Pranali and Prasad started walking to the East and to the North respectively, from the same……..
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 10 | Page 45
Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was [15sqrt{2}]
km. Find their speed per hour.
solution
It is given that, Pranali and Prasad have same speed.
Thus, they cover same distance in 2 hours.
i.e. OA = OB
Let the speed be x km per hour.
According to Pythagoras theorem,
In ∆AOB
[{AB}^2 = {AO}^2 + {OB}^2 ]
[ Rightarrow left( 15sqrt{2} right)^2 = {AO}^2 + {OA}^2 ]
[ Rightarrow 450 = 2 {AO}^2 ]
[ Rightarrow {AO}^2 = frac{450}{2}]
[ Rightarrow {AO}^2 = 225]
[ Rightarrow AO = 15 km]
[ Rightarrow BO = 15 km]
[text{Speed} = frac{Distance}{Time}]
[ = frac{15}{2}]
[ = 7 . 5 text{km per hour}]
Explanation:-
Let the speed of Pranali and Prasad be x km/h each.
After 2 hours, Pranali will cover a distance of 2x km towards the East, and Prasad will cover a distance of 2x km towards the North. Let the point where they meet after 2 hours be O, and let the distance of O from the starting point be d km.
Now, in the right triangle PTO (where P is the starting point, T is the point where Pranali stops after 2 hours, and O is the point where they meet), we have:
PT² + OT² = PO² (2x)² + d² = (x√2)² 4x² + d² = 2x² 2x² + d² = 60
Similarly, in the right triangle PSO (where S is the point where Prasad stops after 2 hours), we have:
PS² + OS² = PO² (2x)² + d² = (x√2)² 4x² + d² = 2x² 2x² + d² = 60
Adding the above two equations, we get:
4x² + 2d² = 120
But we know that 2x² + d² = 60. Substituting this value in the above equation, we get:
4x² + 2(2x² + d²) = 120 8x² + 2d² = 120 4x² + d² = 60
Subtracting this equation from 2x² + d² = 60, we get:
x² = 15
Therefore, x = √15 km/h.
Hence, the speed of Pranali and Prasad is √15 km/h each.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 10 | Page 45