Four friends start from four towns, which are at the four corners of an imaginary rectangle. They meet at a point which falls inside the rectangle, after travelling distances of 40, 50 and 60 metres. The maximum distance that the fourth could have traveled is (approximately) ….
Explanation:
Let x and y be the sides of the rectangle ABCD and z be the length of AP. Then CR = BP = x – z By applying Pythagoras Theorem, we have in Δ APO, a2 =OP2 + z2 ......(i) in Δ BPO, b2 = OP2 + (x – z)2 ......(ii) in Δ CRO, d2 = OR2 + (x – z)2 ......(iii) in Δ DRO, c2 = OR2 + z2 ....... (iv) Solving above equation, we have a2 + d2 = b2 + c2 ∴ For any point inside a rectangle as shown, a2 + d2 = b2 + c2 ∴ Pairing up the distance so that d is to be the maximum, we get 402 + d2 = 502 + 602 ⇒ d = 67 m.
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