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 \[15\sqrt{2}\]
km. Find their speed per hour.
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
\[{AB}^2 = {AO}^2 + {OB}^2 \]
\[ \Rightarrow \left( 15\sqrt{2} \right)^2 = {AO}^2 + {OA}^2 \]
\[ \Rightarrow 450 = 2 {AO}^2 \]
\[ \Rightarrow {AO}^2 = \frac{450}{2}\]
\[ \Rightarrow {AO}^2 = 225\]
\[ \Rightarrow AO = 15 km\]
\[ \Rightarrow BO = 15 km\]
\[\text{Speed} = \frac{Distance}{Time}\]
\[ = \frac{15}{2}\]
\[ = 7 . 5 \text{km per hour}\]
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