A tree is broken at a height of 5m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. find the original height of the tree
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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 = 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 = 13 m
Hence, the total height of the tree = AC + CB = 13 + 5 = 18 m.
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 = 13 m
Hence, the total height of the tree = AC + CB = 13 + 5 = 18 m.
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