A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution :

Let ladder be AB, B be the window and CB be the wall. Then, ABC is a right triangle, angled at C.triangle

\(\therefore\)  \({AB}^2\) = \({AC}^2\) + \({BC}^2\)

So,  \({10}^2\) = \({AC}^2\) + \(8^2\)

or  \({AC}^2\) = 100 – 64

\(\implies\) \({AC}^2\) = 36

\(\implies\)  AC = 6 m

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