Question 993681
Let {{{ s }}} = the speed of the slower vehicle in km/hr
{{{ s + 4 }}} km/hr = the speed of the faster vehicle in km/hr
{{{ t = 1 }}} hr ( given )
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Now find the sides of the right triangle in terms of {{{ s }}}
Shorter side:
{{{ d[1] = s*t }}}
{{{ d[1] = s*1 }}} km
Longer side:
{{{ d[2] = ( s+4 )*t }}}
{{{ d[2] = ( s+4 )*1 }}} km
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The hypotenuse of the right triangle is given as:
{{{ d = 20 }}} km
Use Pythagoerean theorem:
{{{ d^2 = d[1]^2 + d[2]^2 }}}
{{{ 20^2 = s^2 + ( s+4 )^2 }}}
{{{ 400 = s^2 + s^2 + 8s + 16 }}}
{{{ 2s^2 + 8s -384 = 0 }}}
{{{ s^2 + 4s = 192 }}}
complete the square:
{{{ s^2 + 4s + (4/2)^2 = 192 + (4/2)^2 }}}
{{{ s^2 + 4s + 4 = 192 + 4 }}}
{{{ ( s+ 2 )^2 = 14^2 }}}
{{{ s + 2 = 14 }}}
{{{ s = 12 }}}
ans
{{{ s+ 4 = 16 }}}
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The speed of the slower vehicle is 12 km/hr
The speed of the faster vehicle is 16 km/hr
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check:
{{{ 20^2 = s^2 + ( s+4 )^2 }}}
{{{ 400 = 12^2 + 16^2 }}}
{{{ 400 = 144 + 256 }}}
{{{ 400 = 400 }}}
OK