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If a 30 meter ladder is placed against a wall such that it just reaches the top of the wall, if the horizontal distance between the wall and the base of the ladder is 1/3rd of the length of ladder, then the height of wall is :
AC is ladder, C is base of ladder, AB is wall
Given AC = 30 m and BC = $$\frac{1}{3}AC$$ = $$\frac{1}{3}\left(30\right)$$ = 10 m
Applying pythagoras theorem,
$$AB^2+\ BC^2=\ AC^2$$
$$AB^2$$ = $$30^2-10^2$$ = 800
AB = $$20\sqrt{\ 2}$$
Answer is option B.
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