My 10 class mate this is very imp. question for exam pls help me A peacock is sitting on top of a pillar, which is 9m high from a 27m away from the bottom of pillar a snake is coming to its hole at base of pillar. Seeing the snake The peacock bounces on it. If their speeds are equal, at what distance from the hole is the snake caught?
Answers
Answer:
12 m
Step-by-step explanation:
In the figure, AB is the pillar and C is the position of snake from bottom of the pillar.
AB = 9 m. BD = 27 m.
Let peacock catch the snake at a distance x from the pillar at position D.
BD = x ; CD = 27-x
Since it is given the speed of peacock and snake are equal,
Distance covered by Peacock = Distance covered by snake.
So in ΔABD, by Pythagoras theorem
AD² = AB² + BD²
=> AD = √(AB² + BD²) = √(9)² + (x)² = √81 + x²
Distance covered by peacock = Distance covered by snanke = √81 + x²
But distance covered by snake is 27-x. i.e. length CD.
=> 27 - x should be equal to √81 + x²
27 - x = √81 + x²
Squaring on both sides
(27 - x)² = (√81 + x²)²
=> 729 - 2 * 27*x + x² = 81 + x² (∵(a-b)² = a² - 2ab + b²)
=> 54x = 729 - 81 (∵ x² will be eliminated)
=> 54x = 648
=> x = 12 m.
Thus snake is caught at 12 m.
Given :
★ A peacock is sitting on the top of the pillar, which is 9m high. From a point 27m away from the bottom of the pillar, A snake is coming to its hole at the base of the pillar. Seeing the snake the peacock ounces on it. If their speeds are equal.
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Find :
★ What distance from the hole is the snake caught.
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Concept :
★ As per the given question, the concept is finding the distance from a point to a certain point, so let us assume "x" as the difference between them.
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★ Remember that, the variable x is used here instead of word DISTANCE in our equations.
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Calculations :
→ 9² + (27)²
→ 81 + 729 + x - 54
→ 810/54
→ 15
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Finding the distance :
→ x = 15 - 3
→ x = 12 m
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★ Therefore, 12 meter is the distance from the hole.