Answer:
[tex]Solution,\\Area~of~triangle=\frac{1}{2} (base)(height) = \frac{1}{2}(2x+\frac{1}{4}) (x-8)\\~~~~~~~~~~~~~~~~~~~~~~~~ =\frac{1}{2}[2x^2-16x+\frac{1}{4}x-\frac{1}{4}(8)]\\ ~~~~~~~~~~~~~~~~~~~~~~~~ =\frac{1}{2}[2x^2-\frac{63}{4}x - 2]\\~~~~~~~~~~~~~~~~~~~~~~~~=x^2-\frac{63}{8}x-1[/tex]
