Premium Online Home Tutors

3 Tutor System

Starting just at 265/hour

Answer :

Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the
horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.

To find AC, we have to use Pythagoras theorem in \(\triangle\) ABC, is such way;

\(\Rightarrow AC^2 = AB^2+ BC^2\)

\(\Rightarrow AB^2 = (1.8 m)^2 + (2.4 m)^2\)

\(\Rightarrow AB^2 = (3.24 + 5.76) m^2\)

\(\Rightarrow AB^2 = 9.00 m^2\)

\(\Rightarrow AB = \sqrt{9} m = 3 m\)

Thus, the length of the string out is 3 m.

As its given, she pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds

= 12 × 5 = 60 cm = 0.6 m

Let us say now, the fly is at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC - String pulled by Nazima in 12 seconds

= (3.00 - 0.6) m

= 2.4 m

In \(\triangle\) ADB, by Pythagoras Theorem,

\(\Rightarrow AB^2 + BD^2 = AD^2\)

\(\Rightarrow (1.8 m)^2 + BD2 = (2.4 m)^2\)

\(\Rightarrow BD^2 = (5.76 - 3.24) m^2 = 2.52 m^2\)

\(\Rightarrow \) BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m = 2.787 m

= 2.79 m

- In Figure, PS is the bisector of \(\angle\) QPR of \(\triangle\) PQR. Prove that \(\frac{QS}{PQ} = \frac{SR}{PR} \)
- In Fig. 6.57, D is a point on hypotenuse AC of \(\triangle\) ABC, such that BD \(\perp\) AC, DM \(\perp\) BC and DN \(\perp\) AB. Prove that: (i) \(DM^2 = DN . MC\) (ii) \(DN^2 = DM . AN.\)
- In Figure, ABC is a triangle in which \(\angle\) ABC > 90° and AD \(\perp\) CB produced. Prove that \(AC^2= AB^2+ BC^2+ 2 BC.BD.\)
- In Figure, ABC is a triangle in which \(\angle\) ABC < 90° and AD \(\perp\) BC. Prove that \(AC^2= AB^2+ BC^2 – 2 BC.BD.\)
- In Figure, AD is a median of a triangle ABC and AM \(\perp\) BC. Prove that : \( (i) AC^2 = AD^2 + BC.DM + 2 (\frac{BC}{2} )^2\) \( (ii) AB^2 = AD^2 – BC.DM + 2 (\frac{BC}{2} )^2\) \( (iii) AC^2 + AB^2 = 2 AD^2 + {{1} \over {2}} BC^2\)
- Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
- In Figure, two chords AB and CD intersect each other at the point P. Prove that : (i) \(\triangle\) APC ~ \(\triangle\) DPB (ii) AP . PB = CP . DP
- In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that: (i) \(\triangle\) PAC ~ \(\triangle\) PDB (ii) PA . PB = PC . PD.
- In Figure, D is a point on side BC of \(\triangle\) ABC such that \(\frac{BD}{CD} = \frac{AB}{AC} \) . Prove that AD is the bisector of \(\angle\) BAC.

- NCERT solutions for class 10 maths chapter 1 Real Numbers
- NCERT solutions for class 10 maths chapter 1 Electricity, Light , Carbon and it's compounds
- NCERT solutions for class 10 maths chapter 2 Polynomials
- NCERT solutions for class 10 maths chapter 3 Pair of linear equations in two variables
- NCERT solutions for class 10 maths chapter 4 Quadratic Equations
- NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions
- NCERT solutions for class 10 maths chapter 6 Triangles
- NCERT solutions for class 10 maths chapter 7 Coordinate Geometry
- NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry
- NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry
- NCERT solutions for class 10 maths chapter 10 Circles
- NCERT solutions for class 10 maths chapter 11 Constructions
- NCERT solutions for class 10 maths chapter 12 Areas related to circles
- NCERT solutions for class 10 maths chapter 13 Surface Areas and Volumes
- NCERT solutions for class 10 maths chapter 14 Statistics
- NCERT solutions for class 10 maths chapter 15 Probability

- NCERT solutions for class 10 science chapter 1 Chemical Reactions and Equations
- NCERT solutions for class 10 science chapter 2 Acids, Bases and Salts
- NCERT solutions for class 10 science chapter 3 Metals and Non Metals
- NCERT solutions for class 10 science chapter 4 Carbon and its Compounds
- NCERT solutions for class 10 science chapter 5 Periodic Classification of Elements
- NCERT solutions for class 10 science chapter 6 Life Processes
- NCERT solutions for class 10 science chapter 7 Control and Coordination
- NCERT solutions for class 10 science chapter 8 How do Organisms Reproduce
- NCERT solutions for class 10 science chapter 9 Heredity and Evolution
- NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction
- NCERT solutions for class 10 science chapter 11 Human Eye and Colorful World
- NCERT solutions for class 10 science chapter 12 Electricity
- NCERT solutions for class 10 science chapter 13 Magnetic Effect of Electric Current
- NCERT solutions for class 10 science chapter 14 Sources of Energy
- NCERT solutions for class 10 science chapter 15 Our Environment
- NCERT solutions for class 10 science chapter 16 Management of Natural Resources