Question

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.

Answer 1

Let AB be the given tree of height h m, which is broken at D which is 12 m away from its base and the height of remaining part, i.e. CB is 5 m

Now, AB = AC + BC
$\Rightarrow$ AC = AB - BC = h - 5
$\Rightarrow$ AC = CD = h - 5
In right angled $\Delta BDC$, $C{D}^2=C{B}^2+B{D}^2$ [By Pythagoras theorem]
$\Rightarrow (h-5{)}^2=(5{)}^2+(12{)}^2$ [from Eq.(i)]
$\Rightarrow (h-5{)}^2=25+144\Rightarrow (h-5{)}^2=169\Rightarrow h-5=\sqrt{169}=13\Rightarrow h=13+5\Rightarrow h=18m$
Hence, the height of the tree is 18 m