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In the figure above, the triangle ABC is inscribed in a semicircle, wi

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In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 30 May 2017, 06:39
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In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9

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In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post Updated on: 30 May 2017, 20:29
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guireif wrote:
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9


Lets join O to C in the diagram.

Let radius of semicircle = x . So , OA= OB = OC = x
Also angle BOC = 180° - angle COA = 60°
Since OC = OB
angle BCO = angle COB = (180 - angle BOC)/2 = 60°

So triangle BOC is an equilateral triangle . Hence BC = x

Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle.

-> AC^2 + BC^2 = AB^2
-> (3sqrt(3))^2 + x^2 = (2x)^2
-> 3x^2 = 27
-> x^2 = 9
-> x =3

Hence radius = 3.

Answer B.

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Originally posted by shashankism on 30 May 2017, 06:57.
Last edited by shashankism on 30 May 2017, 20:29, edited 1 time in total.
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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 30 May 2017, 07:00
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guireif wrote:
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9


I forgot to attach the diagram.. I have edited the diagram provided by you for the solution..
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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 30 May 2017, 07:17
shashankism wrote:
guireif wrote:
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9


Lest join O to C in the diagram.

Let radius of semicircle = x . So , OA= OB = OC = x
Also angle BOC = 180° - angle COA = 60°
Since OC = OB
angle BCO = angle COB = (180 - angle BOC)/2 = 60°

So triangle BOC is an equilateral triangle . Hence BC = x

Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle.

-> AC^2 + BC^2 = AB^2
-> (3sqrt(3))^2 + x^2 = (2x)^2
-> 3x^2 = 27
-> x^2 = 9
-> x =3

Hence radius = 3.

Answer B.


Since OC = OB
angle BCO = angle COB = (180 - angle BOC)/2 = 60°


That's what I was missing!
You're a genius. Thank you so much.

Best,
Gui.
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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 30 May 2017, 19:32
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guireif wrote:
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9


Triangle inscribed in a semicircle will always have one angle as \(90^{\circ}\).

From the diagram \(\angle\)ACB = \(90^{\circ}\)

Connect OC.
Given; \(\angle\)CÔA = 120°
\(\angle\)COB + \(\angle\)COA = 180
Therefore \(\angle\)COB = \(60^{\circ}\)
OB = OC (Both are radius).
Hence \(\angle\)OCB = \(\angle\)OBC = \(60^{\circ}\)

\(\triangle\) ABC is 30:60:90 Triangle. Sides of 30:60:90 triangle are in ratio = x : x\(\sqrt{3}\) : 2x

BC : AC : AB = x : x\(\sqrt{3}\) : 2x

AC = 3\(\sqrt{3}\)

Therefore BC = 3 and AB = 6

AB is diameter of semicircle. Radius is half of diameter. Therefore Radius = 3.
Answer B...
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In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 31 May 2017, 06:19
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I tried solving it, but since I was taking more than 1 minute to figure it, I guessed right answer assuming that it has to be a 30:60:90 triangle.
Only the 30:60:90 triangle has a √3 component x : x√3 : 2x.

So, I guessed the diameter should be 2x=6 and hence the radius 3.
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In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 10 Mar 2018, 05:31
shashankism wrote:
guireif wrote:
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9


Lets join O to C in the diagram.

Let radius of semicircle = x . So , OA= OB = OC = x
Also angle BOC = 180° - angle COA = 60°
Since OC = OB
angle BCO = angle COB = (180 - angle BOC)/2 = 60°

So triangle BOC is an equilateral triangle . Hence BC = x

Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle.

-> AC^2 + BC^2 = AB^2
-> (3sqrt(3))^2 + x^2 = (2x)^2
-> 3x^2 = 27
-> x^2 = 9
-> x =3

Hence radius = 3.

Answer B.


shashankism why do you asssume that point O divides diameter of circle equally ? :? it could divide not equally ? no ?:)
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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 10 Mar 2018, 08:37
1
dave13 wrote:
shashankism wrote:
guireif wrote:
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9


Lets join O to C in the diagram.

Let radius of semicircle = x . So , OA= OB = OC = x
Also angle BOC = 180° - angle COA = 60°
Since OC = OB
angle BCO = angle COB = (180 - angle BOC)/2 = 60°

So triangle BOC is an equilateral triangle . Hence BC = x

Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle.

-> AC^2 + BC^2 = AB^2
-> (3sqrt(3))^2 + x^2 = (2x)^2
-> 3x^2 = 27
-> x^2 = 9
-> x =3

Hence radius = 3.

Answer B.


shashankism why do you asssume that point O divides diameter of circle equally ? :? it could divide not equally ? no ?:)


dave13 Since the question states "In the figure above, the triangle ABC is inscribed in a semicircle, with center O" therefore point O has to divide the diameter of the circle equal as OA and OB are radii of the semicircle shown.

Hope it clears!
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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 11 Mar 2018, 05:44
Hi ujjwal80, thanks for an explanation:) i have a couple of questions :) can you please explan. see below. have a great weekend :)

Question 1.

Based on which rule do we assume that the length of OC equals OA and OB?
OC could be of any length, no?

Question 2. regarding this --- > "Angle BCO = angle COB = (180 - angle BOC)/2 = 60° "

How can we assume that angle BCO is 60° ? it could be any degree… no ?
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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 11 Mar 2018, 06:03
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dave13 wrote:
Question 1.

Based on which rule do we assume that the length of OC equals OA and OB?
OC could be of any length, no?

Question 2. regarding this --- > "Angle BCO = angle COB = (180 - angle BOC)/2 = 60° "

How can we assume that angle BCO is 60° ? it could be any degree… no ?


Hey dave13 ,

Here are the answers:

Based on which rule do we assume that the length of OC equals OA and OB? --> We are not assuming. It is the rule. If the given figure is a semi circle with center as O, all the line segments joining the center and any point on the circle are equal and are also known as radii. Hence, OC = OA = OB

How can we assume that angle BCO is 60° ? --> Since OC = OA, that means triangle COA is isosceles.

We are given that angle COA = 120. => the sum of angles OAC and OCA = 180-120 = 60.

=> angle OCA = 30 (since triangle COA is isosceles).

Now, every triangle drawn on the semi circle is right angled triangle. This angle ACB = 90.

Therefore, BCO = ACN - OCA = 90-30 = 60

Does that make sense?
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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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New post 11 Mar 2018, 19:13
abhimahna wrote:
dave13 wrote:
Question 1.

Based on which rule do we assume that the length of OC equals OA and OB?
OC could be of any length, no?

Question 2. regarding this --- > "Angle BCO = angle COB = (180 - angle BOC)/2 = 60° "

How can we assume that angle BCO is 60° ? it could be any degree… no ?


Hey dave13 ,

Here are the answers:

Based on which rule do we assume that the length of OC equals OA and OB? --> We are not assuming. It is the rule. If the given figure is a semi circle with center as O, all the line segments joining the center and any point on the circle are equal and are also known as radii. Hence, OC = OA = OB

How can we assume that angle BCO is 60° ? --> Since OC = OA, that means triangle COA is isosceles.

We are given that angle COA = 120. => the sum of angles OAC and OCA = 180-120 = 60.

=> angle OCA = 30 (since triangle COA is isosceles).

Now, every triangle drawn on the semi circle is right angled triangle. This angle ACB = 90.

Therefore, BCO = ACN - OCA = 90-30 = 60

Does that make sense?


Thanks abhimahna for the explanation, dave13 refer to triangle concepts https://gmatclub.com/forum/math-triangles-87197.html if you have more doubts feel free to post.
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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi &nbs [#permalink] 11 Mar 2018, 19:13
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