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Question Number 202276 by professorleiciano last updated on 23/Dec/23

	Answered by professorleiciano last updated on 23/Dec/23

	$A r e a (r e t a n g u l o) = 4 \times 6 = 24 m^{2}$ $A r e a (t r i a n g u l o I) = 3 \times 6 = 18 / 2 = 9 m^{2}$ $A r e a (t r i a n g u l o I I) = 3 \times 4 = 12 / 2 = 6 m^{2}$ $A r e a (t r i a n g u l o I I I) = 3 \times 4 = 12 / 2 = 6 m^{2}$ $A r e a (t o t a l) = 24 m^{2} + 9 m^{2} + 6 m^{2} + 6 m^{2}$ $= 45 m^{2}$ $A l t e r n a t i v a (a)$
	Commented by professorleiciano last updated on 23/Dec/23

	Commented by mr W last updated on 24/Dec/23

	$a b e t t e r a n d m o r e g e n e r a l m e t h o d$ $s e e b e l o w .$
	Answered by mr W last updated on 24/Dec/23

	Commented by mr W last updated on 24/Dec/23

	$A_{k} = \frac{(x_{k + 1} - x_{k}) (y_{k} + y_{k + 1})}{2}$ $\begin{array}{cccccccc} k & x_{k} & y_{k} & x_{k + 1} - x_{k} & y_{k} + y_{k + 1} & A_{k} \\ 1 & 1 & 1 & 0 & 8 & 0 \\ 2 & 1 & 7 & 4 & 14 & 28 \\ 3 & 5 & 7 & 2 & 18 & 18 \\ 4 & 7 & 11 & 3 & 22 & 33 \\ 5 & 10 & 11 & - 5 & 12 & - 30 \\ 6 & 5 & 1 & - 4 & 2 & - 4 \\ 7 (= 1) & 1 & 1 & ╱ & ╱ & 45 \end{array}$ $A r e a o f p o l y g o n = Σ A_{k} = 45 ✓$ $\Rightarrow a n s w e r (a)$
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