Case Study Question 08


Chapter 4: Quadratic Equations

Class 10

A peacock is sitting on the top of a 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. Their speeds are equal and the peacock catches the snake finally.

Peacock and snake

Distance is equal to.

(a) `Speed + time`
(b) `speed \times time`
(c) `\frac{Speed}{time}`
(d) `\frac{time}{Speed}`

  • (b) speed `\times` time

If `ST=PT=(x) m` ,Then what is distance `TQ` in terms of `x` ?

  • `(27-x )` m

Form the equation and solve it.

Let distance decided by peacock `= x`
snake will be at as distance of `(27-x)` peacock distance would be equal to `x^{2}=92+(27-x)^{2}`
distance covered by both `= 15` m

At what distance from the hole is the snake caught?

  • Distance from hole `= ( 27-15) m = 12 m`

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