Question 1179618


The diagonal of a rectangle is {{{2 ft}}} longer than its {{{length}}} and {{{5 ft}}} longer than twice its {{{width}}}. 
What are the dimensions of the rectangle?

Since the {{{diagonal}}} divides rectangle into two {{{right}}}{{{trianles}}}, we will use {{{Pythagoras' }}}{{{Theorem }}} in order to find the dimensions of the rectangle.

In algebraic terms, {{{a^2 + b^2 = c^2}}}where {{{c}}} is the hypotenuse ( in your case the {{{diagonal}}}) while {{{a}}} and {{{b }}}are the legs of the triangle (in your case the {{{lenght}}} and {{{width}}}).

given:

{{{c = a + 2}}}   => {{{a = c  -  2}}}

{{{c = 2b + 5}}}  =>  {{{b = (c  -  5)/2}}}


we will substitute it in:

{{{c^2 = a^2 + b^2}}}

{{{c^2 = (c-2)^2 + ((c  -  5)/2)^2}}}

{{{c^2 = (5c^2)/4 - (13 c)/2 + 41/4}}}   

{{{4c^2 = 5c^2  -  26c +41}}}
{{{0 = 5c^2-4c^2  -  26c +41}}}
{{{c^2 -26c + 41 = 0}}}

{{{c=(-(-26) +- sqrt ((-26)^2 -4*1*41 )) / (2*1)}}}

{{{c=(26 +- sqrt (512 )) /2}}}

{{{c=(26 +- 16sqrt (2 )) /2}}}

We need only positive root =>

{{{c=(26 + 16sqrt (2 )) /2}}}

{{{c = 13 + 8 sqrt(2)}}}

{{{c=24.3ft}}}



{{{a = c  -  2ft}}}
=> {{{a = 24.3ft  -  2ft = 22.3ft}}}

{{{b = (c  -  5ft)/2}}}
{{{b = (24.3ft  -  5ft)/2}}}
=> {{{b = 9.7ft}}}

Check:

{{{(24.3)^2 = (22.3)^2 + (9.7)^2}}}

{{{590.49 = 591.3}}}..round to whole number

  {{{591 = 591}}}