GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 06 Aug 2020, 02:59

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The two lines are tangent to the circle.

Author Message
TAGS:

### Hide Tags

GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
The two lines are tangent to the circle.  [#permalink]

### Show Tags

24 Jul 2018, 04:45
1
Top Contributor
14
00:00

Difficulty:

95% (hard)

Question Stats:

44% (02:58) correct 56% (02:33) wrong based on 118 sessions

### HideShow timer Statistics

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

GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
Re: The two lines are tangent to the circle.  [#permalink]

### Show Tags

26 Jul 2018, 06:01
2
1
Top Contributor
4
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
_________________
If you enjoy my solutions, you'll love my GMAT prep course.

##### General Discussion
Senior Manager
Joined: 01 Feb 2017
Posts: 273
The two lines are tangent to the circle.  [#permalink]

### Show Tags

24 Jul 2018, 06:00
2
2
*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

Posted from my mobile device
Math Expert
Joined: 02 Aug 2009
Posts: 8795
Re: The two lines are tangent to the circle.  [#permalink]

### Show Tags

24 Jul 2018, 06:02
4
1
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
Attachments

PicsArt_07-24-07.26.21.png [ 17.66 KiB | Viewed 2657 times ]

_________________
Senior Manager
Joined: 22 Feb 2018
Posts: 404
Re: The two lines are tangent to the circle.  [#permalink]

### Show Tags

24 Jul 2018, 06:24
2
1
OA:E
Attachment:

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$$= \pi * 10\sqrt{3}*10\sqrt{3} =300\pi$$
Director
Joined: 20 Feb 2015
Posts: 721
Concentration: Strategy, General Management
Re: The two lines are tangent to the circle.  [#permalink]

### Show Tags

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 ?
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
Re: The two lines are tangent to the circle.  [#permalink]

### Show Tags

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

Director
Joined: 20 Feb 2015
Posts: 721
Concentration: Strategy, General Management
Re: The two lines are tangent to the circle.  [#permalink]

### Show Tags

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 !!
Intern
Joined: 01 May 2020
Posts: 3
The two lines are tangent to the circle.  [#permalink]

### Show Tags

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

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!
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
Re: The two lines are tangent to the circle.  [#permalink]

### Show Tags

26 Jul 2020, 06:06
Top Contributor

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

Re: The two lines are tangent to the circle.   [#permalink] 26 Jul 2020, 06:06