Question

molly is putting a triangular sandbox in her backyard. she draws the plan for the sandbox on a coordinate - plane. vertex a is at (-4,6), vertex b is at (6,1), and vertex c is at (0, - 9). calculate the lengths of the sides and put them in order from shortest to longest. side bc side ab side ac

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate length of side AB

Let $(x_1,y_1)=(- 4,6)$ and $(x_2,y_2)=(6,1)$. Then $AB=\sqrt{(6 + 4)^2+(1 - 6)^2}=\sqrt{100 + 25}=\sqrt{125}=5\sqrt{5}$.

Step3: Calculate length of side AC

Let $(x_1,y_1)=(-4,6)$ and $(x_2,y_2)=(0,-9)$. Then $AC=\sqrt{(0 + 4)^2+(-9 - 6)^2}=\sqrt{16+225}=\sqrt{241}$.

Step4: Calculate length of side BC

Let $(x_1,y_1)=(6,1)$ and $(x_2,y_2)=(0,-9)$. Then $BC=\sqrt{(0 - 6)^2+(-9 - 1)^2}=\sqrt{36 + 100}=\sqrt{136}=2\sqrt{34}$.

Step5: Compare the lengths

We know that $\sqrt{125}\approx11.18$, $\sqrt{136}\approx11.66$, $\sqrt{241}\approx15.52$. So, $AB = 5\sqrt{5}\approx11.18$, $BC = 2\sqrt{34}\approx11.66$, $AC=\sqrt{241}\approx15.52$.

Answer:

Shortest to longest: Side AB, Side BC, Side AC