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yaar triangle bnti hai na ...
3rd side nikalo pythagoras theorem sai
......
10 aur 3rd side ko add kro .....result is the answer.....
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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 = 10 m
Using Pythagoras theorem,
In ΔABC
(AC)²=(AB)²+(BC)²
(AC)²=(12)²+(10)²
(AC)²=144+100
(AC)²=244
AC = √244
AC= 15.62(approx) m
Hence, the total height of the tree(A’B) = A’C + CB = 15.62(approx) + 5 = 20.62(approx) m.
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