Question

In the figure, the ratio of AD to DC is 3 to 2. If the area of $\Delta$ ABC is $40c{m}^2$, what is the area of $\Delta BDC$?

• A. $16c{m}^2$
• B. $24c{m}^2$
• C. $30c{m}^2$
• D. $36c{m}^2$

Answer 1

The correct option is A

$16c{m}^2$

Ratio in AD:DC = 3:2
and area $\Delta ABC=40c{m}^2$
Let h be the height, AD = 3x and DC = 2x
Area $\Delta ABD=\frac{1}{2}h\times AD$
= $\frac{1}{2}h\times 3x=\frac{3}{2}hx$
and area $\Delta BDC=\frac{1}{2}h\times DC$
= $\frac{1}{2}h\times 2x=hx$
$\therefore$ Ratio their area = $\frac{3}{2}hx:hx$
$\therefore$ Area of $\Delta BDC=\frac{Totalarea\times 2}{3+2}$
=$\frac{40\times 2}{5}=16c{m}^2$