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# Sides AB and AC of equilateral triangle ABC are tangent to a circle at

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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
ayushkumar22941 wrote:
tough to understand without proper reference image.
i have doubt on sides ratio of triangle OHB as gmat doesnot base question on sine and cosine base

I know it is difficult to understand.

There could be a better way to solve this problem, but this how I could solve it.
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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
Bunuel wrote:
Sides AB and AC of equilateral triangle ABC are tangent to a circle at points B and C respectively. What fraction of the area of triangle ABC lies outside the circle?

(A) $$\frac{4\sqrt{3}\pi}{27} - \frac{1}{3}$$

(B) $$\frac{\sqrt{3}}{2} - \frac{\pi}{8}$$

(C) $$\frac{1}{2}$$

(D) $$\sqrt{3} - \frac{2\sqrt{3}\pi}{27}$$

(E) $$\frac{4}{3} - \frac{4\sqrt{3}\pi}{27}$$

Not understood the question....can somebody draw?
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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
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Bunuel wrote:
Sides AB and AC of equilateral triangle ABC are tangent to a circle at points B and C respectively. What fraction of the area of triangle ABC lies outside the circle?

(A) $$\frac{4\sqrt{3}\pi}{27} - \frac{1}{3}$$

(B) $$\frac{\sqrt{3}}{2} - \frac{\pi}{8}$$

(C) $$\frac{1}{2}$$

(D) $$\sqrt{3} - \frac{2\sqrt{3}\pi}{27}$$

(E) $$\frac{4}{3} - \frac{4\sqrt{3}\pi}{27}$$

Hi

I have attached a Jpeg image.

PS - I have a bad hand writing.
Attachments

Solution.jpg [ 1.84 MiB | Viewed 5747 times ]

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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
sach24x7 wrote:
Bunuel wrote:
Sides AB and AC of equilateral triangle ABC are tangent to a circle at points B and C respectively. What fraction of the area of triangle ABC lies outside the circle?

(A) $$\frac{4\sqrt{3}\pi}{27} - \frac{1}{3}$$

(B) $$\frac{\sqrt{3}}{2} - \frac{\pi}{8}$$

(C) $$\frac{1}{2}$$

(D) $$\sqrt{3} - \frac{2\sqrt{3}\pi}{27}$$

(E) $$\frac{4}{3} - \frac{4\sqrt{3}\pi}{27}$$

Not understood the question....can somebody draw?

I have added an image to my solution.

You can refer to the image and see the solution.

Hope it helps
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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
Ajiteshmathur wrote:
Refer to the attached image.

Triangle ABC is an equilateral with side = s

OB and OC are the radius of the circle = r

A perpendicular AO is drawn from A to the centre of the circle which bisects the line BC at H.

BH=HC= $$\frac{s}{2}$$

In triangle OBC, angle BOC =120 deg. (as in the quadrilateral OBAC $$\angle OBA = \angle OCA = 90^ {\circ} and \angle BAC= 60^ {\circ}. Thus \angle BOC=360-90-90-60 = 120^{\circ}$$ )

OH is perpendicular bisector on BC. (as OBC is isosceles triangle) BH= $$\frac{s}{2}$$.

As the triangle OHB is 30-60-90 triangle, the sides are in the ratio 1:$$\sqrt{3}$$:2

thus, side OH= $$\frac{s}{\sqrt{3}*2}$$ and side OB= $$\frac{s}{\sqrt{3}}$$

Also, OB = radius of the circle (r) ; thus r=$$\frac{s}{\sqrt{3}}$$

Now Area of the sector OBC= $$\frac{120}{360} * Area of the circle$$ = $$\frac{120}{360}*\pi*r^2$$= $$\frac{1}{3}*\frac{\pi s^2}{3}$$

Area of the sector OBC= $$\frac{\pi s^2}{9}$$

and Area of $$\triangle$$ OBC = $$\frac{1}{2}*s*\frac{s}{\sqrt{3}*2}$$

Part of the $$\triangle$$ ABC inside the circle= Area of the sector OBC-Area of $$\triangle$$ OBC

And Fraction of the triangle ABC inside the circle= $$\frac{Area of the sector OBC-Area of \triangle OBC}{Area of \triangle ABC}$$

Fraction of the triangle ABC inside the circle= $$\frac {\frac{\pi s^2}{9}- \frac{1}{2}*s*\frac{s}{\sqrt{3}*2}} {\frac{\sqrt {3}}{4} * s^2}$$

= $$\frac{4}{\sqrt {3}} * ( \frac{\pi}{9} - \frac{1}{4*\sqrt {3}} )$$

=$$\frac{4 \pi}{9 \sqrt {3}} - \frac{1}{3}$$

Thus fraction of the triangle ABC outside the circle = 1-( $$\frac{4 \pi}{9 \sqrt {3}} - \frac{1}{3}$$)

= $$\frac{4}{3} - \frac{4 \pi}{9 \sqrt {3}}$$

which can be written as = $$\frac{4}{3} - \frac{4 \pi \sqrt {3}}{27}$$

Kindly reach out if you find any issues with the solution.

Consider Kudos if you like the solution.

Thanks

For this part " Now Area of the sector OBC= 120360∗Areaofthecircle120360∗Areaofthecircle = 120360∗π∗r2120360∗π∗r2= 13∗πs23"

Could you explain why you chose 120 degress. I didn't quite understand it.
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Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
menonrit wrote:
Ajiteshmathur wrote:
Refer to the attached image.

Triangle ABC is an equilateral with side = s

OB and OC are the radius of the circle = r

A perpendicular AO is drawn from A to the centre of the circle which bisects the line BC at H.

BH=HC= $$\frac{s}{2}$$

In triangle OBC, angle BOC =120 deg. (as in the quadrilateral OBAC $$\angle OBA = \angle OCA = 90^ {\circ} and \angle BAC= 60^ {\circ}. Thus \angle BOC=360-90-90-60 = 120^{\circ}$$ )

OH is perpendicular bisector on BC. (as OBC is isosceles triangle) BH= $$\frac{s}{2}$$.

As the triangle OHB is 30-60-90 triangle, the sides are in the ratio 1:$$\sqrt{3}$$:2

thus, side OH= $$\frac{s}{\sqrt{3}*2}$$ and side OB= $$\frac{s}{\sqrt{3}}$$

Also, OB = radius of the circle (r) ; thus r=$$\frac{s}{\sqrt{3}}$$

Now Area of the sector OBC= $$\frac{120}{360} * Area of the circle$$ = $$\frac{120}{360}*\pi*r^2$$= $$\frac{1}{3}*\frac{\pi s^2}{3}$$

Area of the sector OBC= $$\frac{\pi s^2}{9}$$

and Area of $$\triangle$$ OBC = $$\frac{1}{2}*s*\frac{s}{\sqrt{3}*2}$$

Part of the $$\triangle$$ ABC inside the circle= Area of the sector OBC-Area of $$\triangle$$ OBC

And Fraction of the triangle ABC inside the circle= $$\frac{Area of the sector OBC-Area of \triangle OBC}{Area of \triangle ABC}$$

Fraction of the triangle ABC inside the circle= $$\frac {\frac{\pi s^2}{9}- \frac{1}{2}*s*\frac{s}{\sqrt{3}*2}} {\frac{\sqrt {3}}{4} * s^2}$$

= $$\frac{4}{\sqrt {3}} * ( \frac{\pi}{9} - \frac{1}{4*\sqrt {3}} )$$

=$$\frac{4 \pi}{9 \sqrt {3}} - \frac{1}{3}$$

Thus fraction of the triangle ABC outside the circle = 1-( $$\frac{4 \pi}{9 \sqrt {3}} - \frac{1}{3}$$)

= $$\frac{4}{3} - \frac{4 \pi}{9 \sqrt {3}}$$

which can be written as = $$\frac{4}{3} - \frac{4 \pi \sqrt {3}}{27}$$

Kindly reach out if you find any issues with the solution.

Consider Kudos if you like the solution.

Thanks

For this part " Now Area of the sector OBC= 120360∗Areaofthecircle120360∗Areaofthecircle = 120360∗π∗r2120360∗π∗r2= 13∗πs23"

Could you explain why you chose 120 degress. I didn't quite understand it.

I have applied the formula of Area of Sector =$$\frac{CentralAngle}{360}$$x Area of the Circle.
So the central angle here is $$\angle$$BOC = 120 degrees (Sum of a quadrilateral is 360, so $$\angle$$BOC = 360 -90-90-60 =120 )
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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
Ajiteshmathur wrote:
Refer to the attached image.

Triangle ABC is an equilateral with side = s

OB and OC are the radius of the circle = r

A perpendicular AO is drawn from A to the centre of the circle which bisects the line BC at H.

BH=HC= $$\frac{s}{2}$$

In triangle OBC, angle BOC =120 deg. (as in the quadrilateral OBAC $$\angle OBA = \angle OCA = 90^ {\circ} and \angle BAC= 60^ {\circ}. Thus \angle BOC=360-90-90-60 = 120^{\circ}$$ )

OH is perpendicular bisector on BC. (as OBC is isosceles triangle) BH= $$\frac{s}{2}$$.

As the triangle OHB is 30-60-90 triangle, the sides are in the ratio 1:$$\sqrt{3}$$:2

thus, side OH= $$\frac{s}{\sqrt{3}*2}$$ and side OB= $$\frac{s}{\sqrt{3}}$$

Also, OB = radius of the circle (r) ; thus r=$$\frac{s}{\sqrt{3}}$$

Now Area of the sector OBC= $$\frac{120}{360} * Area of the circle$$ = $$\frac{120}{360}*\pi*r^2$$= $$\frac{1}{3}*\frac{\pi s^2}{3}$$

Area of the sector OBC= $$\frac{\pi s^2}{9}$$

and Area of $$\triangle$$ OBC = $$\frac{1}{2}*s*\frac{s}{\sqrt{3}*2}$$

Part of the $$\triangle$$ ABC inside the circle= Area of the sector OBC-Area of $$\triangle$$ OBC

And Fraction of the triangle ABC inside the circle= $$\frac{Area of the sector OBC-Area of \triangle OBC}{Area of \triangle ABC}$$

Fraction of the triangle ABC inside the circle= $$\frac {\frac{\pi s^2}{9}- \frac{1}{2}*s*\frac{s}{\sqrt{3}*2}} {\frac{\sqrt {3}}{4} * s^2}$$

= $$\frac{4}{\sqrt {3}} * ( \frac{\pi}{9} - \frac{1}{4*\sqrt {3}} )$$

=$$\frac{4 \pi}{9 \sqrt {3}} - \frac{1}{3}$$

Thus fraction of the triangle ABC outside the circle = 1-( $$\frac{4 \pi}{9 \sqrt {3}} - \frac{1}{3}$$)

= $$\frac{4}{3} - \frac{4 \pi}{9 \sqrt {3}}$$

which can be written as = $$\frac{4}{3} - \frac{4 \pi \sqrt {3}}{27}$$

Kindly reach out if you find any issues with the solution.

Consider Kudos if you like the solution.

Thanks

This is a really tough question to crack under two min, considering we have to draw and understand the question and then this whole long calculations, do we have any simpler way to solve it? coz is it wise enough to spend too much time on a single question during the test.
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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
Bunuel wrote:
Sides AB and AC of equilateral triangle ABC are tangent to a circle at points B and C respectively. What fraction of the area of triangle ABC lies outside the circle?

(A) $$\frac{4\sqrt{3}\pi}{27} - \frac{1}{3}$$

(B) $$\frac{\sqrt{3}}{2} - \frac{\pi}{8}$$

(C) $$\frac{1}{2}$$

(D) $$\sqrt{3} - \frac{2\sqrt{3}\pi}{27}$$

(E) $$\frac{4}{3} - \frac{4\sqrt{3}\pi}{27}$$

Bunuel Kindly post the solution of this problem.
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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
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Re: Sides AB and AC of equilateral triangle ABC are tangent to a circle at [#permalink]
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