A tree is broken at a height 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 height of the tree.
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Answer:
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.
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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.
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