When a ladder leans on a wall to reach a heig | vedclass

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(A) Let the length of the ladder be $L$. The ladder, the wall, and the ground form a right-angled triangle.According to the Pythagorean theorem, $L^2

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$8$

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(A) Let the length of the ladder be $L$. The ladder, the wall, and the ground form a right-angled triangle.According to the Pythagorean theorem, $L^2 = (	ext{height})^2 + (	ext{base})^2$.In the first case, height $= 8 \, m$ and base $= 6 \, m$.So, $L^2 = 8^2 + 6^2 = 64 + 36 = 100$.Thus, $L = \sqrt{100} = 10 \, m$.In the second case, the same ladder of length $L = 10 \, m$ is used, and the new height is $6 \, m$.Let the new base be $x$.Using the Pythagorean theorem again: $L^2 = (	ext{new height})^2 + x^2$.$10^2 = 6^2 + x^2$.$100 = 36 + x^2$.$x^2 = 100 - 36 = 64$.$x = \sqrt{64} = 8 \, m$.Therefore, the distance of the lower end from the base of the wall is $8 \, m$.

When a ladder leans on a wall to reach a height of $8 \, m$, its lower end rests $6 \, m$ away from the base of the wall. If the same ladder leans on the wall to reach a height of $6 \, m$, find the distance of its lower end from the base of the wall. (in $m$)

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