Question

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

Answer 1

Step-by-step explanation:

Let A'CB be the tree before it broken at the point C and let the top A' touches the ground at A after it broke. Then ΔABC is a right angled triangle, at B. Hence, the total height of the tree(A'B) = A'C + CB = 13 + 5 = 18 m

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

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

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

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

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

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

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