# Ex. 6.6 Q10 Triangles Solution - NCERT Maths Class 10

## Question

Nazima is fly fishing in a stream. The tip of her fishing rod is \(1.8\,\rm m\) above the surface of the water and the fly at the end of the string rests on the water \(3.6\,\rm m\) away and \(2.4\,\rm m\) 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. )? If she pulls in the string at the rate of \(5\,\rm cm\) per second, what will be the horizontal distance of the fly from her after \(12\,\rm seconds\)?

## Text Solution

**Reasoning:**

Pythagoras Theorem

**Steps:**

To find \(AB\) and \(ED \)

\(BD = 3.6\, {\rm m} ,\\ BC = 2.4 \,{\rm m} , \\ CD = 1.2\,{\rm m},\, \\ AC = 1.8\,\rm cm\)

In \(\Delta ACB\)

\[\begin{align} A{B^2} &= A{C^2} + B{C^2}\\ &= {\left( {1.8} \right)^2} + {\left( {2.4} \right)^2}\\ &= 3.24 + 5.76\\ A{B^2} &= 9.00\\AB &= 3 \end{align}\]

Length of the string out \(AB = \rm{3\,cm}\)

Let the fly at \(E\) after \(12\,\rm{seconds}\)

String pulled in \(12\,\rm{ seconds}\)

\[\begin{align} &= 12 \times 5\\&= 60 \rm{cm}\\ &= 0.6 \rm{m}\\ AE &= 3\,\rm{m}-0.6\, \rm{m}\\ &= 2.4 \,\rm{m} \end{align}\]

Now in \(\Delta ACE\)

\[\begin{align} C{E^2}& = A{E^2} - A{C^2}\\ &= {\left( {2.4} \right)^2} - {\left( {1.8} \right)^2}\\ C{E^2} &= 5.76 - 3.24\\ &= 2.52\\ CE& = 1.587\;\rm{}m \end{align}\]

\[\begin{align} DE &= CE + CD\\ &= 1.587 + 1.2\\ &= 2.787\\ DE& = 2.79\;\rm{}m \end{align}\]

Horizontal distance of the fly after \(12\,\rm seconds\) **\( = 2.79\, \rm m\)**