Rs aggarwal solutions
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# Question 4

A 15-m-long ladder is placed against a wall to reach a window 12 m high. Find the distance of the foot of the ladder from the wall.

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Let BC be the wall and AB be the ladder.

Then, AB = 15 m and BC = 12 m.

Now, ΔABC being right-angled at C, we have:

\begin{array}{l} A B^{2}=B C^{2}+A C^{2} \\ A C^{2}=\left(A B^{2}-B C^{2}\right) \\ A C^{2}=\left(15^{2}-12^{2}\right) \\ A C^{2}=(225-144) \\ A C^{2}=(81) \end{array}

Sending power2 from LHS to RHS it becomes square root

AC = √(81)

AC = 9 cm

∴The distance of the foot of the ladder from the wall is 9 cm.

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