Question

the top of a broken tree touches the ground at a distance of 12m from its base.if the tree is broken at a height of 5m from the ground then the actual height of the tree is

Answer 1

Let ΔABC is a right angled triangle, right angled at B.

AB=12m and BC=5m

Using Pythagoras theorem, In ΔABC

(AC)²+(AB)²+(BC)²

⇒(AC)²

=(12)²+(5)²

⇒(AC)²

=144+25

⇒(AC)²

=169

⇒AC=13m

Hence, the total height of the tree=AC+CB=13+5=18m.

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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

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

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

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

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

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

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