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.

geoNewbie21 Jan 27, 2021

#1**+2 **

Look at this link and see if that helps: https://web2.0calc.com/questions/thanks-for-any-help_4

-hihihi

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hihihi Jan 27, 2021

#2**+1 **

Looks like the solution, but I was hoping for an explanation on what the numbers in "Area" formula are refering to?

geoNewbie21 Jan 27, 2021

#3**+1 **

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

\(1/2(b1.h1) + 1/2(b2.h2) \)

\(1/2(3.1) + 1/2(3.3) = 3/2 +9/2 =12/2 =6\)

geoNewbie21 Jan 27, 2021