What is the area of a triangle with sides of length 2, 4, and 5?

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Answer 1 · 2018-03-26T07:17:05

$Area of the triangle A_{t} = 3.8 sq units$ $color(blue)("Area of the triangle " A_t = 3.8 " sq units"$

$Given : a = 2, b = 4, c = 5$ $"Given : " a = 2, b = 4, c = 5$

![https://www.onlinemathlearning.com/http://area-triangle.html](https://useruploads.socratic.org/ad1HaSmRQkmT98jNRSEH_area%20of%20triangle.png)

Having known three sides, we can calculate the area of the triangle using the formula,

$A_{t} = \sqrt{s \cdot (s - a) \cdot (s - b) \cdot (s - c)}$ $A_t = sqrt(s * (s-a) * (s-b) * (s-c))$

where s is the semi-perimeter of the triangle and a,b,c are the sides.

$s = \frac{a + b + c}{2} = \frac{2 + 4 + 5}{2} = 5.5$ $s = (a + b + c) / 2 = (2 + 4 + 5) / 2 = 5.5$

$A_{t} = \sqrt{5.5 \cdot (5.5 - 2) \cdot (5.5 - 4) \cdot (5.5 - 5)}$ $A_t = sqrt(5.5 * (5.5 - 2) * (5.5 - 4) * (5.5 - 5))$

$\Rightarrow \sqrt{5.5 \cdot 3.5 \cdot 1.5 \cdot 0.5} \approx 3.8 sq units$ $=> sqrt(5.5 * 3.5 * 1.5 * 0.5) ~~ 3.8 " sq units"$