+0  
 
+1
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avatar+110 

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.

 Jan 27, 2021
 #1
avatar+310 
+2

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

 

 

 

-hihihi

 

 

🔮🔮🔮

 Jan 27, 2021
 #2
avatar+110 
+1

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

 Jan 27, 2021
 #7
avatar
0

[ABD] = 1/2(BD * AO) = 1.5

 

[BCD] = 1/2(BD * CO) = 4.5

 

[ABCD] = 1.5 + 4.5 = 6

Guest Jan 28, 2021
 #3
avatar+110 
+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

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

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

 Jan 27, 2021
 #4
avatar+112523 
0

What a nice response!

 

Thanks for the link hihihi and thanks for your response GeoNewbie21

 

Welcome to our Web2.0calc forum  laugh

 

It would be nice if you both went back and gave Omi a point for her answer though.   wink

Melody  Jan 27, 2021
edited by Melody  Jan 27, 2021
 #5
avatar+110 
+1

Hi Melody, Thank you and Will do!!

 Jan 28, 2021
 #6
avatar+112523 
0

Thanks Geo,

I am sure Hihihi would like a point from you too  wink

Melody  Jan 28, 2021

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