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How do you use Heron's formula to determine the area of a triangle with sides of that are 35, 28, and 21 units in length?

1 Answer

Jan 16, 2016

$294$ $294$ ${units}^{2}$ $"units"^2$

Explanation:

First, determine the semiperimeter $s$ $s$ of the triangle (which has sides $a, b, c$ $a,b,c$ ).

$s = \frac{a + b + c}{2}$ $s=(a+b+c)/2$

We know that $a = 35, b = 28, c = 21$ $a=35,b=28,c=21$ so

$s = \frac{35 + 28 + 21}{2} = 42$ $s=(35+28+21)/2=42$

Plug these into Heron's formula, which determines the area of a triangle:

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

$A = \sqrt{42 (42 - 35) (42 - 28) (42 - 21)}$ $A=sqrt(42(42-35)(42-28)(42-21))$

$A = \sqrt{42 \times 7 \times 14 \times 21}$ $A=sqrt(42xx7xx14xx21)$

$A = \sqrt{2^{2} \times 3^{2} \times 7^{4}}$ $A=sqrt(2^2xx3^2xx7^4)$

$A = 2 \times 3 \times 7^{2}$ $A=2xx3xx7^2$

$A = 294$ $A=294$

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