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The foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

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Answer : [8 m]

Here    AC = 10 m is a ladder

BC = 6 m distance from the base of the wall.

In right-angle triangle ABC use Pythagoras theorem

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

\left ( 10 \right )^{2}=\left ( AB \right )^{2}+\left ( 6 \right )^{2}

100= AB^{2}+36

100-36= AB^{2}

64= AB^{2}

AB=\sqrt{64}

AB=8\; m

Hence, the height of the point on the wall where the top of the ladder reaches is 8 m.

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