1. A 6 meter ladder rests against a vertical wall. The midpoint of the ladder is twice as far from the ground as it is from the wall.
At what height on the wall does the ladder reach? Simplify this as much as possible, and explain your reasoning.
Soln.
Let $\mathrm{x}$ be the distance from the midpoint of the ladder to the wall. Using congruent triangles and Pythagoras, $6^{2}=(2 x)^{2}+(4 x)^{2}, \quad[1 \mathrm{mk}]$ whence $\mathrm{x}=\frac{3}{\sqrt{5}}(1.34 \mathrm{~m}) \quad[1 \mathrm{mk}]$
So the desired height is $\frac{12}{\sqrt{5}}(5.37 \mathrm{~m}) \quad$ [1 mk]