Question

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

Answer 1

Let height of tree AD and the bending point be. B ,BD bent to make BC and the distance between foot of tree and bend is 5m and on ground is 12m applying (Pythagoras theorem)

The answer become 13 =BC =BD .

Now add AB &BD = AD =18m

Answer 2

Let A'CB represent the tree before it is broken at the point C and let the top A' touches the ground at A after it broke Then $\triangle$ ABC is the right angled triangle, right angled at B.

Then,

${ \underline{\purple{ \sf{AB = 12m \: and \: BC = 5m}}}}$

Using Pythagoras Theorem in $\triangle$ ABC

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

$\sf{(AC)^{2} = {12}^{2} + {5}^{2} } \\ \sf{(AC)^{2} = 144 + 25}$

$\sf{(AC)^{2} = 169}$

$\sf \blue{AC= \sqrt{169} }$

$\sf { \underline{\fbox{\pink{AC = 13m}}}}$

$\sf{Original \: height \: of \: tree \: = 5 +13 =}$ $\sf{\Large{\underline{\fbox{\red{18m}}}}}$