Question 1041173
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The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. 
Find the dimension of the triangle, correct to one decimal place.
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{{{x^2 + y^2}}} = {{{12^2}}},       (1)
{{{(1/2)*xy}}}  = {{{24}}},       (2)

or

{{{x^2 + y^2}}} = {{{144}}},        (1')
{{{xy}}}     = {{{48}}}.         (2')

Multiply eqn. (2') by 2 and add to eqn(1') (both sides). You will get

{{{x^2 + 2xy + y^2}}} = {{{144 + 2*48}}},   or

{{{(x+y)^2}}} = {{{240}}},   or

{{{x+ y}}} = {{{sqrt(240)}}} = {{{4*sqrt(15)}}}.

Thus from (1) and (2) we got 

x + y = {{{4*sqrt(15)}}},         (3)
xy    = 48.             (4)

   Notice that in this way we decreased the degree of the original equations/system.

   It makes it easier to work in future with (3),(4) instead of (1),(2).


Next, from (3)  x = {{{4*sqrt(15) - y}}}. Substitute it into (4) and get

{{{(4*sqrt(15)-y)*y}}} = {{{48}}}.     (5)

{{{y^2 - (4*sqrt(15))*y + 48}}} = {{{0}}},

Solve this quadratic equation by applying the quadratic formula. You will get

{{{y[1,2]}}} = {{{(4*sqrt(15) +- sqrt(240-4*48))/2}}} = {{{(4*sqrt(15) +- sqrt(48))/2}}} = {{{2*sqrt(15) +- 2*sqrt(3)}}}.

Then {{{x[1,2]}}} = {{{4*sqrt(5)}}} - {{{(2*sqrt(15) +- 2*sqrt(3))}}}.

<U>Answer</U>. There are two solutions:

        a)  x = {{{2*sqrt(15)-2*sqrt(3)}}},  y = {{{2*sqrt(15) + 2*sqrt(3)}}},  and

        b)  x = {{{2*sqrt(15)+2*sqrt(3)}}},  y = {{{2*sqrt(15) - 2*sqrt(3)}}}.

<U>Check</U>.  a) {{{x^2 +y^2}}} = {{{4*15 + 4*3 + 4*15 + 4*3}}} = 144,  xy = 4*15 - 4*3 = 60 - 12 = 48.   OK ! ! !
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