Question

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the graph shows triangles bcd and efg.

Explanation:

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

The distance formula is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For triangle \(BCD\):

• \(BD\): \(B(8,4)\), \(D(- 6,2)\)

\(BD=\sqrt{(8 + 6)^2+(4 - 2)^2}=\sqrt{196 + 4}=\sqrt{200}=10\sqrt{2}\)

• \(BC\): \(B(8,4)\), \(C(4,2)\)

\(BC=\sqrt{(8 - 4)^2+(4 - 2)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\)

• \(CD\): \(C(4,2)\), \(D(-6,2)\)

\(CD=\sqrt{(4 + 6)^2+(2 - 2)^2}=\sqrt{100}=10\)

For triangle \(EFG\):

• \(EG\): \(E(4,2)\), \(G(-3,1)\)

\(EG=\sqrt{(4 + 3)^2+(2 - 1)^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}\)

• \(EF\): \(E(4,2)\), \(F(2,1)\)

\(EF=\sqrt{(4 - 2)^2+(2 - 1)^2}=\sqrt{4 + 1}=\sqrt{5}\)

• \(FG\): \(F(2,1)\), \(G(-3,1)\)

\(FG=\sqrt{(2 + 3)^2+(1 - 1)^2}=\sqrt{25}=5\)

Step2: Check the ratios of the corresponding sides

\(\frac{BD}{EG}=\frac{10\sqrt{2}}{5\sqrt{2}} = 2\), \(\frac{BC}{EF}=\frac{2\sqrt{5}}{\sqrt{5}}=2\), \(\frac{CD}{FG}=\frac{10}{5}=2\)

Since the ratios of the corresponding sides of \(\triangle BCD\) and \(\triangle EFG\) are equal (\(\frac{BD}{EG}=\frac{BC}{EF}=\frac{CD}{FG} = 2\)), by the Side - Side - Side (SSS) similarity criterion, the two triangles are similar.

Answer:

Yes, \(\triangle BCD\) is similar to \(\triangle EFG\) because the ratios of their corresponding sides are equal.