A man goes northwards and travels 5 km and then goes 5 km towards the east, then travels 15 km towards the right and finally travels 17 km towards the right. Approximately how far is he from the original place?
Answers
He is approximately at a distance of 15.6 km from his initial position.
Given:
A man goes northwards and travels 5 km and then goes 5 km towards the east, then travels 15 km towards the right, and finally travels 17 km towards the right.
To Find:
The final position of the man with respect to his initial position.
Solution:
Let us assume that his initial position was O.
He travels 5 km towards the north, say to a position A.
⇒ OA = 5 km.
From A, he travels 5 km towards the east, say to a position B.
⇒ AB = 5 km
From B, he goes 15 km towards the right, say to a position D.
⇒ BD = 15 km.
From the figure, BC = OA = 5 km and CD = BD - BC = 15-5 = 10 km.
Finally from D, he goes 17 km towards the right, say to a position F.
⇒ DF = 17 km.
From the figure, DE = AB = 5 km and EF = DF - DE = 17 - 5 = 12 km.
His distance between the initial and final positions = OF, and is calculated as follows:
From the figure, we see that Δ OEF is a right-angled triangle.
⇒ OF² = OE² + EF²
⇒ OF = √(OE² + EF²) = √(10² + 12²) = 15.60 km.
∴ He is approximately at a distance of 15.6 km from his initial position.
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