A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

In the given figure, BC represents the unbroken part of the tree. Point C represents the point where the tree broke and CA represents the broken part of the tree. Triangle ABC, thus formed, is right-angled at B.

Applying Pythagoras theorem in ΔABC,

AC^{2}
= BC^{2} + AB^{2}

AC^{2}
= (5 m)^{2} + (12 m)^{2}

AC^{2}
= 25 m^{2} + 144 m^{2} = 169 m^{2}

AC = 13 m

Thus, original height of the tree = AC + CB = 13 m + 5 m = 18 m

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