# Thread: area of a triangle

1. ## area of a triangle

Find the area of the triangle with vertices , , and .

2. Originally Posted by wvlilgurl
Find the area of the triangle with vertices , , and .
are you aware of Heron's formula?

Of course, there are other ways to approach this, like the (1/2)base*height formula

3. Originally Posted by wvlilgurl
Find the area of the triangle with vertices , , and .
Let A(-1, -4) , B(3, -3), and C(-4, 4) be the coordinates of triangle. Please see attached diagram.

Using distance formula find

$BC = a = \sqrt{(3+4)^2+(-3-4)^2}= \sqrt{98}$

$CA = b = \sqrt{(-4+1)^2+(4+4)^2}= \sqrt{73}$

$AB = c = \sqrt{(-1-3)^2+(-4+3)^2}= \sqrt{17}$

Now, use cos law to find angle C

$\cos C = \frac{a^2+b^2-c^2}{2ab}$

$\cos C = \frac{98+73-17}{2\sqrt{98}\sqrt{73}}$

$\cos C = 0.910366$

C = 24.44 degrees

Area of triangle ABC $= \frac{1}{2}ab \;\sin C$

$= \frac{1}{2}\sqrt{98}\sqrt{73} \;\sin 24.44$

= 17.5 square units.

4. The formula for arrea of a triangle is

Area of triangle ABC $= \frac{1}{2}ab \;\sin C$

Area of triangle ABC $= \frac{1}{2}bc \;\sin A$

Area of triangle ABC $= \frac{1}{2}ca \;\sin B$