What is the area of a triangle with sides of length 2, 6, and 9?

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What is the area of a triangle with sides of length 2, 6, and 9?

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Answer 1 · 2018-05-06T06:38:28

You cannot make a triangle with these three lengths.

Explanation:

Use Heron's formula where $A = \sqrt{s (s - a) (s - b) (s - c)}$ $A=sqrt(s(s-a)(s-b)(s-c))$
$a, b and c$ $a,b and c$ are the three sides and $s = \frac{a + b + c}{2}$ $s=[a+b+c]/2$

$s = \frac{2 + 6 + 9}{2}$ $s=[2+6+9)/2$ =8.5

$A = \sqrt{8.5 (8.5 - 2) (8.5 - 6) (8.5 - 9)}$ $A=sqrt(8.5(8.5-2)(8.5-6)(8.5-9))$

$A = \sqrt{8.5 \times 6.5 \times 2.5 \times - 0.5}$ $A=sqrt(8.5xx6.5xx2.5xx-0.5)$

$A = \sqrt{- 69.0625}$ $A=sqrt(-69.0625)$
Cannot take the square root of a negative number.

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What is the area of a triangle with sides of length 2, 6, and 9?

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