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

Answer 1

Let $A^{\prime }CB$ be the tree before it broken at the point C and let the top $A^{\prime }$ touches the ground at A after it broke$.$ Then $\Delta ABC$ is a right angled triangle$,$ at $B.$

$AB=12m$ and $BC=5m$

Using Pythagoras theorem$,$

In $\Delta ABC$

$(AC{)}^2=(AB{)}^2+(BC{)}^2$

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

$(AC{)}^2=144+25$

$(AC{)}^2=169$

$AC=\sqrt{169}$

$AC=13m$

Hence$,$ the total height of the tree$\left(A^{\prime }B\right)$ $=A^{\prime }C+CB=13+5=18m.$