+126 How do I find the are of a convex quadrialater with vertices (-1,0),(0,1),(2,0) and (0-3)?
I used the distance formula and found the side lengths to be \(\sqrt{2}, \sqrt{5}, \sqrt{13}, and \sqrt{10}.\)
Not sure how to proceed beyond this.
+285 Look at this link and see if that helps: https://web2.0calc.com/questions/thanks-for-any-help_4
-hihihi
🔮🔮🔮
+126 Looks like the solution, but I was hoping for an explanation on what the numbers in "Area" formula are refering to?
+126 Got it thanks The figure in the link https://web2.0calc.com/questions/thanks-for-any-help_4 helped.
Area of Triangle1 + Area of Triangle2
\(1/2(b1.h1) + 1/2(b2.h2) \)
\(1/2(3.1) + 1/2(3.3) = 3/2 +9/2 =12/2 =6\)
