Question

Can anyone figure this out?

Answer 1

`\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ NA=√((6+3)^2+(3-10)^2)\implies NA=√(130) \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=√((6-6)^2+(-1-3)^2)\implies AD=4 \\\\[-0.35em] ~\dotfill`

`\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=√((-3-6)^2+(10+1)^2)\implies DN=√(202)`

now that we know how long each one is, let's plug those in Heron's Area formula.

`\bf \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=√(130)\\ b=4\\ c=√(202)\\[1em] s=(√(130)+4+√(202))/(2)\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-√(130))(14.81-4)(14.81-√(202))} \\\\\\ A=√(324)\implies A=18`