Question

warm-up
what are the side lengths of triangle abc?
type the correct answer in each box. if necessary, round any decimal to the nearest tenth.
ab = units
bc = units
ac = units

Explanation:

Step1: Find coordinates of points

From the graph, \(A(-6,3)\), \(B(-3,6)\), \(C(5,0)\)

Step2: Use distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

For \(AB\):
\(x_1=-6,y_1 = 3,x_2=-3,y_2 = 6\)
\(AB=\sqrt{(-3+6)^2+(6 - 3)^2}=\sqrt{3^2+3^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\approx4.2\)

Step3: For \(BC\):

\(x_1=-3,y_1 = 6,x_2=5,y_2 = 0\)
\(BC=\sqrt{(5 + 3)^2+(0 - 6)^2}=\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10\)

Step4: For \(AC\):

\(x_1=-6,y_1 = 3,x_2=5,y_2 = 0\)
\(AC=\sqrt{(5 + 6)^2+(0 - 3)^2}=\sqrt{11^2+(-3)^2}=\sqrt{121 + 9}=\sqrt{130}\approx11.4\)

Answer:

\(AB\approx4.2\) units, \(BC = 10\) units, \(AC\approx11.4\) units