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# A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are th

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are th [#permalink]

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Updated on: 24 Nov 2016, 18:26
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A triangle’s area is S=$$\sqrt{( p(p-a )(p-b )(p-c))}$$, where a, b and c are the three side lengths of the triangle and p=(a+b+c)/2. When a=5, b=6 and c=7, what is the area of the triangle?

A. 4√6 B. 5√6 C. 6√6 D. 7√6 E. 8√6

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"Only $79 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Originally posted by MathRevolution on 22 Nov 2016, 01:41. Last edited by MathRevolution on 24 Nov 2016, 18:26, edited 1 time in total. SVP Joined: 12 Sep 2015 Posts: 2477 Location: Canada Re: A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are th [#permalink] ### Show Tags 22 Nov 2016, 07:39 Expert's post Top Contributor MathRevolution wrote: A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are the three side lengths of the triangle and p=(a+b+c)/2. When a=5, b=6 and c=7, what is the area of the triangle? A. 4√6 B. 5√6 C. 6√6 D. 7√6 E. 8√6 You're missing a few key variables in your Heron's Law formula. S=√[(p)(p-a)(p-b)(p-c)], where a, b and c are the three side lengths of the triangle and p=(a+b+c)/2 _________________ Brent Hanneson – Founder of gmatprepnow.com Senior CR Moderator Status: Long way to go! Joined: 10 Oct 2016 Posts: 1378 Location: Viet Nam Re: A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are th [#permalink] ### Show Tags 22 Nov 2016, 08:11 MathRevolution wrote: A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are the three side lengths of the triangle and p=(a+b+c)/2. When a=5, b=6 and c=7, what is the area of the triangle? A. 4√6 B. 5√6 C. 6√6 D. 7√6 E. 8√6 Heron's formula: $$S=\sqrt{p(p-a)(p-b)(p-c)}$$ with $$p=\frac{a+b+c}{2}$$ a=5, b=6, c=7 then $$p=\frac{5+6+7}{2}=9$$ $$S=\sqrt{9 \times (9-5) \times (9-6) \times (9-7)}=\sqrt{9 \times 4 \times 3 \times 2}=\sqrt{6^2 \times 6}=6\sqrt{6}$$ The answer is C _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5452 GMAT 1: 800 Q59 V59 GPA: 3.82 Re: A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are th [#permalink] ### Show Tags 24 Nov 2016, 02:46 ==> If you substitute p=(5+6+7)/2=9 into S=$$\sqrt{√(p(p-a)(p-b)(p-c))}$$, it becomes S=$$\sqrt{(9*4*3*2)}$$=6√6. Therefore, the answer is C. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 3 month Online Course"
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Re: A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are th [#permalink]

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27 Oct 2017, 10:36
Hello Modertors/MathRevolution

The highlighted part of the question stem is incorrect here. It is not possible to get the answer with whats written. Please take and look and correct the same.

MathRevolution wrote:
A triangle’s area is S=$$\sqrt{( ((p-) )(p-) )(p-c))}$$, where a, b and c are the three side lengths of the triangle and p=(a+b+c)/2. When a=5, b=6 and c=7, what is the area of the triangle?

A. 4√6 B. 5√6 C. 6√6 D. 7√6 E. 8√6

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Re: A triangle’s area is S=√( ((p-) )(p-) )(p-c)), where a, b and c are th   [#permalink] 27 Oct 2017, 10:36
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