Question
In figure, $PQ || BC$ and $AP : PB = 1 : 2.$
Find: $\frac{\text{area}(\triangle\text{APQ})}{\text{area}(\triangle\text{ABC})}$
Find: $\frac{\text{area}(\triangle\text{APQ})}{\text{area}(\triangle\text{ABC})}$
✓
Answer
In $\triangle\text{ABC} PQ || BC$
$\triangle\text{APQ}\sim\triangle\text{ABC}$
But $AP : PB = 1 : 2$
The ratio of the areas of two similar triangles are proportional to the square of their corresponding sides
$\therefore\frac{\text{area}(\triangle\text{APQ})}{\text{area}(\triangle\text{ABC})}=\frac{\text{AP}^2}{\text{AB}^2}=\frac{(1)^2}{(1+2)^2}=\frac{(1)^2}{(3)^2}$
$=\frac{(1)^2}{(3)^2}=\frac{1}{9}$
$\therefore$ Ratio $= 1 : 9.$
$\triangle\text{APQ}\sim\triangle\text{ABC}$
But $AP : PB = 1 : 2$
The ratio of the areas of two similar triangles are proportional to the square of their corresponding sides
$\therefore\frac{\text{area}(\triangle\text{APQ})}{\text{area}(\triangle\text{ABC})}=\frac{\text{AP}^2}{\text{AB}^2}=\frac{(1)^2}{(1+2)^2}=\frac{(1)^2}{(3)^2}$
$=\frac{(1)^2}{(3)^2}=\frac{1}{9}$
$\therefore$ Ratio $= 1 : 9.$
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