Question 11
The top of a broken tree touches the ground at a distance of 12 m from its base. If the tree is broken at a height of 5 m from the ground, then the actual height of the tree is
a) 25 m
b) 13 m
c) 18 m
d) 17 m
Let AB be the given tree of height h m, which is broken at D which is 12 m away from its base and the height of remaining part, i.e. CB is 5 m
Now, AB = AC + BC
⇒ AC = AB - BC = h - 5
⇒ AC = CD = h - 5
In right angled ΔBDC, CD2=CB2+BD2 [By Pythagoras theorem]
⇒(h−5)2=(5)2+(12)2 [from Eq.(i)]
⇒(h−5)2=25+144⇒(h−5)2=169⇒h−5=√169=13⇒h=13+5⇒h=18 m
Hence, the height of the tree is 18 m.