+136 Question: Solve for the length of segment AB using an equation based on the phythagorean theorem
My attempt was:
AB^2= AC^2+BC^2
AB^2=Z^2+6^2
AB=Z^2+36
And I feel that I'm not doing it right...so please help! I'm struggling a bit with this one
+26367 Solve for the length of segment AB using an equation based on the phythagorean theorem:
\(\begin{array}{|rcll|} \hline AB^2 &=& AC^2+BC^2 \quad | \quad AB = z+2,\ AC = z,\ BC=6 \\ \left(z+2\right)^2 &=& z^2+6^2 \\ \not{z^2}+4z+4 &=& \not{z^2}+ 36\\ 4z+4 &=& 36 \quad | \quad -4 \\ 4z &=& 32 \quad | \quad :4 \\\\ z &=& \dfrac{32}{4} \\\\ \mathbf{ z } &=& \mathbf{8} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline AB &=& z+2 \quad | \quad z=8 \\ AB &=& 8+2 \\ \mathbf{ AB } &=& \mathbf{10} \\ \hline \end{array}\)
check:
\(\begin{array}{|rcll|} \hline (z+2)^2 &=& z^2+6^2 \\ (8+2)^2 &=& 8^2+6^2\\ 10^2 &=& 8^2 +6^2 \\ 100 &=& 64 + 36 \\ 100 &=& 100\ \checkmark \\ \hline \end{array}\)
+26367 Solve for the length of segment AB using an equation based on the phythagorean theorem:
\(\begin{array}{|rcll|} \hline AB^2 &=& AC^2+BC^2 \quad | \quad AB = z+2,\ AC = z,\ BC=6 \\ \left(z+2\right)^2 &=& z^2+6^2 \\ \not{z^2}+4z+4 &=& \not{z^2}+ 36\\ 4z+4 &=& 36 \quad | \quad -4 \\ 4z &=& 32 \quad | \quad :4 \\\\ z &=& \dfrac{32}{4} \\\\ \mathbf{ z } &=& \mathbf{8} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline AB &=& z+2 \quad | \quad z=8 \\ AB &=& 8+2 \\ \mathbf{ AB } &=& \mathbf{10} \\ \hline \end{array}\)
check:
\(\begin{array}{|rcll|} \hline (z+2)^2 &=& z^2+6^2 \\ (8+2)^2 &=& 8^2+6^2\\ 10^2 &=& 8^2 +6^2 \\ 100 &=& 64 + 36 \\ 100 &=& 100\ \checkmark \\ \hline \end{array}\)
