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Re: what is the radius (in cm) of the biggest possible circle [#permalink]
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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\(\sqrt{3}\)



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

\(x = \frac{10*(\sqrt{3} - 2)}{3 - 4}\)

\(x = 10*(2 - \sqrt{3})\)
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Re: what is the radius (in cm) of the biggest possible circle [#permalink]
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 \(\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})\)

Answer (C)
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Re: what is the radius (in cm) of the biggest possible circle [#permalink]
Expert Reply
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 \(\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})\)

Answer (C)


Check: https://gmatclub.com/forum/what-is-the- ... l#p1993435
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Re: what is the radius (in cm) of the biggest possible circle [#permalink]
Expert Reply
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 \(2x + \sqrt{3}x = 10\)
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what is the radius (in cm) of the biggest possible circle [#permalink]
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)



Hi karishma,

How did you find the bisector of angle BDC?
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Re: what is the radius (in cm) of the biggest possible circle [#permalink]
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