Question 170043
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I think you mean as the "height is 5 inches less than the base?"
We'll base the computation as per condition:
{{{A[T]=(1/2)bh}}}
{{{b=x(in)}}}
{{{h=x-5(in)}}}
So, {{{6=(1/2)(x)(x-5)}}}
{{{12=x^2-5x}}} ----> {{{x^2-5x-12=0}}}
Remember ---{{{system(a=1,b=-5,c=-12)}}}
By Quadratic Eqn: {{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-(-5)+-sqrt(-5^2-4*1*-12))/(2*1)}}}
{{{x=(5+-sqrt(25+48))/2}}} --->{{{x=(5+-sqrt(73))/2=(5+-8.544)/2}}}
2 values:
{{{x=(5+8.544)/2=13.544/2=highlight(6.772inches=base)}}}
{{{x=(5-8.544)/2=-3.544/2=-1.772}}}, don't use
Height: {{{6.772-5=highlight(1.772inches=h)}}}
Check,
{{{6in^2=(1/2)(6.772)(1.772)}}}
{{{6in^2=(1/2)(12)}}}
{{{6in^2=6in^2}}}
Thank you,
Jojo</pre>