Maheep starts walking from his office and walks 120 m towards south. Then he turns left and walks a certain distance A m. Then turns right and walks 80 m. Then he turns left and walks another certain distance B m. He there by reaches his home. If the shortest distance between his office and home is 290 m, then what is the sum of the distances A and B that Maheep covered/walked?
Maheep starts walking from his office and walks 120 m towards south. Then he turns left and walks a certain distance A m. Then turns right and walks 80 m. Then he turns left and walks another certain distance B m. He there by reaches his home. If the shortest distance between his office and home is 290 m, then what is the sum of the distances A and B that Maheep covered/walked? Correct Answer 210 m
According to the given information :
[ alt="F1 Prashant Ravi 13.03.21 D30" src="//storage.googleapis.com/tb-img/production/21/03/F1_Prashant_Ravi_13.03.21_D30.png" style="width: 422px; height: 293px;">
Shortest distance between office and home = OP = 290 m.
OQ = 120 + 80 = 200 m.
PQ = A + B.
In right angled ΔOPQ, by Pythagoras Theorem :
(OP)2 = (OQ)2 + (PQ)2
2902 = 2002 + (A + B)2
84100 = 40000 + (A + B)2
84100 - 40000 = (A + B)2
44100 = (A + B)2
(A + B) = √44100
(A + B) = 210.
Hence, the correct answer is "210 m".
