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The two lines are tangent to the circle.

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BrentGMATPrepNow
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The two lines are tangent to the circle. [#permalink] New post   24 Jul 2018, 05:45
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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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BrentGMATPrepNow
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Re: The two lines are tangent to the circle. [#permalink] New post   26 Jul 2018, 07:01
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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
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Re: The two lines are tangent to the circle. [#permalink] New post   24 Jul 2018, 07:02
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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


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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Shobhit7
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The two lines are tangent to the circle. [#permalink] New post   24 Jul 2018, 07:00
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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] New post   24 Jul 2018, 07:24
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OA:E
Attachment:
gmatprepnow.PNG
gmatprepnow.PNG [ 69.75 KiB | Viewed 7738 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] New post   24 Jul 2018, 10:37
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] New post   24 Jul 2018, 10:43
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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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Re: The two lines are tangent to the circle. [#permalink] New post   24 Jul 2018, 11:08
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] New post   25 Jul 2020, 19:01
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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BrentGMATPrepNow
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Re: The two lines are tangent to the circle. [#permalink] New post   26 Jul 2020, 07:06
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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, 07:06
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