What is the area of a triangle with sides of length 2, 6, and 5?

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What is the area of a triangle with sides of length 2, 6, and 5?

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Answer 1 · 2016-03-14T01:27:46

Apply Heron's formula to find the area to be

$\frac{\sqrt{351}}{4} \approx 4.6837$ $sqrt(351)/4~~4.6837$

Explanation:

Heron's formula states that, given a triangle with side lengths $a, b, c$ $a, b, c$ and semiperimeter $s = \frac{a + b + c}{2}$ $s = (a+b+c)/2$ the area $A$ $A$ of the triangle is

$A = \sqrt{s (s - a) (s - b) (s - c)}$ $A= sqrt(s(s-a)(s-b)(s-c))$

In this case, we have $a = 2$ $a = 2$ , $b = 6$ $b = 6$ , and $c = 5$ $c = 5$ . Then, for this triangle we have $s = \frac{2 + 6 + 5}{2} = \frac{13}{2}$ $s = (2+6+5)/2 = 13/2$ . Applying Heron's formula gives us

$A = \sqrt{\frac{13}{2} (\frac{13}{2} - 2) (\frac{13}{2} - 6) (\frac{13}{2} - 5)}$ $A = sqrt(13/2(13/2-2)(13/2-6)(13/2-5))$

$= \sqrt{\frac{13}{2} \cdot \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{3}{2}}$ $=sqrt(13/2*9/2*1/2*3/2)$

$= \sqrt{\frac{351}{16}}$ $=sqrt(351/16)$

$= \frac{\sqrt{351}}{4} \approx 4.6837$ $=sqrt(351)/4~~4.6837$

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What is the area of a triangle with sides of length 2, 6, and 5?

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