Question

a peacock is sitting on a top of the pillar which is 9 metre high from a point 27 metre away from the bottom of the pillar .A snake is coming to the hole at the base of the pillar seeing the snake the peacock pounces on it if this speeds are equal at ,what distance from the hole is the snake caught ?

Answer 1

HELLO DEAR,


GIVEN that:-

The speed of peacock and snake are same

Let the distance covered by them is x


PQ = 9m
PS = Xm
QS = (27 - x)

NOW,

IN ∆ PQS

PS² = PQ² + QS²


=> X² = 9² + (27 - x)²

=> X² = 81 + 729 + 54x + x²

=> X² - x² - 54x - 810 = 0

=> 54x = 810

=> X = 810/54

=> X = 15m

QS = (27 –x) m = (27 – 15) m = 12 m

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


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

heya..!!!

let the distance the snake travels before being caught be x

Height = 9 m = PQ

PS = X

QS = (27 - X)

now IN∆PQS

Using Pythagoras

81 + (27 - x)²= x²

54x = 81 + 27²

x = 15

HENCE, the distance from the hole is 27 - 15 = 12m