Question

In the given figure, point D is on side BC of $△ABC$ such that $△ABC\sim △DAC$. If $AD=4,AC=5$, then $A\left(△ADC\right):A\left(△BAC\right)$ is equal to:

• A. $16:25$
• B. $9:4$
• C. $4:9$
• D. $16:18$

Answer 1

Answer: (A)

$16:25$

Given that $△ABC\sim △DAC$ and $\frac{AD}{AC}=\frac{4}{5}$

So, $\frac{AB}{DA}=\frac{BC}{AC}=\frac{AC}{DC}$ ....CSST

Hence, $\frac{AB}{BC}=\frac{AD}{AC}=\frac{4}{5}$

$\frac{A\left(△ABC\right)}{A\left(△DAC\right)}=\frac{A{B}^2}{B{C}^2}$ ...Theorem on ratio of areas of similar triangles

$\frac{A\left(△ABC\right)}{A\left(△DAC\right)}=\frac{4^2}{5^2}$

$\frac{A\left(△ABC\right)}{A\left(△DAC\right)}=\frac{16}{25}$