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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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30 May 2017, 06:39
1
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Difficulty:

35% (medium)

Question Stats:

71% (02:26) correct 29% (02:51) wrong based on 224 sessions

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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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gmat_question.png [ 44.86 KiB | Viewed 5303 times ]

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

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Updated on: 30 May 2017, 20:29
2
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

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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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30 May 2017, 07:00
2
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..
Attachments

gmat_question.png [ 45.17 KiB | Viewed 5268 times ]

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

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30 May 2017, 07:17
1
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

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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30 May 2017, 19:32
2
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.
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In the figure above, the triangle ABC is inscribed in a semicircle, wi  [#permalink]

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31 May 2017, 06:19
1
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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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

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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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

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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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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11 Mar 2018, 06:03
1
1
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 ,

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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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 ,

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  [#permalink]

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Re: In the figure above, the triangle ABC is inscribed in a semicircle, wi   [#permalink] 11 Apr 2019, 12:05
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