Question

Exercise 6.6 Que. 10. Cbse NCERT. Solve it​

Answer 1

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: QUESTION \: \: } \mid}}}}}$

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

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\mathfrak{Let,} \\ \: \: \: \: \: \: \: \: \: \text{ \red{AB} be the height of the rod tip from the} \\ \text{surface of water and \red{BC} be the horizontal } \\ \text{distance between fly to tip of the rod. Then, } \\ \text{\red{AC} will be the length of the string.}$

$\: \: \: \: \: \: \: \underline{ \text{ \: In Figure. 1, \: }} \\ \\ \underline{\bold{ \: \: By \: \: using \: \: Pythagoras \: \: theorem, \: \: in \: \:}} \\ \underline{ \bold{ \: the \: \red{ \triangle \: ABC}, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \bold{AC {}^{2} = AB {}^{2} + BC {}^{2}} \\ \\ \bold{\implies AC {}^{2} = (1.8 \: m) {}^{2} + (2.4 \: m) {}^{2} } \\ \\ \bold{ \implies AC {}^{2} = (3.24 + 5.76) \: m {}^{2} } \\ \\ \bold{\implies AC {}^{2} = 9 \: m {}^{2} } \\ \\ \bold{\implies AC = \sqrt{ 9 \: m {}^{2} } } \\ \\ \: \: \: \: \: \: \bold{ \therefore \: \red{AC = 3 \: m }} \\ \\ \bold{ \therefore \: The \: \: length \: \: of \: \: the \: \: string, \red{AC \: = 3 \: m}.}$

$\: \: \: \: \: \: \tt{ If \: \: Nazima \: \: pulls \: \: in \: \: the \: \: string \: \: at \: \: the \: \: } \\ \tt{rate \: \: of \: \: 5 cm/s, \: then \: \: the \: \: distance \: \: } \\ \tt{ travelled \: \: by \: \: fly \: \: in \: \: 12 \: \: seconds \: \: will \: \: be } \\ \\ \tt{= \red{( 5 \times 12) \: cm }= \red{60 \: cm} = \red{0.6 \: m}}$

$\mathfrak{Let,} \\ \: \: \: \: \: \: \: \: \text{ \red{D} be the position of fly after \red{12 seconds}. } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{[Figure. 2]} \\ \text{ Hence, } \\ \: \: \: \text{ \red{AD} will be the length of string which is out} \\ \text{ after \red{12 seconds}.} \\ \\ \tt{ \therefore \: Length \: of \: \: the \: \: string \: \: pulls \: \: by \: \: Nazima} \\ \\ \: \: \: \: \: \: \: \: \: \tt{ = \red{AD} = \red{3 \: m - 0.6 \: m} = \red{2.4 \: m}}$

$\mathfrak{Now,} \\ \: \: \: \: \: \: \: \: \: \: \underline{ \text{ \: In \: Figure. 2, \: }} \\ \\ \underline{ \bold{ \: By \: \: using \: \: Pythagoras \: \: theorem \: \: in \: \: }} \\ \underline{ \bold{ \: the \: \triangle ADB, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \bold{AD {}^{2} = AB {}^{2} + BD {}^{2} } \\ \\ \bold{ \implies (2.4 \: m) {}^{2} =(1.8 \: m) {}^{2} + BD {}^{2} } \\ \\ \bold{\implies 5.76 \: m {}^{2} = 3.24 \: m {}^{2} + BD {}^{2} } \\ \\ \bold{\implies BD {}^{2} = (5.76 - 3.24) \: m {}^{2} } \\ \\ \bold{\implies BD {}^{2} = 2.52 \: m {}^{2}} \\ \\ \bold{ \implies \: BD = \sqrt{2.52 \: m {}^{2} }} \\ \\ \: \: \: \: \: \bold{\therefore \: \red{BD = 1.59 \: m }\: \: \: (Approx.)}$

$\sf{Hence, \: the \: \: horizontal \: \: distance \: \: of \: \: the \: \: fly \: \: } \\ \sf{ from \: \: Nazima \: \: after \: \: 12 \: seconds \:= (1.59 + 1.2 ) \: m } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{= \red{2.79 \: m} \: (Approx.)}$