Question 4129
Let a and b be sides of a rectangle, and d a diagonal. Let {{{b<a}}}.
We have following equations:
{{{a=b+7}}}
{{{d=b+8}}}
By using Pythagorean theorem you can calculate that {{{d=sqrt(a^2+b^2)}}}
So our system of equations becomes:
{{{a=b+7}}}
{{{sqrt(a^2+b^2)=b+8}}}
If we put {{{a=b+7}}} in second equation, we get:
{{{sqrt((b+7)^2+b^2)=b+8}}}
{{{(b+7)^2+b^2=(b+8)^2}}}
{{{b^2+14b+49+b^2=64+16b+b^2}}}
{{{b^2-2b-15=0}}}
{{{b=(2+-sqrt(4-4*1*(-15)))/2}}}
{{{b=(2+-sqrt(64))/2}}}
{{{b=(2+-8)/2}}}
{{{b=5}}} or {{{b=-3}}}
b cannot be a negative number, so b=5.
First eqaution gives us a:
{{{a=b+7}}}
{{{a=5+7}}}
{{{a=12}}}
So, sides of a rectangle are: (5,12).