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# The two lines are tangent to the circle.

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CEO
Joined: 11 Sep 2015
Posts: 3235
The two lines are tangent to the circle.  [#permalink]

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24 Jul 2018, 04:45
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Difficulty:

95% (hard)

Question Stats:

46% (02:43) correct 54% (02:23) wrong based on 74 sessions

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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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Manager
Joined: 01 Feb 2017
Posts: 171
The two lines are tangent to the circle.  [#permalink]

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24 Jul 2018, 06:00
*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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Math Expert
Joined: 02 Aug 2009
Posts: 7108
Re: The two lines are tangent to the circle.  [#permalink]

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24 Jul 2018, 06:02
3
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..
Area =π*(10√3)^2=300π

E
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PicsArt_07-24-07.26.21.png [ 17.66 KiB | Viewed 906 times ]

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Senior Manager
Joined: 22 Feb 2018
Posts: 411
Re: The two lines are tangent to the circle.  [#permalink]

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24 Jul 2018, 06:24
2
OA:E
Attachment:

gmatprepnow.PNG [ 69.75 KiB | Viewed 882 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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Director
Joined: 20 Feb 2015
Posts: 796
Concentration: Strategy, General Management
Re: The two lines are tangent to the circle.  [#permalink]

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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 ?
CEO
Joined: 11 Sep 2015
Posts: 3235
Re: The two lines are tangent to the circle.  [#permalink]

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24 Jul 2018, 09:43
1
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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Director
Joined: 20 Feb 2015
Posts: 796
Concentration: Strategy, General Management
Re: The two lines are tangent to the circle.  [#permalink]

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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 !!
CEO
Joined: 11 Sep 2015
Posts: 3235
Re: The two lines are tangent to the circle.  [#permalink]

### Show Tags

26 Jul 2018, 06:01
1
Top Contributor
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

Cheers,
Brent
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Re: The two lines are tangent to the circle. &nbs [#permalink] 26 Jul 2018, 06:01
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