A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?

Solution :

Let AB = 24 m be guy wire attached to a vertical pole BC of height 18 m. To keep the wire taut, let it be fixed to stake at A. Then, ABC is a right angled triangle at C.

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

So, \({24}^2\) = \({AC}^2\) + \({18}^2\)

\(\implies\) \({AC}^2\) = 576 – 324

\(\implies\)  \({AC}^2\) = 252

\(\implies\)  AC = \(6\sqrt{7}\)

Hence, the stake may be placed at a distance of \(6\sqrt{7}\) m from the base of pole.

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