Consider a square ABCD. EFGH is another square obtained by joining the…
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Consider a square ABCD. EFGH is another square obtained by joining the midpoints of the sides of the square ABCD where E, F , G amd H are the midpoints of AB, BC, CD and DA respectively. Lakshman and Kanika start from points B and D respectively at speeds ‘l’ kmph and ‘k’ kmph respectively and travel towards each other along the sides of the square ABCD. Jagadeesh starts from Point E and travels along the Square EFGH in the anti-clockwise direction at ‘j’ kmph. Lakshman and Kanika meet for the second time at H where Jagadeesh also meets them for the first time. If l : k : j is 1: 3 : 5√2 then the distance travelled by Jagadeesh is
- A.
7.5 × √2 times the side of the square ABCD
- B.
7.5 × √2 times the side of the square EFGH
- C.
7.5 times the side of the square ABCD
- D.
7.5 times the side of the square EFGH
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Correct answer: A
Let the side of square ABCD be S.
Side of inner square EFGH = S/√2, so its perimeter = 4·(S/√2) = 2√2·S.
Treat the outer square perimeter as a loop of length 4S. The two outer travellers start at opposite vertices, so their initial separation along the perimeter is 2S. They meet when (l+k)·t = 2S (first meeting) and the next time when (l+k)·t = 6S (second meeting). Thus the combined distance covered by them at the second meeting is 6S.
With l:k = 1:3, the first outer traveller covers (1/4) of 6S = 1.5S by that time, which matches the path from B to H (B→A is S and A→H is S/2).
Time until this meeting is t = 6S/(l+k). Given l:k:j = 1:3:5√2, we have j/(l+k) = (5√2)/(1+3) = (5√2)/4.
Distance travelled by the inner traveller = j·t = 6S · (5√2)/4 = (30√2/4)·S = (15√2/2)·S = 7.5√2·S.
Answer: 7.5 × √2 times the side of the square ABCD.