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The speed of the peacock and snake is equal; they cover an equal distance in the same time limit. Let the distance covered each of them be ‘x’ m. Here, PS and RS represent the distance covered by the peacock and the snake respectively. Triangle PQR forms a right angled triangle, ⇒ PQ2 + QS2 = PS2 ⇒ 92 + (27 – x)2 = x2 ⇒ 81 + (729 – 54x + x2) = x2 ⇒ 54x = 810 ⇒ x = 15 QS = (27 –x) m = (27 – 15) m = 12 m Hence, the snake is caught at a distance of 12 m from the hole
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why BC²*2
Answer. Let the distance covered by the peacock be AC = x m Hence the distance covered by the snake is DC = x m In right ΔABC, by Pythagoras theorem we have AC2 = AB2 + BC2 x2 = 92 + (27 – x)2 ⇒ x2 = 81 + 729 – 54x + x2 ⇒ 54x = 810 ∴ x = 15
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