Question

Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

A= units^2

Answer 1

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( x_2- x_1)^2 + ( y_2- y_1)^2)}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=√([ 15- 10]^2 + [ 15- 5]^2) \\\\\\ ~\hfill \boxed{a=√(125)} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=√([ 30- 15]^2 + [ 9- 15]^2) \\\\\\ ~\hfill \boxed{b=√(261)}`

`(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=√([ 10- 30]^2 + [ 5- 9]^2) \\\\\\ ~\hfill \boxed{c=√(416)} \\\\[-0.35em] ~\dotfill`

`\qquad \textit{Heron's area formula} \\\\ A=√(s(s-a)(s-b)(s-c))\qquad \begin{cases} s=(a+b+c)/(2)\\[-0.5em] \hrulefill\\ a=√(125)\\ b=√(261)\\ c=√(416)\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-√(125))(23.87-√(261))(23.87-√(416))}\implies \boxed{A\approx 90}`