# CASE STUDY QUESTION 3-Class X-Maths

CASE STUDY QUESTION 3-Class X-Maths

**Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above thesurface of the water and the fly at the end of the string rests on the water 3.6 maway and 2.4 m from a point directly under the tip of the rod. She is pulling thestring at the rate of 5 cm per second. Nazima’s friend observe her position anddraw a rough sketch by using A, B, C and D positions of tip, point directly underthe tip of the rod, fish and Nazima’s position (see the below figure). Assuming thather string (from the tip of her rod to the fly) is taut, answer the followingquestions**

**(a) What is the length AC?**

(i) 2 m (ii) 3 m (iii) 4 m (iv) 5 m

**(b) What is the length of string pulled in 12 seconds?**

(i) 6 m (ii) 0.3 m (iii) 0.6 m (iv) 3 m

**(c) What is the length of string after 12 seconds?**

(i) 2.4 m (ii) 2.7 m (iii) 2 m (iv) 2.2 m

**(d) What will be the horizontal distance of the fly from her after 12 seconds?**

(i) 2.7 m (ii) 2.78 m (iii) 2.58 m (iv) 2.2 m

**(e) The given problem is based on which concept?**

(i) Triangles

(ii) Co-ordinate geometry

(iii) Height and Distance

(iv) None of these

**Solutions:**

**a. **AC^{2} = AB^{2} + BC^{2}

[ By Pythagoras theorem ]

⇒ AC^{2} = (1.8)^{2 }+ (2.4)^{2}

⇒ AC^{2} = 3.24 + 5.76

⇒ AC^{2} = 9

⇒ AC = 3m

**b.** She pulls the string at the rate of 5cm/s

∴ String pulled in 12 second = 12 × 5 = 60cm

= 0.6m

**c.** Length of string out after 12 second is AP.

⇒ AP = AC – String pulled by Nazima in 12 seconds.

⇒ AP = (3 − 0.6)m

=2.4m

**d.** In △ADB, AB^{2 }+ BP^{2}=AP^{2} ⇒ (1.8)^{2} + BP^{2} = (2.4)^{2}

⇒ BP^{2} = 5.76 − 3.24 ⇒ BP2 = 5.76 − 3.24

⇒ BP^{2} = 2.52 ⇒ BP=1.58 m

Horizontal distance of fly = BP + 1.2m

Horizontal distance of fly =1.58m + 1.2m

∴ Horizontal distance of fly = 2.78m

**e.** Triangles