Question 288985
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A uniform-width boardwalk is built around the inside edge of a rectangular parkland that is 10m by 15m. 
If the boardwalk takes 20% of the lot, how wide is the boardwalk to the nearest centimetre?
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<pre>
Let x be the uniform wide of the boardwalk, in meters.


The area inside the boardwalk is  0.75*10*15 = 120 square meters.


Therefore, our equation for this area is

    (10-2x)*(15-2x) = 120.


Simplify and reduce it to the standard form quadratic equation

    150 - 50x + 4x^2 = 120,

    4x^2 - 50x + 30 = 0,

    2x^2 - 25x + 15 = 0.

    {{{x[1,2]}}|} = {{{(25 +- sqrt((-25)^2 -4*2*15))/(2*2)}}} = {{{(25 +- sqrt(505))/4}}}.


One root  {{{(25 + sqrt(505))/4}}}  is  about  11.86 meters, and it is too big to be a solution 
to the problem, so we reject it.


The other root  {{{(25 - sqrt(505))/4}}}  is about  0.63 meters, and it makes sense as a solution 
to the problem, so we accept it.


<U>ANSWER</U>.  The width of the boardwalk is about 63 centimeters.
</pre>

Solved.