A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments. Determine the sum of the squares of the lengths of the three paths.
Explanation:
Here D is midpoint of AC, E is the midpoint of AD and F is the midpoint of CD. Hence, AE = ED = DF = CF = 20 Let AB = c and BC = a Applying Apollonius theorem in ∆ABC, we get, BD2 + 402 = 1/2 × (c2 + a2) = 1/2 × 802 BD2 = 402 … (i) Now, applying Apollonius theorem in ∆ABD, we get, BE2 + 202 = 1/2 × (c2 + BD2) … (ii) Similarly, applying Apollonius in ∆CDB, we get, BF2 + 202 = 1/2 × (a2 + BD2) … (iii) Adding (ii) and (iii), and substituting value from (i), we get, BE2 + BF2 + 2 × 202 = 1/2 (a2 + c2 + 2 × BD2) = 1/2 (802 + 2 × 402) = 3 × 402 Hence, BE2 + BF2 + BD2 = 3 × 402 + 402 – 2 × 202 = 5600 Hence, option (c).
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