Question 165069
Given: 
{{{h1+h2=h}}}
.
{{{h1=(x+2)}}}
.
{{{h2=(4/(x(x+2)))}}}
.
{{{b=(x+4)}}}
.
Formula: {{{(A)rea=(1/2)*(b)ase*(h)eight}}}
{{{A=(1/2)(x+4)(4/(x(x+2)))}}}
multiply across and you get {{{A=((1)(x+4)(4))/(2x(x+2))}}}
reduce the 4 in the numerator with the 2 in the denominator and you get:
{{{A=(2(x+4))/(x(x+2))}}}
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