How do you find the area of the triangle given a=28.16 cm, b=60.15 cm, c=51.17 cm?

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How do you find the area of the triangle given a=28.16 cm, b=60.15 cm, c=51.17 cm?

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Answer 1 · 2016-06-27T06:08:09

$11.81016511$ $11.81016511$ $C m^{2}$ $Cm^2$

Explanation:

Area of a triangle whose sides have lengths a, b, and c is

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

where s is the semiperimeter of the triangle; that is,

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

$s = \frac{28.16 + 60.15 + 51.17}{2}$ $s = (28.16 + 60.15 + 51.17)/2$ $= 69.74$ $= 69.74$

$A = \sqrt{69.74 \cdot (69.74 - 28.16) \cdot (69.74 - 60.15) \cdot (69.74 - 51.17)}$ $A = sqrt (69.74* (69.74 - 28.16)*(69.74 - 60.15)*(69.74 - 51.17)$

$A = \sqrt{69.74 \cdot 41.58 \cdot 9.59 \cdot 18.57}$ $A = sqrt( 69.74*41.58*9.59*18.57)$ = $11.81016511$ $11.81016511$

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How do you find the area of the triangle given a=28.16 cm, b=60.15 cm, c=51.17 cm?

1 Answer

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