in the diagram, △abc ≅ △wrs. what is the perimeter of △wrs?\n10 units\n11 units\n12 units\n13 units

Explanation:

Step1: Find side - lengths using distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For side $WR$, with $W(-1, - 1)$ and $R(2,1)$: [ \begin{align*} d_{WR}&=\sqrt{(2+1)^2+(1 + 1)^2}\ &=\sqrt{3^2+2^2}\ &=\sqrt{9 + 4}\ &=\sqrt{13}\approx3.6 \end{align*} ] For side $RS$, with $R(2,1)$ and $S(5,0)$: [ \begin{align*} d_{RS}&=\sqrt{(5 - 2)^2+(0 - 1)^2}\ &=\sqrt{3^2+( - 1)^2}\ &=\sqrt{9+1}\ &=\sqrt{10}\approx3.2 \end{align*} ] For side $WS$, with $W(-1,-1)$ and $S(5,0)$: [ \begin{align*} d_{WS}&=\sqrt{(5 + 1)^2+(0 + 1)^2}\ &=\sqrt{6^2+1^2}\ &=\sqrt{36 + 1}\ &=\sqrt{37}\approx6.1 \end{align*} ] Or we can use the fact that since $\triangle ABC\cong\triangle WRS$, we can find side - lengths of $\triangle ABC$ and use congruence. For $\triangle ABC$, $AB = 3$, $BC=4$, and using the Pythagorean theorem for $AC$ (where $A(-4,4),C(-1,1)$), $AC=\sqrt{( - 4+1)^2+(4 - 1)^2}=\sqrt{(-3)^2+3^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\approx4.2$. The perimeter of $\triangle ABC$: $P = 3+4 + 5=12$ (by Pythagorean triple for right - triangle with sides 3, 4, 5). Since $\triangle ABC\cong\triangle WRS$, the perimeter of $\triangle WRS$ is the same as the perimeter of $\triangle ABC$.

Answer:

12 units