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The length of one of the sides of a triangle is 13 units. If the area

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The length of one of the sides of a triangle is 13 units. If the area  [#permalink]

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New post 26 Feb 2018, 10:25
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The length of one of the sides of a triangle is 13 units. If the area of the triangle is 90 units^2 and the length of the another side of the triangle is 15 units. Find the length of the third side.


A) √124
B) √134
C) √224
D) √234
E) √244

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The length of one of the sides of a triangle is 13 units. If the area  [#permalink]

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New post 26 Feb 2018, 10:53
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saswata4s wrote:
The length of one of the sides of a triangle is 13 units. If the area of the triangle is 90 units^2 and the length of the another side of the triangle is 15 units. Find the length of the third side.


A) √124
B) √134
C) √224
D) √234
E) √244

Two Simple Approach for doing this Problem:

First Approach:

We have area of Triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\). Let c be the third side.
s = \(\frac{a+b+c}{2} = \frac{28+c}{2}\)
\(s(s-a)(s-b)(s-c) = (\frac{28+c}{2})(\frac{28-c}{2})(\frac{c+2}{2})(\frac{c-2}{2}) = 90*90 = 8100\)

\(c = \sqrt{244}\)

Second Approach:

Area of Triangle = \(\frac{1}{2}abSinC = \frac{1}{2}bcSinA = \frac{1}{2}acSinB\)
Let a,b be the side known to us i.e. a = 13, b =15
\(90 = \frac{1}{2}*13*15*SinC\)
\(SinC = \frac{12}{13}\) or \(CosC = \frac{5}{13}\)

We have this Formula between angle and Sides of a Triangle

\(CosC = \frac{a^2 + b^2 - c^2}{2ab}\)
\(c^2 = 244\) or \(c = \sqrt{244}\)
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The length of one of the sides of a triangle is 13 units. If the area  [#permalink]

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New post 26 Feb 2018, 11:30
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Formula used:
Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) where semiperimeter s = \(\frac{a+b+c}{2}\)

If the third side is x, the semiperimter is \(\frac{13+15+x}{2} = \frac{28+x}{2}\)

\(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{(\frac{28+x}{2})(\frac{28+x}{2} - 15)(\frac{28+x}{2} - 13)(\frac{28+x}{2} - x)}\) = \(\sqrt{(\frac{28+x}{2})(\frac{x-2}{2})(\frac{x+2}{2})(\frac{28-x}{2})} = 90\)

Here, \(\sqrt{\frac{(28^2 - x^2)(x^2 - 4)}{16}} = 90\) because \(a^2 - b^2 = (a-b)(a+b)\)

Squaring on both sides, \((784 - x^2)(x^2 - 4) = 8100*16 = 16 * 81 * 100\)

Evaluating the answer options,

A) x = √124 | (784 - 124)(124 - 4) = (660)(120)
B) x = √134 | (784 - 134)(134 - 4) = (650)(130)
C) x = √224 | (784 - 224)(224 - 4) = (560)(220)
D) x = √234 | (784 - 234)(234 - 4) = (550)(230)

E) x = √244 | (784 - 244)(244 - 4) = (540)(240) = (2*270)(3*80) = 16*81*100


Therefore, Option E(√244) is the length of the third side
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Re: The length of one of the sides of a triangle is 13 units. If the area  [#permalink]

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New post 01 Mar 2018, 04:18
Is there no easier way to solve this question? Are we supposed to know this formula for the GMAT?
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Re: The length of one of the sides of a triangle is 13 units. If the area  [#permalink]

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New post 01 Mar 2018, 07:53
nuclearbattery wrote:
Is there no easier way to solve this question? Are we supposed to know this formula for the GMAT?


I couldn't find easier way to solve this... But Yes, these Formulas are of Great help.. !!

--== Message from the GMAT Club Team ==--

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This discussion does not meet community quality standards. It has been retired.


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Re: The length of one of the sides of a triangle is 13 units. If the area &nbs [#permalink] 01 Mar 2018, 07:53
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