+0

+1
120
7

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
+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
+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
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
+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
0

What a nice response!

Welcome to our Web2.0calc forum It would be nice if you both went back and gave Omi a point for her answer though. Melody  Jan 27, 2021
edited by Melody  Jan 27, 2021
#5
+1

Hi Melody, Thank you and Will do!!

Jan 28, 2021
#6
0

Thanks Geo,

I am sure Hihihi would like a point from you too Melody  Jan 28, 2021