Question

Q.10 of NCERT book. page number 153. chapter triangles. optional excercise

Answer 1

Ans. I. To find The length of AC.
By Pythagoras theorem,
AC
2
= (2.4)
2
+ (1.8)
2
AC
2
= 5.76 + 3.24 = 9.00
AC = 3 m
Length of string she has out= 3 m
Length of the string pulled at the rate of 5 cm/sec in 12 seconds

= (5 x 12) cm = 60 cm = 0.60 m
Remaining string left out = 3 – 0.6 = 2.4 m
II. To find: The length of PB
PB
2
Ans. I. To find The length of AC.
By Pythagoras theorem,
AC 2 = (2.4) 2
+ (1.8) 2+ AC 2
= 5.76 + 3.24 = 9.00
AC = 3 m
Length of string she has out= 3 m
Length of the string pulled at the rate of 5 cm/sec in 12 seconds
= PC 2 – BC 2 = (2.4) 2 – (1.8) 2
= 5.76 – 3.24 = 2.52
PB = = 1.59 (approx.)
Hence, the horizontal distance of the fly from Nazima after 12 seconds
= 1.59 + 1.2 = 2.79 m (approx.)