Question 204677
Let c = the length of the hypotenuse. a and b are the other two legs.
{{{a = c-4}}} and...
{{{b = c-2}}} Using the Pythagorean theorem: {{{c^2 = a^2+b^2}}} and making the appropriate substitutions, we get:
{{{c^2 = (c-4)^2+(c-2)^2}}} Expanding:
{{{c^2 = (c^2-8c+16)+(c^2-4c+4)}}} Simplifying:
{{{c^2 = 2c^2-12c+20}}} Subtracting {{{c^2}}} from both sides.
{{{c^2-12c+20 = 0}}} Factor the trinomial.
{{{(c-10)(c-2) = 0}}} Applying the zero product rule, we have:
{{{c-10 = 0}}} or {{{c-2 = 0}}} which leads to:
{{{c = 10}}} or {{{c = 2}}} Discard the second solution as c, the hypotenuse cannot be smaller than either of the other two legs, so...
{{{highlight(c = 10)}}}m is the length of the hypotenuse.