# Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

**Solution:**

Using Pythagoras Theorem to find AB and ED

BD = 3.6 m, BC = 2.4 m, CD = 1.2 m

AC = 1.8 cm

In ΔACB

AB² = AC² + BC²

= (1.8)² + (2.4)²

= 3.24 + 5.76

AB² = 9

AB = 3

Length of the string out AB = 3cm

Let the fly at E after 12 seconds

String pulled in 12 seconds = 12 × 5

= 60 cm

=0.6 m

AE = 3m - 0.6 m

= 2.4 m

Now In

ΔACE

CE^{2} = AE² - AC²

= (2.4)² - (1.8)²

CE² = 5.76 - 3.24

= 2.5²

CE = 1.587 m

DE = CE + CD

= 1.587 + 1.2

= 2.787

DE = 2.79 m

The horizontal distance of the fly after 12 seconds = 2.79 m

**Video Solution:**

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.6 Question 10:

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. If she pulls in the string at the rate of 5 cm per second, then the horizontal distance of the fly after 12 seconds = 2.79 m