Given : Rohit and Rahul start at the same point and they each other move in a right angle. They are separated by 80 km after 4 hours.



Rahul’s speed is 4 kmph less than Rohit’s.



Let speed of Rahul be x.



Then, Distance covered by Rahul = Speed × Time



\begin{gathered}\begin{aligned} & = 4 \times x \\\\ = & 4 x k m p h \end{aligned}\end{gathered}

=

?



=4×x

4xkmph

?



?





& speed of Rohit is 4kmph more than Rahul. Hence,



Rohit covered 4(x + 4)kmph



Now, by Pythagoras theorem,



a^2+b^2+c^2a

2

+b

2

+c

2

whereas a = 4x, b = 4 (x + 4) and c = 80kmph



\begin{gathered}\begin{aligned} ( 4 x ) ^ { 2 } + [ 4 ( x + 4 ) ] ^ { 2 } & = ( 80 ) ^ { 2 } \\\\ 16 x ^ { 2 } + 16 \left( x ^ { 2 } + 8 x + 16 \right) & = 80 \times 80 \\\\ 16 \left( x ^ { 2 } + x ^ { 2 } + 8 x + 16 \right) & = 16 ( 5 ) \times 80 \\\\ \left( 2 x ^ { 2 } + 8 x + 16 \right) & = 400 \end{aligned}\end{gathered}

(4x)

2

+[4(x+4)]

2



16x

2

+16(x

2

+8x+16)

16(x

2

+x

2

+8x+16)

(2x

2

+8x+16)

?



=(80)

2



=80×80

=16(5)×80

=400

?



?





\begin{gathered}\begin{array} { c } { \left( x ^ { 2 } + 4 x + 8 \right) = 200 } \\\\ { x ^ { 2 } + 4 x + 8 - 200 = 0 } \\\\ { x ^ { 2 } + 4 x - 192 = 0 } \\\\ { x ^ { 2 } + 16 x - 12 x - 192 = 0 } \\\\ { x ( x + 16 ) - 12 ( x + 16 ) = 0 } \\\\ { ( x - 12 ) ( x + 16 ) = 0 } \end{array}\end{gathered}

(x

2

+4x+8)=200

x

2

+4x+8-200=0

x

2

+4x-192=0

x

2

+16x-12x-192=0

x(x+16)-12(x+16)=0

(x-12)(x+16)=0

?



?





Hence, x = -16 or x = 12.



Now speed cannot be minus, thus



x = 12 kmph i.e. Speed of Rahul



Speed of Rohit = x + 4 = 12 + 4 = 16 kmph