Question

A TREE IS BROKEN AT A HEIGHT OF 5m FROM THE GROUND AND IT'S TOP TOUCHES THE GROUND AT A DISTANCE OF 12m FROM THE BASE OF THE TREE. FIND THE ORIGINAL HEIGHT OF THE TREE.​

Answer 1

Answer:

Use pythagoras theorem

${h}^{2} = {p}^{2} + {b}^{2} \\ \\ {h }^{2} = {5}^{2} + {12}^{2} \\ \\ {h }^{2} = 25 + 144 \\ \\ {h}^{2} = 169 \\ \\ {h}^{2} = {13}^{2} \\ \\ square \: and \: square \: cancelled \: out.. \\ \\ h = 13$

The real height of the tree is 5+13 m= 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

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

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

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

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

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

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