Question

the vertices of triangle rst are r(-1,-2), s(2,-1), and t(4,2). if △jkl ~ △rst and the length of kl is \frac{\sqrt{13}}{2} units, what is the length of \overline{jk}? \frac{\sqrt{10}}{2} \sqrt{10} \sqrt{13} 2\sqrt{13}

Explanation:

Step1: Calculate the length of \( \overline{RS} \)

Use the distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \).
For \( R(-1,-2) \) and \( S(2,-1) \), \( x_1=-1,y_1 = - 2,x_2=2,y_2=-1 \).
\( RS=\sqrt{(2-(-1))^2+(-1-(-2))^2}=\sqrt{(2 + 1)^2+(-1 + 2)^2}=\sqrt{9 + 1}=\sqrt{10} \).

Step2: Use the property of similar triangles

Since \( \triangle{JKL}\sim\triangle{RST} \), we have the proportion \( \frac{JK}{RS}=\frac{KL}{ST} \).
We know \( KL=\frac{\sqrt{13}}{2} \). First, find \( ST \):
For \( S(2,-1) \) and \( T(4,2) \), \( x_1 = 2,y_1=-1,x_2=4,y_2 = 2 \).
\( ST=\sqrt{(4 - 2)^2+(2-(-1))^2}=\sqrt{4+9}=\sqrt{13} \).
From \( \frac{JK}{RS}=\frac{KL}{ST} \), substitute \( RS=\sqrt{10} \), \( KL=\frac{\sqrt{13}}{2} \), \( ST=\sqrt{13} \).
\( JK=\frac{RS\times KL}{ST}=\frac{\sqrt{10}\times\frac{\sqrt{13}}{2}}{\sqrt{13}}=\frac{\sqrt{10}}{2} \).

Answer:

\(\frac{\sqrt{10}}{2}\)