Question

A tree is broken at a height of 5 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

Answer:

18 cm

Step-by-step explanation:

Let BD be the actual height of the tree

BD = AB + AC

so we can say that AC = AD

Given: AB = 5cm , BC = 12cm

Lets consider ΔABC

by Pythagoras theorem,

AC² = AB²+BC²

AC² = 5² + 12²

AC² = 25 + 144

AC² = 169

∴AC = √169

∴AC = 13cm

we know AC = BD

∴ BD = 13cm

Orginal height of tree = 13 + 5 = 18cm

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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

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

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

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

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

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

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