# 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 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. She is pulling the
string at the rate of 5 cm per second. Nazima’s friend observe her position and
draw a rough sketch by using A, B, C and D positions of tip, point directly under
the tip of the rod, fish and Nazima’s position (see the below figure). Assuming that
her string (from the tip of her rod to the fly) is taut, answer the following
questions

(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. AC2 = AB2 + BC2
[ By Pythagoras theorem ]
⇒ AC2 = (1.8)2 + (2.4)2
⇒ AC2 = 3.24 + 5.76
⇒ AC2 = 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, AB2 + BP2=AP2 ⇒ (1.8)2 + BP2 = (2.4)2
⇒ BP2 = 5.76 − 3.24 ⇒ BP2 = 5.76 − 3.24
⇒ BP2 = 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

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