Q.5 Solve the following. 2. A tree is broken at a height of 4m from the ground. Its top touches the ground at a distance of 3m from the base of the tree. Find the original height of the tree.
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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 ∆ABC is a right angled triangle, at B.
AB = 12m and BC = 5m
Using Pythagoras theorem
In ∆ABC
(AC)²= (AB)²+(BC)²
(AC)²= (12)²+(5)²
(AC)²= 169
AC= √169
AC= 13m
Hence, the total height of the tree
(A' B)= A'C+CB= 13+5= 18m
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