Question 236398
Let L = the length of the rectangle and W = its width.
{{{L = 2W-1}}} "The length (L) of a rectangle is 1m less than twice its width (W)."
From the Pythagorean theorem,...
{{{sqrt(L^2+W^2) = 17}}} "...a diagonal of the rectangle is 17m,..."
{{{(sqrt(L^2+W^2))^2 = 17^2}}} Simplify.
{{{L^2+W^2 = 289}}} Substitute {{{L = 2W-1}}} ansd solve for W.
{{{(2W-1)^2+W^2 = 289}}}
{{{(4W^2-4W+1)+W^2 = 289}}} Simplify and subtract 289 from both sides.
{{{5W^2-4W-288 = 0}}} Solve using the quadratic formula:{{{W = (-b+-sqrt(b^2-4ac))/2a}}} where: a = 5, b = -4 and c = -288.
{{{W = (-(-4)+-sqrt((-4)^2-4(5)(-288)))/2(4)}}}
{{{W = (4+-sqrt(16-(-5760)))/8}}}
{{{W = (4+-sqrt(5776))/8}}}
{{{highlight(W = 8)}}} or {{{cross(W = -7.2)}}} Discard the negative solution.
{{{L = 2W-1}}}
{{{L = 2(8)-1}}}
{{{L = 15}}}
The length is 15 meters and the width is 8 meters.