A 8 , m long bamboo tree standing erect on th | vedclass

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(B) The total height of the bamboo tree is $8 \, m$.It breaks at a height of $3 \, m$ from the ground.Let the total height be $H = 8 \, m$ and the hei

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$4$

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(B) The total height of the bamboo tree is $8 \, m$.It breaks at a height of $3 \, m$ from the ground.Let the total height be $H = 8 \, m$ and the height at which it breaks be $h = 3 \, m$.The length of the broken part (hypotenuse of the right-angled triangle formed) is $L = H - h = 8 - 3 = 5 \, m$.The height of the remaining trunk is $h = 3 \, m$.Let $x$ be the distance between the top of the tree on the ground and the base of the tree.According to the Pythagorean theorem in the right-angled triangle formed:$x^2 + h^2 = L^2$$x^2 + 3^2 = 5^2$$x^2 + 9 = 25$$x^2 = 25 - 9 = 16$$x = \sqrt{16} = 4 \, m$.Thus,the distance is $4 \, m$.

$A$ $8 \, m$ long bamboo tree standing erect on the ground breaks at the height of $3 \, m$ from the ground. The broken part of the tree remains attached to the trunk. Find the distance in $m$ between the top of the tree on the ground and the base of the tree.

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