Answer:
[tex]\sqrt{29} , \sqrt{20} , \sqrt{17}[/tex]
Step-by-step explanation:
Consider ΔABC with vertices [tex]A\left ( 0,0 \right )\,,\,B\left ( 6,0 \right )\,,\,C\left ( 4,4 \right )[/tex] such that P , Q , R are midpoints of sides BC , AC and AB .
We know that midpoint of line segment joining points [tex]\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )[/tex] is equal to [tex]\left ( \frac{x_1+x_2}{2}\,,\,\frac{y_1+y_2}{2} \right )[/tex]
Midpoints P , Q , R :
[tex]P\left ( \frac{6+4}{2}\,,\,\frac{0+4}{2} \right )=P\left ( 5\,,\,2 \right )\\Q\left ( \frac{0+4}{2}\,,\,\frac{0+4}{2} \right )=Q\left ( 2\,,\,2 \right )\\R\left ( \frac{6+0}{2}\,,\,\frac{0+0}{2} \right )=R\left ( 3\,,\,0 \right )[/tex]
We know that distance between points [tex]\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )[/tex] is given by [tex]\sqrt{\left ( x_2-x_1 \right )^2+\left ( y_2-y_1 \right )^2}[/tex]
Length of AP :
AP = [tex]\sqrt{\left ( 5-0 \right )^2+\left ( 2-0\right )^2}=\sqrt{25+4}=\sqrt{29}[/tex]
Length of BQ :
BQ = [tex]\sqrt{\left (2-6 \right )^2+\left ( 2-0 \right )^2}=\sqrt{16+4}=\sqrt{20}[/tex]
Length of CR :
[tex]\sqrt{\left (3-4\right )^2+\left ( 0-4 \right )^2}=\sqrt{1+16}=\sqrt{17}[/tex]
