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

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New post Updated on: 24 Nov 2016, 17:26
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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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Originally posted by MathRevolution on 22 Nov 2016, 00:41.
Last edited by MathRevolution on 24 Nov 2016, 17:26, edited 1 time in total.
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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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New post 22 Nov 2016, 06:39
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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
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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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New post 22 Nov 2016, 07:11
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
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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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New post 24 Nov 2016, 01:46
==> 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
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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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New post 27 Oct 2017, 09:36
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.

Thanking you in advance.

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 &nbs [#permalink] 27 Oct 2017, 09:36
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