Question 122745

The length {{{L}}} of a rectangle is {{{1 cm}}} longer than it's width {{{W}}}

=>…..{{{L = W + 1cm}}}

the diagonal {{{d}}} of the rectangle is {{{4 cm}}}

find: {{{L}}} and {{{W}}}

use Pythagorean Theorem 

{{{d^2 = L^2 + W^2}}}

{{{(4 cm )^2 = (W + 1cm )^2 + W^2}}}

{{{(4 cm )^2 = W^2 + 2W + 1cm^2 + W^2}}}

{{{16cm^2 = 2W^2 + 2W + 1cm^2 }}}

{{{ 2W^2 + 2W + 1cm^2 - 16cm^2 = 0 }}}

{{{ 2W^2 + 2W - 15cm^2 = 0 }}}

{{{ W [1,2]=(-2 +- sqrt (2^2 -4*2*(- 15) )) / (2*2)}}}

{{{ W [1,2]=(-2 +- sqrt (4 + 120 )) / 4}}}

{{{ W [1,2]=(-2 +- sqrt (124 )) / 4}}}

{{{ W [1,2]=(-2 +- 11.14) / 4}}}………you need only positive root

{{{ W [1]=(-2 + 11.14) / 4}}}

{{{ W [1]= 9.14 / 4}}}

{{{ W [1]= 2.285}}}

{{{ W [1]= 2.28cm}}}

=>…..{{{L = W + 1cm}}}

=>…..{{{L = 3.28cm}}}


Check:

{{{(4 cm )^2 = (3.28cm)^2 + (2.28cm)^2}}}

{{{16cm^2 = 10.75cm^2 + 5.19cm^2}}}

{{{16cm^2 = 15.94 cm^2}}}

{{{16cm^2 = 16 cm^2}}}