Answer
The length of the side AB is\[21\text{ ft}\].
Work Step by Step
\[\Delta CDA\]and\[\Delta CDB\]are right triangles. In\[\Delta CDA\], \[AC=13\]and\[CD=12\].
Compute the value of \[c\]by using Pythagorean Theorem in \[\Delta CDA\] and substitute the value of a and b into\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\].
\[\begin{align}
& {{c}^{2}}={{a}^{2}}+{{b}^{2}} \\
& A{{C}^{2}}=C{{D}^{2}}+A{{D}^{2}} \\
& {{\left( 13 \right)}^{2}}={{\left( 12 \right)}^{2}}+A{{D}^{2}} \\
& 169=144+A{{D}^{2}}
\end{align}\]
\[\begin{align}
& 169-144=A{{D}^{2}} \\
& 25=A{{D}^{2}} \\
& \sqrt{25}=A{{D}^{2}} \\
& 5ft=AD
\end{align}\]
In \[\Delta CDB\], \[CB=20\]and\[CD=12\].
Compute the value of \[c\]by using Pythagorean Theorem in \[\Delta CDB\] and substitute the value of a and b into\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\].
\[\begin{align}
& {{c}^{2}}={{a}^{2}}+{{b}^{2}} \\
& C{{B}^{2}}=C{{D}^{2}}+B{{D}^{2}} \\
& {{\left( 20 \right)}^{2}}={{\left( 12 \right)}^{2}}+B{{D}^{2}} \\
& 400=144+B{{D}^{2}}
\end{align}\]
Re arrange the equation as follows:
\[\begin{align}
& B{{D}^{2}}=400-144 \\
& B{{D}^{2}}=256 \\
& \sqrt{B{{D}^{2}}}=\sqrt{256} \\
& BD=16ft
\end{align}\]
Distance between point A and point B is computed as follows:
\[\begin{align}
& AB=AD+BD \\
& =5\text{ ft}+16\text{ ft} \\
& =21\text{ ft}
\end{align}\]
Hence, the length of the side AB is\[21\text{ ft}\].
