Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 18 Jul 2019, 14:43 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the figure above (not drawn to scale), triangle ABC is inscribed in

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56257
In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags 00:00

Difficulty:   35% (medium)

Question Stats: 78% (02:40) correct 22% (03:12) wrong based on 156 sessions

### HideShow timer Statistics  In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment: circle_inscribed_triangle.gif [ 8.86 KiB | Viewed 2703 times ]

_________________
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2959
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags

1
Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

AC = OB = Radius (r)
and AB = 2*r

ABC is right angle triangle as traingle drawn in Semicircle is always the Right angle triangle with right angle at vertex touching circumference and Hypotenuse at Diameter of circle

Also BC^2 = AB^2 - AC^2 = (2r)^2 - r^2 = 3r^2

i.e. BC = r√3

Area of ABC = (1/2)*BC*AC = (1/2)*r√3*r = 8√3
i.e. r = 4
Area of Circle = π*r^2 = π*4^2 = 16π

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Originally posted by GMATinsight on 28 Jul 2015, 04:10.
Last edited by GMATinsight on 28 Jul 2015, 04:12, edited 1 time in total.
Manager  B
Joined: 14 Mar 2014
Posts: 145
GMAT 1: 710 Q50 V34 Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags

1
1
Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

IMO: C

Angle in a semi circle = 90. Thus Angle ACB = 90.
AC = OB = r
So angles of the triangle will be ABC = 30 , ACB = 90 , BAC = 60.
So sides of the triangle will be AB = 2r AC = r BC = r√3

Area of triangle is 8√3 = 1/2 b*h
base and heights will be the sides of the triangle since it is right angled triangle
8√3 = 1/2 * r * r√3
16 = r^2

Area of circle = πr^2
= 16π
_________________
I'm happy, if I make math for you slightly clearer
And yes, I like kudos
¯\_(ツ)_/¯ Manager  B
Joined: 12 Nov 2014
Posts: 62
Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags

1
From the figure it can be seen that the triangle ACB is inscribed within a semicircle. So angle C is 90 degree.
Lets take OB = x , so OA =x (both radius) 2x, Also it is given that AC = OB = x.
For triangle OAC (draw a line ) all sides are equal (OA = AC= OC) so it is an equilateral triangle. So <A = 60degree

The right triangle ACB is thus a 30 60 90 triangle so sides are x, x√3and 2x.
Area of triangle ACB = 0.5 * x * x√3 = 8√3 So x = 4 i.e. radius of the circle is 4

Area of circle = π x * x = 16π

_________________
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Kindly press Kudos if the explanation is clear.
Thank you
Ambarish
CEO  S
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags

Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

Given, AB is the diameter of the circle= 2r and AC = OB =r

Let BC = h

Thus is right triangle ABC, right angled at C,

$$r^2+h^2 = (2r)^2$$ and from area of the triangle = $$8\sqrt{3}$$ = $$0.5*h*r$$ ----> $$h = \frac{16\sqrt{3}}{r}$$

Thus, $$r^2+ (\frac{16\sqrt{3}}{r})^2 = 2r^2$$ ----> $$r^4 = 256$$ ---> $$r = 4$$

Thus the area of the circle = $$\pi*4^2$$ = $$16\pi$$. C is the correct answer.
Current Student Affiliations: Scrum Alliance
Joined: 09 Feb 2010
Posts: 75
Location: United States (MI)
Concentration: Strategy, General Management
GMAT 1: 600 Q48 V25 GMAT 2: 710 Q48 V38 WE: Information Technology (Retail)
Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags

1
Here, $$AC = OB = OA = r$$, we can say that $$AB = 2r$$.

Given the inscribed triangle property, we can get that triangle ABC is a right triangle and so $$AB^2 = AC^2 + BC^2$$.

So, $$(2r)^2 = r^2 + BC^2$$, which means that $$BC = \sqrt{3}r$$.

Area of the triangle ABC = $$\frac{1}{2}(AC)(BC) = 8\sqrt{3}$$, means $$\frac{1}{2}(r)(\sqrt{3}r) = 8\sqrt{3}$$, which simplifies to $$r^2 = 16$$, so Area of the circle = 16$$\pi$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 56257
Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags

1
Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

800score Official Solution:

There are four main concepts that must be understood in order to solve this problem. The first concept is that any triangle inscribed within a circle such that any one side of the triangle is a diameter of the circle, is a right triangle. The second concept is that the proportions of the sides of a 30⁰-60⁰-90⁰ triangle are x, x√3, 2x. The third concept is that the area of a right triangle is equal to one half the product of its legs: (1/2)bh. Finally, the fourth concept is that the area of a circle is π × radius².

Since the triangle is inscribed in the circle and one of its sides constitutes a diameter of the circle, the triangle must be a right triangle. AB is the diameter so AB = 2 × OB. We know that OB = AC. Therefore AB = 2 × AC. Therefore it must be a 30°-60°-90° triangle where AC = x, AB = 2x and CB = √3 × x (This can also be calculated using the Pythagorean Theorem).

We are now ready to solve this problem. Using the third concept from above, the area must be equal to (1/2) × √3 × x × x = 8√3.
The solution of the equation is x = 4. So the radius of the circle is 4. Therefore the area of the circle is π × 4² = 16π. The correct answer is C.
_________________
Board of Directors P
Joined: 17 Jul 2014
Posts: 2539
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags

Bunuel wrote: In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

OB is the radius, thus, AC has the same length as the radius.

since we have a triangle inscribed in a circle, and one side is the diagonal, the triangle MUST be a right triangle.
since the hypotenuse is 2R,and since one leg is R, it must be true that the triangle is 30-60-90. It must also be true, from the property of 30-60-90 triangle, that the other leg must be R*sqrt(3).
we then have the area of the triangle:
R*R(sqrt3)/2 = 8*sqrt(3)
r^2(sqrt3) = 16*sqrt(3)
r=4.
Area of the circle:
pi*r^2 = 16pi.
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 11702
Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: In the figure above (not drawn to scale), triangle ABC is inscribed in   [#permalink] 21 Dec 2018, 17:08
Display posts from previous: Sort by

# In the figure above (not drawn to scale), triangle ABC is inscribed in  