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What is the area of the composite figure whose vertices have the following coordinates?

(−2, −2)(−2, −2) , (4, −2)(4, −2) , (5, 1)(5, 1) , (2, 3)(2, 3) , (−1, 1)

 

Enter your answer in the box.

___  units²

 

Sorry for the easy question, I'm just dumb

 Feb 18, 2019
 #1
avatar-18 
-2

Area of a rectangle = W X L Area of a rectangle = 5 X 2 Area of a rectangle = 10 Area of triangle 1= 1/2 X B X H Area of triangle 1= 1/2 X 2 X 2 Area of triangle 1= 1/2 X 4

 

This is someting that might help you!

 Feb 18, 2019
edited by Pigglepoox2007  Feb 18, 2019
 #5
avatar-18 
0

Sorry so so sorry got carried away well if it helps let it rip then!!!!!!!!!

Pigglepoox2007  Feb 18, 2019
 #6
avatar+1268 
+2

I the answer I believe is 30 squared units. and your not dumb you won't get far discouraging yourself think more positive and you will be real good at whatever you aim for.

 Feb 18, 2019
edited by HiylinLink  Feb 18, 2019
 #8
avatar+99604 
+2

******

 

 

cool cool cool

 Feb 18, 2019
edited by CPhill  Feb 19, 2019
 #9
avatar+21980 
+4

What is the area of the composite figure whose vertices have the following coordinates?

(−2, −2) , (4, −2) , (5, 1) , (2, 3) , (−1, 1)

 

\(\begin{array}{|r|r|r|r|r|} \hline \text{Point} & x & y & \\ \hline 1 & -2 & -2 &\\ & & & (-2)(-2) - 4(-2) & = 12 \\ 2 & 4 & -2 & \\ & & & 4\cdot 1 - 5 (-2) & = 14 \\ 3 & 5 & 1 &\\ & & & 5\cdot 3 - 2 (1) & = 13 \\ 4 & 2 & 3 &\\ & & & 2\cdot 1 - (-1)3 & = 5 \\ 5 & -1 & 1 &\\ & & & (-1)(-2) - (-2)1 & = 4 \\ 1 & -2 & -2 & \\ \hline & & & & \text{sum} = 48 \\ & & & & \text{area of the composite figure } = \dfrac{\text{sum}}{2} = \dfrac{\text{48}}{2} = \mathbf{24} \\ \hline \end{array}\)

 

laugh

 Feb 19, 2019
edited by heureka  Feb 19, 2019

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