# How do you solve the triangle given a = 5, b = 8, c = 12?

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2
Mar 25, 2018

color(red)(A_t = sqrt(s * (s-a) * (s-b) * (s-c)) = 14.52 " sq units"

#### Explanation:

We will use semi-perimeter formula to find the area of the triangle.

$\text{Semi-perimeter } s = \frac{a + b + c}{2} = \frac{5 + 8 + 12}{2} = 12.5$

Area of the triangle is given by the formula

${A}_{t} = \sqrt{s \cdot \left(s - a\right) \cdot \left(s - b\right) \cdot \left(s - c\right)}$

${A}_{t} = \sqrt{12. 5 \cdot \left(12.5 - 5\right) \cdot \left(12.5 - 8\right) \cdot \left(12.5 - 12\right)}$

color(red)(A_t = sqrt(12.5 * 7.5 * 4.5 * 0.5) = 14.52 " sq units"

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1
Mar 25, 2018

color(green)(A_t = (1/2) a b sin C ~~ 14.52 " sq units)

#### Explanation:

We can solve the sum in two ways to find the area of the triangle :

1) Using semi-perimeter formula, having known three sides.

2) By applying law of cosines, we can find one angle and using law of sines we can find the area of the triangle.

Here, we will use Law of cosines and sines and arrive at the area of the triangle.

cos C = (a^2 + b^2 - c^2 ) / (2 * a * b)

$\cos C = \frac{{5}^{2} + {8}^{2} - {12}^{2}}{2 \cdot 5 \cdot 8}$

$\cos C = - \frac{55}{80} = - \frac{11}{16}$

$\hat{C} = {133.43}^{\circ}$

Now, using Law of sines, we calculate the area.

color(green)(A_t = (1/2) a b sin C = (1/2) * 5 * 8 * sin 133.43 ~~ 14.52 " sq units)#

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