ $$x = 10*(2 - \sqrt{3})$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Joined: 12 Sep 2017
Posts: 15
Re: what is the radius (in cm) of the biggest possible circle [#permalink]

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08 Jan 2018, 07:09
VeritasPrepKarishma wrote:
Rocky1304 wrote:
what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$

A)10($$\sqrt{3}$$ -1)
B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$
C) 10(2$$\sqrt{3}$$-3)
D) 5
E) 10$$\sqrt{3}$$

Since I would like to see how various sides are related, I need to first draw the diagram.
The moment I look at the options and see $$\sqrt{3}$$ everywhere and a play of 2 on 10 cm (the radius), I start thinking about 30-60-90 triangle.

Attachment:
IMG_8275.jpg

So this is what I have. Since the circle is inscribed in the sector, it will touch the radii at one point. So the radius of the inscribed (smaller) circle will be tangential to radius of the large circle (the one to which the sector belongs). SO I find the 30-60-90 triangle I was looking for.

$$\sqrt{3}x + 2x = 10$$
$$x = 10*(2 - \sqrt{3})$$ I'm confused.. How does this make sense?

The radius of the small circle is $$\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})$$

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8030
Location: Pune, India
Re: what is the radius (in cm) of the biggest possible circle [#permalink]

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09 Jan 2018, 11:38
HarryAxel wrote:
VeritasPrepKarishma wrote:
Rocky1304 wrote:
what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$

A)10($$\sqrt{3}$$ -1)
B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$
C) 10(2$$\sqrt{3}$$-3)
D) 5
E) 10$$\sqrt{3}$$

Since I would like to see how various sides are related, I need to first draw the diagram.
The moment I look at the options and see $$\sqrt{3}$$ everywhere and a play of 2 on 10 cm (the radius), I start thinking about 30-60-90 triangle.

Attachment:
IMG_8275.jpg

So this is what I have. Since the circle is inscribed in the sector, it will touch the radii at one point. So the radius of the inscribed (smaller) circle will be tangential to radius of the large circle (the one to which the sector belongs). SO I find the 30-60-90 triangle I was looking for.

$$\sqrt{3}x + 2x = 10$$
$$x = 10*(2 - \sqrt{3})$$ I'm confused.. How does this make sense?

The radius of the small circle is $$\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})$$

Check: https://gmatclub.com/forum/what-is-the- ... l#p1993435
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8030 Location: Pune, India Re: what is the radius (in cm) of the biggest possible circle [#permalink] Show Tags 09 Jan 2018, 11:41 VeritasPrepKarishma wrote: Rocky1304 wrote: what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$ A)10($$\sqrt{3}$$ -1) B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$ C) 10(2$$\sqrt{3}$$-3) D) 5 E) 10$$\sqrt{3}$$ Since I would like to see how various sides are related, I need to first draw the diagram. The moment I look at the options and see $$\sqrt{3}$$ everywhere and a play of 2 on 10 cm (the radius), I start thinking about 30-60-90 triangle. Attachment: IMG_8275.jpg So this is what I have. Since the circle is inscribed in the sector, it will touch the radii at one point. So the radius of the inscribed (smaller) circle will be tangential to radius of the large circle (the one to which the sector belongs). SO I find the 30-60-90 triangle I was looking for. $$\sqrt{3}x + 2x = 10$$ $$x = 10*(2 - \sqrt{3})$$ The radius of the small circle is $$\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})$$ Answer (C) Quote: I did not get how did you take 3‾√x+2x=10 . So far I understand that it is because of the c^2=a^2+b^2, but it is not mentioned that C^2 will be 10. I am missing something. Look at the diagram above. DO + AO is the radius of the sector. This is given to be 10. So $$2x + \sqrt{3}x = 10$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Status: Turning my handicaps into assets
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what is the radius (in cm) of the biggest possible circle [#permalink]

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10 Jan 2018, 06:31
VeritasPrepKarishma wrote:
Rocky1304 wrote:
what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120$$^{\circ}$$

A)10($$\sqrt{3}$$ -1)
B)20($$\sqrt{3}$$-1)/$$\sqrt{3}$$
C) 10(2$$\sqrt{3}$$-3)
D) 5
E) 10$$\sqrt{3}$$

Since I would like to see how various sides are related, I need to first draw the diagram.
The moment I look at the options and see $$\sqrt{3}$$ everywhere and a play of 2 on 10 cm (the radius), I start thinking about 30-60-90 triangle.

Attachment:
IMG_8275.jpg

So this is what I have. Since the circle is inscribed in the sector, it will touch the radii at one point. So the radius of the inscribed (smaller) circle will be tangential to radius of the large circle (the one to which the sector belongs). SO I find the 30-60-90 triangle I was looking for.

$$\sqrt{3}x + 2x = 10$$
$$x = 10*(2 - \sqrt{3})$$

The radius of the small circle is $$\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})$$

Hi karishma,

How did you find the bisector of angle BDC?
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what is the radius (in cm) of the biggest possible circle   [#permalink] 10 Jan 2018, 06:31
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