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Given that a fence height 12-ft running parallel to house wall at a distance of 30 \mathrm{ft}. So,Let the horizontal distance from the foot of the tower to fence is b \mathrm{ft} and height of the house is a ft From the above figure the triangles \triangle A C E and \triangle B C D are similar. Therufore,\begin{array}{l}\frac{b}{30+b}=\frac{12}{a} \\\Rightarrow a=\frac{12(30+b)}{b}\end{array}Using the pythogorean theorem\begin{array}{l}E C=l=\sqrt{a^{2}+(30+b)^{2}} \\l=\sqrt{\left(\frac{12(30+b)}{b}\right)^{2}+(30+b)^{2}} \quad\left(\because a=\frac{12(30+b)}{b}\right) \\l=\sqrt{(30+b)^{2}\left(\frac{144}{b^{2}}+1\right)}\end{array}\begin{array}{l}l=\sqrt{(30+b)^{2}\left(\frac{194+b^{2}}{b^{2}}\right)} \\l=\frac{30+b}{b} \sqrt{144+b^{2}}\end{array}To find the shortest length we have to solve l^{\prime}(b)=0\begin{array}{l}l^{\prime}(b)=\left(\frac{30+b}{b}\right)\left(\frac{2 b}{2 \sqrt{144+b^{2}}}\right)+\left(\frac{b-(30+b)}{b^{2}}\right) \sqrt{144+b^{2}}=0 \\l^{\prime}(b)=\frac{30+b}{\sqrt{144+b^{2}}}-\frac{30 \sqrt{144+b^{2}}}{b^{2}}=0 \\l^{\prime}(b)=\frac{30 b^{2}+b^{3}-30\left(144+b^{2}\right)}{b^{2} \sqrt{144+b^{2}}}=0 \\l^{\prime}(b)=\frac{b^{3}-4320}{b^{2} \sqrt{144+b^{2}}}=0 \\\Rightarrow b^{3}=4320 \\b=\sqrt[3]{4320}=6 \sqrt[3]{20}=16.287 \\b=16.287 \\l=\frac{30+16.287}{16.287} \sqrt{144+(16.287)^{2}} \\l=57.49379374125 \approx 57.5\end{array}Hence,The length of the shortest ladder is 57.5 \mathrm{ft}. ...