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# Area of triangle, given ratio of interior angles

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Area of triangle, given ratio of interior angles [#permalink]  01 Nov 2009, 03:06
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Re: Area of triangle, given ratio of interior angles [#permalink]  01 Nov 2009, 03:19
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angles in ratio x:2x:3x so its a 30-60-90 triangle and with shortest side 1, from trigonometry we have hypotenuse =2 and longest side =\sqrt{3},

so area =1/2 * base * height
= 1/2 * 1 *\sqrt{3}
= \sqrt{3}/2 which is D
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Re: Area of triangle, given ratio of interior angles [#permalink]  01 Nov 2009, 03:22
x+2x+3x=180

x=180/6, x=30

so the angles are 30,60,90

and AB is the shortest side and it has to be oppsite to 30.

in 30-60-90 combination the ratio of lengths are 1:\sqrt{3}:2

so the one opp to 60 is \sqrt{3}

one opp to 90 is 2

the area is \sqrt{3}
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Re: Area of triangle, given ratio of interior angles [#permalink]  01 Nov 2009, 03:41
divyakatas wrote:
x+2x+3x=180

x=180/6, x=30

so the angles are 30,60,90

and AB is the shortest side and it has to be oppsite to 30.

in 30-60-90 combination the ratio of lengths are 1:\sqrt{3}:2

so the one opp to 60 is \sqrt{3}

one opp to 90 is 2

the area is \sqrt{3}

kirankp has it right and you are very close, something wrong in the final step...
OA:
[Reveal] Spoiler:
D
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Re: Area of triangle, given ratio of interior angles [#permalink]  01 Nov 2009, 03:56
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Expert's post
$$x+2x+3x=180$$ --> $$x=30$$ degrees, $$2x=60$$ degrees, $$3x=90$$ degrees.

Hence we have right triangle $$30-60-90$$ and the shortest side$$=1$$.

For the right triangle $$30-60-90$$ the ratio of sides are $$1/\sqrt{3}/2$$, as the shortest side is $$1$$: $$Area=1*\sqrt{3}/2=\sqrt{3}/2$$

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Re: Area of triangle, given ratio of interior angles [#permalink]  09 Jul 2014, 05:39
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Re: Area of triangle, given ratio of interior angles   [#permalink] 09 Jul 2014, 05:39
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