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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]

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New post 01 Nov 2009, 03:06
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Re: Area of triangle, given ratio of interior angles [#permalink]

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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]

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New post 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]

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New post 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...
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[Reveal] Spoiler:
D
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Re: Area of triangle, given ratio of interior angles [#permalink]

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New post 01 Nov 2009, 03:56
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\(x+2x+3x=180\) --> \(x=30\) 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 \(1\): \(Area=1*\sqrt{3}/2=\sqrt{3}/2\)

Answer: D.
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Re: Area of triangle, given ratio of interior angles [#permalink]

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Re: Area of triangle, given ratio of interior angles [#permalink]

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New post 18 Jul 2016, 18:02
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Re: Area of triangle, given ratio of interior angles   [#permalink] 18 Jul 2016, 18:02
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