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

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

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Tough and Tricky questions: Geometry.



Attachment:
2014-12-29_1955.png
2014-12-29_1955.png [ 4.36 KiB | Viewed 2894 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

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[Reveal] Spoiler: OA

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

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New post 29 Dec 2014, 09:51
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 \(\frac{1}{2}* 2\sqrt{3}*2=2\sqrt{3}\).

Answer B.
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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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Answer = B. 2√3

\(\triangle\) 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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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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Image
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.

[Reveal] Spoiler:
Attachment:
circle_inscribed_triangle.gif
circle_inscribed_triangle.gif [ 8.86 KiB | Viewed 3055 times ]

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 21 Jul 2015, 03:37
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Bunuel wrote:
Image
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.

[Reveal] Spoiler:
Attachment:
circle_inscribed_triangle.gif


CB = 2√3

OB = radius = 2
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)

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

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New post 21 Jul 2015, 03:53
Bunuel wrote:
Image
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.

[Reveal] Spoiler:
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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New post 21 Jul 2015, 09:45
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Bunuel wrote:
Image
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.

[Reveal] Spoiler:
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

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

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New post 13 Oct 2017, 09:23
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.

Answer: B
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Re: In the figure above (not drawn to scale), the triangle ABC is inscribe   [#permalink] 13 Oct 2017, 09:23
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