Question

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 1

Answer: A

Given that:

$A_{ABC}=6$

And the formula for finding the area of a triangle is

$A=\frac{1}{2}bh$

We can rewrite this as

$A_{ABC}=\frac{1}{2}(AB)(BC)$$6=\frac{1}{2}(AB)(BC)$

*Solve for BC

$\frac{6}{AB}=\frac{\frac{1}{2}(AB)(BC)}{AB}$$\frac{6}{AB}=\frac{1}{2}(BC)$$2\times\lbrack\frac{6}{AB}\rbrack=\lbrack\frac{1}{2}(BC)\rbrack\times2$$\frac{12}{AB}=BC$

Now, given that BC = CD,

$\frac{12}{AB}=BC=CD$

Given that BC and CD are equal, this would mean that:

$BD=BC+CD$$BD=\frac{12}{AB}+\frac{12}{AB}$$BD=\frac{24}{AB}$

To find the Area of ACD, we will subtract the Area of ABC from the Area of ABD

$A_{ACD}=A_{ABD}-A_{ABC}$

We can rewrite this as:

$A_{ACD}=\frac{1}{2}(AB)(\frac{24}{AB})-6$

Evaluate

$A_{ACD}=\frac{1}{2}(24)-6$$A_{ACD}=12-6$$A_{ACD}=6$

Therefore, the Area of ACD is 6.