In the figure above, the area of triangle ABC is 6, if BC = CD what is he area of triangle ACD? A. 6 B.8 C.9 D.10

Answer: A
Given that:
[tex]A_{ABC}=6[/tex]And the formula for finding the area of a triangle is
[tex]A=\frac{1}{2}bh[/tex]We can rewrite this as
[tex]A_{ABC}=\frac{1}{2}(AB)(BC)[/tex][tex]6=\frac{1}{2}(AB)(BC)[/tex]*Solve for BC
[tex]\frac{6}{AB}=\frac{\frac{1}{2}(AB)(BC)}{AB}[/tex][tex]\frac{6}{AB}=\frac{1}{2}(BC)[/tex][tex]2\times\lbrack\frac{6}{AB}\rbrack=\lbrack\frac{1}{2}(BC)\rbrack\times2[/tex][tex]\frac{12}{AB}=BC[/tex]Now, given that BC = CD,
[tex]\frac{12}{AB}=BC=CD[/tex]Given that BC and CD are equal, this would mean that:
[tex]BD=BC+CD[/tex][tex]BD=\frac{12}{AB}+\frac{12}{AB}[/tex][tex]BD=\frac{24}{AB}[/tex]To find the Area of ACD, we will subtract the Area of ABC from the Area of ABD
[tex]A_{ACD}=A_{ABD}-A_{ABC}[/tex]We can rewrite this as:
[tex]A_{ACD}=\frac{1}{2}(AB)(\frac{24}{AB})-6[/tex]Evaluate
[tex]A_{ACD}=\frac{1}{2}(24)-6[/tex][tex]A_{ACD}=12-6[/tex][tex]A_{ACD}=6[/tex]Therefore, the Area of ACD is 6.