A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m 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 pounces on it. If their speeds are equal, at what distance from the whole is the snake caught?

Let PQ be the pole and the peacock is sitting at the top P of the pole. Let the hole be at Q. Initially, the snake is at S when the peacock notices the snake such that QS=27m

Suppose v m/sec is the common speed of both the snake and the peacock and the peacock catches the snake after t seconds at point T. clearly distance travelled by the snake in t seconds is same as the distance flown by peacock

Thus, in right triangle PQT, we have

Using Pythagoras theorem, we have

Hence the snake is caught at a distance of 12m from the hole.

Answer:

Let PQ be the pole and the peacock is sitting at the top P of the pole. Let the hole be at Q. Initially, the snake is at S when the peacock notices the snake such that QS=27m

Suppose v m/sec is the common speed of both the snake and the peacock and the peacock catches the snake after t seconds at point T. clearly distance travelled by the snake in t seconds is same as the distance flown by peacock

\text{∴PT=ST=x}∴PT=ST=x

Thus, in right triangle PQT, we have

\text{QT=27−x,PT=x and PQ=9}QT=27−x,PT=x and PQ=9

Using Pythagoras theorem, we have

⇒ {PT}^{2} = {PQ}^{2} + {TQ}^{2}⇒PT

2

=PQ

2

+TQ

2

⇒ {x}^{2} = {9}^{2} + (27 - x {)}^{2}⇒x

2

=9

2

+(27−x)

2

⇒ {x}^{2} = 81 + 729 - 54x + {x}^{2}⇒x

2

=81+729−54x+x

2

⇒810 - 54x⇒810−54x

⇒54x = 810⇒54x=810

⇒x = 15⇒x=15

∴QT = SQ - ST = (27 - 15)m = 12m∴QT=SQ−ST=(27−15)m=12m

Hence the snake is caught at a distance of 12m from the hole.

Step-by-step explanation:

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