Question

the vertices of (\triangle abc) are (a(1,-2)), (b(1,1)), and (c(5,-2)). which could be the side lengths of a triangle that is similar but not congruent to (\triangle abc)?
(\bigcirc) 3, 4, and 5 units
(\bigcirc) 9, 12, and 15 units
(\bigcirc) 6, 4, and 5 units
(\bigcirc) 9, 12, and 5 units

Explanation:

Step1: Calculate the side lengths of \(\triangle ABC\)

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

• \(AB=\sqrt{(1 - 1)^2+(1+ 2)^2}=\sqrt{0 + 9}=3\)
• \(AC=\sqrt{(5 - 1)^2+(-2 + 2)^2}=\sqrt{16+0}=4\)
• \(BC=\sqrt{(5 - 1)^2+(-2 - 1)^2}=\sqrt{16 + 9}=5\)

Step2: Check the similarity ratio

For similar triangles, the ratio of corresponding side lengths is the same.
If the side lengths of \(\triangle ABC\) are \(3\), \(4\), \(5\) and we consider the option \(9\), \(12\), \(15\)
The ratio is \(\frac{9}{3}=\frac{12}{4}=\frac{15}{5}=3\)

Answer:

B. \(9, 12,\) and \(15\) units