Question

the graph shows triangles bcd and efg. is bcd similar to efg? justify your

Explanation:

Step1: Find the lengths of the sides using the distance formula

The distance formula is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(BC\): \(B(9,3)\), \(C(5,1)\)
\(BC=\sqrt{(9 - 5)^2+(3 - 1)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\)
For \(CD\): \(C(5,1)\), \(D(-7,2)\)
\(CD=\sqrt{(-7 - 5)^2+(2 - 1)^2}=\sqrt{144+1}=\sqrt{145}\)
For \(BD\): \(B(9,3)\), \(D(-7,2)\)
\(BD=\sqrt{(-7 - 9)^2+(2 - 3)^2}=\sqrt{256 + 1}=\sqrt{257}\)
For \(EF\): \(E(5,1)\), \(F(2,1)\)
\(EF=\sqrt{(2 - 5)^2+(1 - 1)^2}=\sqrt{9+0}=3\)
For \(FG\): \(F(2,1)\), \(G(-3,1)\)
\(FG=\sqrt{(-3 - 2)^2+(1 - 1)^2}=\sqrt{25+0}=5\)
For \(EG\): \(E(5,1)\), \(G(-3,1)\)
\(EG=\sqrt{(-3 - 5)^2+(1 - 1)^2}=\sqrt{64+0}=8\)

Step2: Check the ratio of the sides

\(\frac{BC}{EF}=\frac{2\sqrt{5}}{3}\), \(\frac{CD}{FG}=\frac{\sqrt{145}}{5}\), \(\frac{BD}{EG}=\frac{\sqrt{257}}{8}\)
Since \(\frac{BC}{EF}
eq\frac{CD}{FG}
eq\frac{BD}{EG}\)

Answer:

Triangles \(BCD\) and \(EFG\) are not similar.