In square ABCD shown, X is the midpoint of BC and Y is the midpoint of AX. If CZ:ZD=3:1 and the side length of the square is 5 units, find YZ.

Guest Jun 25, 2020

#2**+3 **

**In square ABCD shown, X is the midpoint of BC and Y is the midpoint of AX. If CZ:ZD=3:1 and the side length of the square is 5 units, find YZ.**

\(\begin{array}{|rcll|} \hline \mathbf{ZD} &=& \dfrac{1}{1+3}*5 \\\\ &=& \dfrac{5}{4} \\\\ &=& \mathbf{1.25} \\\\ \mathbf{CZ} &=& \dfrac{3}{1+3}*5 \\\\ &=& \dfrac{15}{4} \\\\ &=& \mathbf{3.75} \\\\ \mathbf{z} &=& 5-ZD-2.5\\ &=& 5-1.25-2.5\\ &=& \mathbf{1.25} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{Intercept theorem:}\\ \hline \mathbf{\dfrac{2.5}{5-y}} &=& \mathbf{\dfrac{5}{2.5}} \\\\ \dfrac{5-y}{2.5} &=& \dfrac{2.5}{5} \\\\ 5-y &=& \dfrac{2.5^2}{5} \\\\ y &=& 5-\dfrac{2.5^2}{5} \\\\ y &=& \dfrac{25-2.5^2}{5} \\\\ \mathbf{ y } &=& \mathbf{3.75} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^2 &=& y^2+z^2 \\ x^2 &=& 3.75^2+1.25^2 \\ x^2 &=& 15.625 \\ \mathbf{x} &=& \mathbf{3.95284707521} \\ \hline \end{array}\)

\(YZ\) is \(\approx \mathbf{3.95}\)

heureka Jun 25, 2020