A tree is broken at a height of $15 m$ from the ground and touched the ground at a distance of $36 m$ from the base of the tree. Find the original height of the tree.

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Solution

The Remaining part of the tree $C D = A C$

Given $B C = 15 m A B = 36 m$

$A C = \sqrt{15^{2} + 36^{2}} = \sqrt{1521} = 39$

Original Height of tree $= 39 + 15 = 54 m$

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A tree is broken at a height of 15m from the ground and touched the ground at a distance of 36m from the base of the tree. Find the original height of the tree.

The Remaining part of the tree CD=ACGiven BC=15 mAB=36 mAC=√152+362=√1521=39Original Height of tree =39+15=54 m

A tree is broken at a height of $15 m$ from the ground and touched the ground at a distance of $36 m$ from the base of the tree. Find the original height of the tree.

The Remaining part of the tree $C D = A C$

Given $B C = 15 m A B = 36 m$

$A C = \sqrt{15^{2} + 36^{2}} = \sqrt{1521} = 39$

Original Height of tree $= 39 + 15 = 54 m$