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