A peacock is sitting on the top of a pillar which is9 m high. From a point27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing snake, the peacock pounces on it. If their speeds are equal, at what distance from hole, the snake caught?

Dear Student!

 

AB is the pillar and C is the position of the hole from the bottom of the pillar.

Here, AB = 9m and BC = 27 m

Let peacock catches the snake at a distance x m from the bottom of the pillar.

∴ BD = x and CD = 27 – x

Given, speed of the peacock and the snake are equal.

∴ Distance covered by peacock = Distance covered by snake

In ΔABD,

AD2 = AB2 + BD2

∴ AD2 = (9)2 + x2

Distance covered by peacock =

Distance covered by snake = (27 – x) m

Squaring on both sides, we get

81 + x2 = (27 – x)2

∴ 81 + x2 = 729 + x2 – 54x

Thus, snake is caught at distance 15 m from the hole.

Cheers!

  • 60

 First form a right angled triangle

Height = 9 m

let the distance the snake travels before being caught be x

The base of the triangle = 27 - x

because the speeds are equal the peacock will also fly x metres

The hypotenuse = x

Using Pythagoras

81 + [27 - x]^2 = x^2

54x = 81 + 27^2

x = 15

So the distance from the hole is 27 - 15 = 12 m

  • 16

grtt..!!

:)

  • -19

thanks

  • -15
What are you looking for?