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

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Posts: 50623
In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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29 Dec 2014, 07:56
00:00

Difficulty:

15% (low)

Question Stats:

85% (01:53) correct 15% (02:46) wrong based on 105 sessions

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Tough and Tricky questions: Geometry.

Attachment:

2014-12-29_1955.png [ 4.36 KiB | Viewed 3693 times ]

In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with center O. AB is a diameter of the circle. If CB equals 2√3 and the segment OB equals 2, what is the area of the triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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29 Dec 2014, 08:51
From the fact that AB is the diameter of the circle and C lies on the circle, we can conclude that it is a right triangle and we can use the Pythagoras theorem:

$$AC^2+BC^2=AB^2=\frac{4}{9}+BC^2=\frac{144}{9}$$

We desolate for BC and get BC=2.

So the area of the triangle is $$\frac{1}{2}* 2\sqrt{3}*2=2\sqrt{3}$$.

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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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29 Dec 2014, 09:37
1
Answer is 2√3, as mentioned AB = Diameter, the triangle becomes right triangle. And then we can simply use pythagoras rule.
After calculating side AC as 2 (which is the base of the triangle), we can use (base x height)/2 = (2 x 2√3)/2 = 2√3.
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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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02 Jan 2015, 00:24
1

$$\triangle$$ ABC is a right angle triangle with AC as base & AB as height (As it's inscribed with one side the diameter)

$$AC = \sqrt{4^2 - (2\sqrt{3})^2} = \sqrt{16-12} = 2$$

Area $$= \frac{1}{2} * 2 * 2\sqrt{3} = 2\sqrt{3}$$
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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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08 Jan 2015, 08:30
2
1
Bunuel wrote:

Tough and Tricky questions: Geometry.

Attachment:
2014-12-29_1955.png

In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with center O. AB is a diameter of the circle. If CB equals 2√3 and the segment OB equals 2, what is the area of the triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

OFFICIAL SOLUTION:

(B) There are three 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. Finally, the third concept is that the area of a triangle is equal to one half the product of its base and height: (1/2)bh.

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. Since we are told that OB, the radius of the circle, is equal to 2, the longest side of the triangle must be equal to 4. And, since it is a right triangle with a hypotenuse equal to 4 and the second longest leg is equal to 2√3, it must be a 30-60-90 triangle where the shortest side is equal to 2 (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)bh. Taking the shortest side to be the base, b and the longest of the two legs to be the height, h, we have:
(1/2)bh = (1/2)(2)(2√3) = 2√3.

So, the correct answer choice is (B).
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In the figure above, the triangle ABC is inscribed in the circle with  [#permalink]

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21 Jul 2015, 02:16
1

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

Attachment:

circle_inscribed_triangle.gif [ 8.86 KiB | Viewed 4095 times ]

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In the figure above, the triangle ABC is inscribed in the circle with  [#permalink]

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21 Jul 2015, 02:37
2
Bunuel wrote:

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

CB = 2√3

i.e. Diameter AB = 2*2 = 4

Property: The Triangle Drawn in the semicircle with Hypotenuse as Diameter of Circle will always be Right angle Triangle (Right angled at the point on the circumference)

i.e. AC^2 = AB^2 - BC^2 = (4)^2 - (2√3)^2 = 16 - 12 = 4

i.e. AC = 2

Area of Triangle ABC = (1/2) AC*BC = (1/2)*2* (2√3) = (2√3)

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Re: In the figure above, the triangle ABC is inscribed in the circle with  [#permalink]

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21 Jul 2015, 02:53
Bunuel wrote:

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

Bunuel, the same question has been discussed at in-the-figure-above-not-drawn-to-scale-the-triangle-abc-is-inscribe-190871.html

For the sake of completeness, we can solve this question in 2 ways:

Method 1:

Triangle ABC is right angled at C (as any triangle drawn in a circle with diameter as one of the sides makes a 90 degree angle on the circumference). Thus ,

AB^2 = AC^2+BC^2 ---> $$(2+2)^2 = (2\sqrt{3})^2+x^2$$ ---> x = 2.

Thus area of the triangle ABC = $$0.5*AC*BC = 0.5*2\sqrt{3}*2 = 2\sqrt{3}$$, B is the correct answer.

Method 2: Ballparking.

Let $$\pi$$=3 ---> area of the circle = $$\pi*(4)^2$$--> $$Area= 12 units^2$$

Now the triangle is less than half the area of the circle (close to 70% of the half)---> 0.5*0.7*12 = approx. 4 units^2. The only option that comes close to this is option B with $$\sqrt{3} = 1.7$$
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Re: In the figure above, the triangle ABC is inscribed in the circle with  [#permalink]

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21 Jul 2015, 08:45
2
Bunuel wrote:

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

This is a classical x, x√3, 2x triangle. We know that AC is the shortest side and therefore must be 2. We can deviate this because we know that CB equals x√3 = 2√3, so x = 2.

Area = Base (2√3) * Height (2) / 2 = 2√3

_________________

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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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13 Oct 2017, 08:23
Bunuel wrote:

Tough and Tricky questions: Geometry.

Attachment:
2014-12-29_1955.png

In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with center O. AB is a diameter of the circle. If CB equals 2√3 and the segment OB equals 2, what is the area of the triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

From geometry, we know that if a triangle is constructed inside a circle using the circle’s diameter (AB) as the hypotenuse and a point on the circumference of the circle (C) as a vertex of the triangle, then the triangle will be a right triangle.Thus, we can see that triangle ABC is a right triangle.

We can determine side AC (which we shall denote as b) using the Pythagorean theorem:

(2√3)^2 + b^2 = 4^2

12 + b^2 = 16

b^2 = 4

b = 2

The area of the triangle is bh/2 = (2√3 x 2)/2 = 2√3.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: In the figure above (not drawn to scale), the triangle ABC is inscribe &nbs [#permalink] 13 Oct 2017, 08:23
Display posts from previous: Sort by
The content structure is as follows: The text presents a geometry problem involving a triangle inscribed in a circle, with specific measurements provided. The problem asks for the area of the triangle, and multiple-choice answers are given.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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# In the figure above (not drawn to scale), the triangle ABC is inscribe

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Math Expert
Joined: 02 Sep 2009
Posts: 50623
In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

### Show Tags

29 Dec 2014, 07:56
00:00

Difficulty:

15% (low)

Question Stats:

85% (01:53) correct 15% (02:46) wrong based on 105 sessions

### HideShow timer Statistics

Tough and Tricky questions: Geometry.

Attachment:

2014-12-29_1955.png [ 4.36 KiB | Viewed 3693 times ]

In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with center O. AB is a diameter of the circle. If CB equals 2√3 and the segment OB equals 2, what is the area of the triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

_________________
Manager
Joined: 22 Oct 2014
Posts: 88
Concentration: General Management, Sustainability
GMAT 1: 770 Q50 V45
GPA: 3.8
WE: General Management (Consulting)
Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

### Show Tags

29 Dec 2014, 08:51
From the fact that AB is the diameter of the circle and C lies on the circle, we can conclude that it is a right triangle and we can use the Pythagoras theorem:

$$AC^2+BC^2=AB^2=\frac{4}{9}+BC^2=\frac{144}{9}$$

We desolate for BC and get BC=2.

So the area of the triangle is $$\frac{1}{2}* 2\sqrt{3}*2=2\sqrt{3}$$.

_________________

$$\sqrt{-1}$$ $$2^3$$ $$\Sigma$$ $$\pi$$ ... and it was delicious!

Please consider giving +1 Kudos if deserved!

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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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29 Dec 2014, 09:37
1
Answer is 2√3, as mentioned AB = Diameter, the triangle becomes right triangle. And then we can simply use pythagoras rule.
After calculating side AC as 2 (which is the base of the triangle), we can use (base x height)/2 = (2 x 2√3)/2 = 2√3.
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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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02 Jan 2015, 00:24
1

$$\triangle$$ ABC is a right angle triangle with AC as base & AB as height (As it's inscribed with one side the diameter)

$$AC = \sqrt{4^2 - (2\sqrt{3})^2} = \sqrt{16-12} = 2$$

Area $$= \frac{1}{2} * 2 * 2\sqrt{3} = 2\sqrt{3}$$
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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

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08 Jan 2015, 08:30
2
1
Bunuel wrote:

Tough and Tricky questions: Geometry.

Attachment:
2014-12-29_1955.png

In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with center O. AB is a diameter of the circle. If CB equals 2√3 and the segment OB equals 2, what is the area of the triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

OFFICIAL SOLUTION:

(B) There are three 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. Finally, the third concept is that the area of a triangle is equal to one half the product of its base and height: (1/2)bh.

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. Since we are told that OB, the radius of the circle, is equal to 2, the longest side of the triangle must be equal to 4. And, since it is a right triangle with a hypotenuse equal to 4 and the second longest leg is equal to 2√3, it must be a 30-60-90 triangle where the shortest side is equal to 2 (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)bh. Taking the shortest side to be the base, b and the longest of the two legs to be the height, h, we have:
(1/2)bh = (1/2)(2)(2√3) = 2√3.

So, the correct answer choice is (B).
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50623
In the figure above, the triangle ABC is inscribed in the circle with  [#permalink]

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21 Jul 2015, 02:16
1

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

Attachment:

circle_inscribed_triangle.gif [ 8.86 KiB | Viewed 4095 times ]

_________________
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Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2700
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In the figure above, the triangle ABC is inscribed in the circle with  [#permalink]

### Show Tags

21 Jul 2015, 02:37
2
Bunuel wrote:

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

CB = 2√3

i.e. Diameter AB = 2*2 = 4

Property: The Triangle Drawn in the semicircle with Hypotenuse as Diameter of Circle will always be Right angle Triangle (Right angled at the point on the circumference)

i.e. AC^2 = AB^2 - BC^2 = (4)^2 - (2√3)^2 = 16 - 12 = 4

i.e. AC = 2

Area of Triangle ABC = (1/2) AC*BC = (1/2)*2* (2√3) = (2√3)

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
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Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
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Re: In the figure above, the triangle ABC is inscribed in the circle with  [#permalink]

### Show Tags

21 Jul 2015, 02:53
Bunuel wrote:

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

Bunuel, the same question has been discussed at in-the-figure-above-not-drawn-to-scale-the-triangle-abc-is-inscribe-190871.html

For the sake of completeness, we can solve this question in 2 ways:

Method 1:

Triangle ABC is right angled at C (as any triangle drawn in a circle with diameter as one of the sides makes a 90 degree angle on the circumference). Thus ,

AB^2 = AC^2+BC^2 ---> $$(2+2)^2 = (2\sqrt{3})^2+x^2$$ ---> x = 2.

Thus area of the triangle ABC = $$0.5*AC*BC = 0.5*2\sqrt{3}*2 = 2\sqrt{3}$$, B is the correct answer.

Method 2: Ballparking.

Let $$\pi$$=3 ---> area of the circle = $$\pi*(4)^2$$--> $$Area= 12 units^2$$

Now the triangle is less than half the area of the circle (close to 70% of the half)---> 0.5*0.7*12 = approx. 4 units^2. The only option that comes close to this is option B with $$\sqrt{3} = 1.7$$
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Re: In the figure above, the triangle ABC is inscribed in the circle with  [#permalink]

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21 Jul 2015, 08:45
2
Bunuel wrote:

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

This is a classical x, x√3, 2x triangle. We know that AC is the shortest side and therefore must be 2. We can deviate this because we know that CB equals x√3 = 2√3, so x = 2.

Area = Base (2√3) * Height (2) / 2 = 2√3

_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe  [#permalink]

### Show Tags

13 Oct 2017, 08:23
Bunuel wrote:

Tough and Tricky questions: Geometry.

Attachment:
2014-12-29_1955.png

In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with center O. AB is a diameter of the circle. If CB equals 2√3 and the segment OB equals 2, what is the area of the triangle ABC?

A. √3
B. 2√3
C. π√3
D. 2π√3
E. 8√3

From geometry, we know that if a triangle is constructed inside a circle using the circle’s diameter (AB) as the hypotenuse and a point on the circumference of the circle (C) as a vertex of the triangle, then the triangle will be a right triangle.Thus, we can see that triangle ABC is a right triangle.

We can determine side AC (which we shall denote as b) using the Pythagorean theorem:

(2√3)^2 + b^2 = 4^2

12 + b^2 = 16

b^2 = 4

b = 2

The area of the triangle is bh/2 = (2√3 x 2)/2 = 2√3.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: In the figure above (not drawn to scale), the triangle ABC is inscribe &nbs [#permalink] 13 Oct 2017, 08:23
Display posts from previous: Sort by
user
The content structure is as follows: The text presents a geometry problem involving a triangle inscribed in a circle, with specific measurements provided. The problem asks for the area of the triangle, and multiple-choice answers are given.

TYPE: SINGLE_QUESTION_WITH_ANSWER