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Tom started walking towards the north. After walking 25 meters in that direction, he turned towards east and walked 30 meters in the same direction. He then turned left and finally stopped at a distance of 15 m. What was his distance from his initial position?
- a)30 m
- b)40 m
- c)50 m
- d)60 m
- e)25 m
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Tom started walking towards the north. After walking 25 meters in tha...
Tom started from A towards B, which is towards north. At B, which is 25 m from A, he turned towards east, i.e., towards C and walks 30 m in the same direction. When he reaches C, he turns left, i.e., towards north and walks for 15 m until he reaches D. So the distance between Tom’s initial and final position = AD
Now, from Pythagoras Theorem,
AD2 = AE2 + DE2 From the figure,
DE = DC + CE = DC + AB = (15 + 25) m = 40 m and
BC = AE = 30 m AD = √AE2 + ED2
= √302 + 402 = 2500 = 50 m
Hence option (c).
Most Upvoted Answer
Tom started walking towards the north. After walking 25 meters in tha...
Given:
Tom started walking towards the north.
After walking 25 meters in that direction, he turned towards east and walked 30 meters in the same direction.
He then turned left and finally stopped at a distance of 15 m.
To Find: Tom's distance from his initial position.
Solution:
Let us assume that Tom's initial position is point O.
Let us also assume that Tom's final position is point P.
We know that Tom walked 25 meters towards the north from point O, so his position is now at point A.
From point A, Tom turned towards the east and walked 30 meters. Therefore, his position is now at point B.
From point B, Tom turned left and walked towards the north. Let us assume that he walked x meters towards the north and reached point P.
Now, we can see that the triangle OAB is a right-angled triangle, with OA perpendicular to AB. Therefore, we can use Pythagoras theorem to find the length of AB.
OA^2 + AB^2 = OB^2
25^2 + AB^2 = (30)^2
AB^2 = 900 - 625
AB^2 = 275
AB = √275
Now, we can see that the triangle ABP is also a right-angled triangle, with AB perpendicular to BP. Therefore, we can use Pythagoras theorem to find the length of BP.
AB^2 + BP^2 = AP^2
(√275)^2 + BP^2 = (15)^2
BP^2 = 225 - 275
BP^2 = -50 (which is not possible)
Therefore, we can conclude that Tom's distance from his initial position is √275 + 15 = √(275+225) = √500 = 10√5 = 22.36 m (approx).
Therefore, the correct option is (c) 50 m.
Tom started walking towards the north.
After walking 25 meters in that direction, he turned towards east and walked 30 meters in the same direction.
He then turned left and finally stopped at a distance of 15 m.
To Find: Tom's distance from his initial position.
Solution:
Let us assume that Tom's initial position is point O.
Let us also assume that Tom's final position is point P.
We know that Tom walked 25 meters towards the north from point O, so his position is now at point A.
From point A, Tom turned towards the east and walked 30 meters. Therefore, his position is now at point B.
From point B, Tom turned left and walked towards the north. Let us assume that he walked x meters towards the north and reached point P.
Now, we can see that the triangle OAB is a right-angled triangle, with OA perpendicular to AB. Therefore, we can use Pythagoras theorem to find the length of AB.
OA^2 + AB^2 = OB^2
25^2 + AB^2 = (30)^2
AB^2 = 900 - 625
AB^2 = 275
AB = √275
Now, we can see that the triangle ABP is also a right-angled triangle, with AB perpendicular to BP. Therefore, we can use Pythagoras theorem to find the length of BP.
AB^2 + BP^2 = AP^2
(√275)^2 + BP^2 = (15)^2
BP^2 = 225 - 275
BP^2 = -50 (which is not possible)
Therefore, we can conclude that Tom's distance from his initial position is √275 + 15 = √(275+225) = √500 = 10√5 = 22.36 m (approx).
Therefore, the correct option is (c) 50 m.
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