Question

a tree is broken at a height of 5 cm from the ground and its top touches the ground at a distance of 12 metres from the base of the tree find the original height of the tree
5cm and 12m​

Answer 1

Answer: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.

AB = 12 m and BC = 5 m

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 = √169

AC= 13 m

Hence, the total height of the tree(A’B) = A’C + CB = 13 + 5 = 18 m.

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Answer 2

Step-by-step explanation:

Let A'CB represents 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, 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.