Question 80865
QUESTION:


The length of a rectangle is 4 cm more than 2 times its width.  If the area of the rectangle is 90 cm^2, find the dimensions of the rectangle to the nearest thousandth.


ANSWER:


Assume that width of the rectangle is  'x' cm.


Then two times width = 2x



4 cm more than 2 times its width = (2x + 4)cm



It is given that its length is  4 cm more than 2 times its width


So length of the rectangle = (2x + 4 ) cm.



Area of a rectangle is given by the formula, A = length * width



Here  area is given that 90 cm^2




So we can write it as,



90 = (2x + 4 ) * x



==> 90 = 2x * x  + 4 * x


==> 90 = 2x^2 + 4x



Subtract 90 from both sides of the equation, then we will obtain a quadratic equation.



==> 90 - 90 = 2x^2 + 4x - 90



==> 0 = 2x^2 + 4x - 90




2x^2 + 4x - 90 = 0



We can solve this equation using quadratic formula.



standard form of a quadratic equation is,



ax^2 + bx + c = 0 ---------------(2)



By quadratic formula, the solution is given by,




{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}



Comparing (1) and (2) we have,



a = 2, b = 4 and c = -90




so the solution is,



{{{x = (-4 +- sqrt( 4^2-4*2*(-90) ))/(2*2) }}}




{{{x = (-4 +- sqrt( 16 + 720  ))/(4) }}}




{{{x = (-4 +- sqrt( 736 ))/(4) }}}




{{{x = (-4 +- 27.129)/4 }}}



x = (-4 + 27.129)/4   or x = (-4 - 27.129 )/4



x = 23.129/4     or x = -41.129/4



( since negative values are not admisible here)

==> x = 5.782 cm



So width of the rectangle is 5.782  cm. 



so width = 5.782 cm.


and length = 2x + 4 = 2 * (5.782) + 4 = 11. 564 + 4 = 15.564 cm



So the dimenstions of the given rectangle:


length = 15. 564 cm   and


Width  = 5.782 cm.


To check your answer, multiply lenth with breadth, then you will get 90 approximately.




Hope you  understood.


Regards.



Praseena.