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The two lines are tangent to the circle.

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The two lines are tangent to the circle.  [#permalink]

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New post 24 Jul 2018, 04:45
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The two lines are tangent to the circle. If AC = 10 and AB = 10√3, what is the area of the circle?

A) 100π
B) 150π
C) 200π
D) 250π
E) 300π

*kudos for all correct solutions

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Re: The two lines are tangent to the circle.  [#permalink]

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New post 26 Jul 2018, 06:01
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GMATPrepNow wrote:
Image

The two lines are tangent to the circle. If AC = 10 and AB = 10√3, what is the area of the circle?

A) 100π
B) 150π
C) 200π
D) 250π
E) 300π

*kudos for all correct solutions


If AC = 10, then BC = 10
Image

Since ABC is an isosceles triangle, the following gray line will create two right triangles...
Image

Now focus on the following blue triangle. Its measurements have a lot in common with the BASE 30-60-90 special triangle
Image

In fact, if we take the BASE 30-60-90 special triangle and multiply all sides by 5 we see that the sides are the same as the sides of the blue triangle.
Image

So, we can now add in the 30-degree and 60-degree angles
Image

Now add a point for the circle's center and draw a line to the point of tangency. The two lines will create a right triangle (circle property)
Image

We can see that the missing angle is 60 degrees
Image

Now create the following right triangle
Image

We already know that one side has length 5√3
Image

Since we have a 30-60-90 special triangle, we know that the hypotenuse is twice as long as the side opposite the 30-degree angle.
Image
So, the hypotenuse must have length 10√3

In other words, the radius has length 10√3

What is the area of the circle?
Area = πr²
= π(10√3)²
= π(10√3)(10√3)
= 300π

Answer: E

Cheers,
Brent
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The two lines are tangent to the circle.  [#permalink]

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New post 24 Jul 2018, 06:00
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*Let O be the centre of circle.
*AC=BC=10 (Power point theorem)
*Let E be perpendicular bisector of line AB.
*AE=BE=5√3
*Triangle AEC: as per side ratio of a:a√3:2a, angle ACE is 60.
*Triangle AOC: angle OAC is 90, angle ACO is 60 and angle AOC is 30. Side AC opposite angle 30 is 10. Therefore, Side AO (Radius) opposite angle 60 is 10√3.
Hence, area of circle= (10√3)^2*π= 300π

Ans E

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Re: The two lines are tangent to the circle.  [#permalink]

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New post 24 Jul 2018, 06:02
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GMATPrepNow wrote:
Image

The two lines are tangent to the circle. If AC = 10 and AB = 10√3, what is the area of the circle?

A) 100π
B) 150π
C) 200π
D) 250π
E) 300π

*kudos for all correct solutions


Look at triangle ACD..
It is 30:60:90 triangle as sides are _:5√3:10
Now join OA, where O is the centre..
angle OAD = OAC-DAC=90-30=60... OAC is 90 as it is tangent..
So OAD becomes 30:60:90.. radius is hypotenuse =2*AD=2*5√3=10√3
Area =π*(10√3)^2=300π

E
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Re: The two lines are tangent to the circle.  [#permalink]

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New post 24 Jul 2018, 06:24
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OA:E
Attachment:
gmatprepnow.PNG
gmatprepnow.PNG [ 69.75 KiB | Viewed 2634 times ]

In above sketch, \(OA=OB=\) radius
\(\angle OAC = \angle OBC =\angle ADC =\angle BDC = 90^{\circ}\)
\(AD= DB = \frac{{10\sqrt[]{3}}}{{2}}=5\sqrt[]{3}\)

\(AD =5\sqrt[]{3}\) and\(\angle ADC = 90^{\circ}\) , along with \(AC= 10\),
We can say that \(\triangle ADC\) is \(30^{\circ}-60^{\circ}-90^{\circ}\) Triangle with \(\angle ACD = 60^{\circ}\)

Now in \(\triangle\) \(OAC\) ,\(\angle AOC =30^{\circ}\)

\(\angle AOB = 2 * \angle AOC = 60^{\circ}\)

This imply that \(\triangle AOB\) is equilateral \(\triangle\) , with all sides equal to \(10\sqrt{3}\).
\(OA=OB=\) radius \(=10\sqrt{3}\)

Area of circle\(= \pi * 10\sqrt{3}*10\sqrt{3} =300\pi\)
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Re: The two lines are tangent to the circle.  [#permalink]

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New post 24 Jul 2018, 09:37
Princ wrote:
OA:E
Attachment:
gmatprepnow.PNG

In above sketch, \(OA=OB=\) radius
\(\angle OAC = \angle OBC =\angle ADC =\angle BDC = 90^{\circ}\)
\(AD= DB = \frac{{10\sqrt[]{3}}}{{2}}=5\sqrt[]{3}\)

\(AD =5\sqrt[]{3}\) and\(\angle ADC = 90^{\circ}\) , along with \(AC= 10\),
We can say that \(\triangle ADC\) is \(30^{\circ}-60^{\circ}-90^{\circ}\) Triangle with \(\angle ACD = 60^{\circ}\)

Now in \(\triangle\) \(OAC\) ,\(\angle AOC =30^{\circ}\)

\(\angle AOB = 2 * \angle AOC = 60^{\circ}\)

This imply that \(\triangle AOB\) is equilateral \(\triangle\) , with all sides equal to \(10\sqrt{3}\).
\(OA=OB=\) radius \(=10\sqrt{3}\)

Area of circle\(= \pi * 10\sqrt{3}*10\sqrt{3} =300\pi\)



I understand that 2 of the sides are 10 and 5\sqrt{3}
How can we conclude that the triangle here is a 30 :60:90 triangle ?
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Re: The two lines are tangent to the circle.  [#permalink]

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New post 24 Jul 2018, 09:43
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Top Contributor
CounterSniper wrote:
Princ wrote:
OA:E
Attachment:
gmatprepnow.PNG

In above sketch, \(OA=OB=\) radius
\(\angle OAC = \angle OBC =\angle ADC =\angle BDC = 90^{\circ}\)
\(AD= DB = \frac{{10\sqrt[]{3}}}{{2}}=5\sqrt[]{3}\)

\(AD =5\sqrt[]{3}\) and\(\angle ADC = 90^{\circ}\) , along with \(AC= 10\),
We can say that \(\triangle ADC\) is \(30^{\circ}-60^{\circ}-90^{\circ}\) Triangle with \(\angle ACD = 60^{\circ}\)

Now in \(\triangle\) \(OAC\) ,\(\angle AOC =30^{\circ}\)

\(\angle AOB = 2 * \angle AOC = 60^{\circ}\)

This imply that \(\triangle AOB\) is equilateral \(\triangle\) , with all sides equal to \(10\sqrt{3}\).
\(OA=OB=\) radius \(=10\sqrt{3}\)

Area of circle\(= \pi * 10\sqrt{3}*10\sqrt{3} =300\pi\)



I understand that 2 of the sides are 10 and 5\sqrt{3}
How can we conclude that the triangle here is a 30 :60:90 triangle ?


HINT: You'll find the answer to your question in the following video on Circle Properties:


Cheers,
Brent
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If you enjoy my solutions, you'll love my GMAT prep course.

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Re: The two lines are tangent to the circle.  [#permalink]

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New post 24 Jul 2018, 10:08
GMATPrepNow wrote:
CounterSniper wrote:
Princ wrote:
OA:E
Attachment:
gmatprepnow.PNG

In above sketch, \(OA=OB=\) radius
\(\angle OAC = \angle OBC =\angle ADC =\angle BDC = 90^{\circ}\)
\(AD= DB = \frac{{10\sqrt[]{3}}}{{2}}=5\sqrt[]{3}\)

\(AD =5\sqrt[]{3}\) and\(\angle ADC = 90^{\circ}\) , along with \(AC= 10\),
We can say that \(\triangle ADC\) is \(30^{\circ}-60^{\circ}-90^{\circ}\) Triangle with \(\angle ACD = 60^{\circ}\)

Now in \(\triangle\) \(OAC\) ,\(\angle AOC =30^{\circ}\)

\(\angle AOB = 2 * \angle AOC = 60^{\circ}\)

This imply that \(\triangle AOB\) is equilateral \(\triangle\) , with all sides equal to \(10\sqrt{3}\).
\(OA=OB=\) radius \(=10\sqrt{3}\)

Area of circle\(= \pi * 10\sqrt{3}*10\sqrt{3} =300\pi\)



I understand that 2 of the sides are 10 and 5\sqrt{3}
How can we conclude that the triangle here is a 30 :60:90 triangle ?


HINT: You'll find the answer to your question in the following video on Circle Properties:


Cheers,
Brent


figured it out from the ratio 1:root3:2

Thanks !!
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The two lines are tangent to the circle.  [#permalink]

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New post 25 Jul 2020, 18:01
BrentGMATPrepNow wrote:
GMATPrepNow wrote:

The two lines are tangent to the circle. If AC = 10 and AB = 10√3, what is the area of the circle?

A) 100π
B) 150π
C) 200π
D) 250π
E) 300π

*kudos for all correct solutions


If AC = 10, then BC = 10

Since ABC is an isosceles triangle, the following gray line will create two right triangles...

Now focus on the following blue triangle. Its measurements have a lot in common with the BASE 30-60-90 special triangle

In fact, if we take the BASE 30-60-90 special triangle and multiply all sides by 5 we see that the sides are the same as the sides of the blue triangle.

So, we can now add in the 30-degree and 60-degree angles

Now add a point for the circle's center and draw a line to the point of tangency. The two lines will create a right triangle (circle property)

We can see that the missing angle is 60 degrees

Now create the following right triangle

We already know that one side has length 5√3

Since we have a 30-60-90 special triangle, we know that the hypotenuse is twice as long as the side opposite the 30-degree angle.

So, the hypotenuse must have length 10√3

In other words, the radius has length 10√3

What is the area of the circle?
Area = πr²
= π(10√3)²
= π(10√3)(10√3)
= 300π

Answer: E

Cheers,
Brent


Hi BrentGMATPrepNow,
Is there any property that states that a perpendicular bisector from C to AB will also pass through the center O? Else we will not be able to confirm the length 5*(root 3) is the same in the two triangles. I guess I am missing a trick here..Thanks!
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Re: The two lines are tangent to the circle.  [#permalink]

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New post 26 Jul 2020, 06:06
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harshbirsingh wrote:
Hi BrentGMATPrepNow,
Is there any property that states that a perpendicular bisector from C to AB will also pass through the center O? Else we will not be able to confirm the length 5*(root 3) is the same in the two triangles. I guess I am missing a trick here..Thanks!


Yes, the perpendicular bisector from C to AB will pass through the center.
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Re: The two lines are tangent to the circle.   [#permalink] 26 Jul 2020, 06:06

The two lines are tangent to the circle.

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