Question
In $\triangle\text{ABC},$ P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of $\triangle\text{APQ},$ and trapezium BPQC.
✓
Answer
In $\triangle\text{ABC}$ P is a point on AB such that AP : PQ = 1 : 2
PQ || BC
Now we have to find the ratio between area $\triangle\text{APQ}$ and area trap BPQC
$\because\text{PQ}||\text{BC}$
$\therefore\triangle\text{APQ}\sim\triangle\text{ABC}$
$\frac{\text{area}(\triangle\text{APQ)}}{\text{area}(\triangle\text{ABC})}=\frac{(\text{AP)}^2}{(\text{AB})^2}$.
$=\frac{(1)^2}{(1+2)^2}=\frac{(1)^2}{(3)^2}=\frac{1}{9}$
$\therefore9\text{ area}(\triangle\text{APQ})=\text{area}(\triangle\text{ABC})$
$\Rightarrow9\text{ area}(\triangle\text{APQ})=\text{area}(\triangle\text{APQ})+\text{area (trap BPQC})$
$\Rightarrow9\text{ area}(\triangle\text{APQ})-\text{are}(\triangle\text{APQ})=\text{are trap BPQC}$
$\Rightarrow8\text{ and}(\triangle\text{APQ})=\text{area (trap BPQC)}$
$\frac{\text{area}(\triangle\text{APQ)}}{\text{area}(\text{trap. }\text{BPQC})}=\frac{1}{8}$
$\therefore\text{Ratio} = 1 : 8$
PQ || BC
Now we have to find the ratio between area $\triangle\text{APQ}$ and area trap BPQC
$\because\text{PQ}||\text{BC}$
$\therefore\triangle\text{APQ}\sim\triangle\text{ABC}$
$\frac{\text{area}(\triangle\text{APQ)}}{\text{area}(\triangle\text{ABC})}=\frac{(\text{AP)}^2}{(\text{AB})^2}$.
$=\frac{(1)^2}{(1+2)^2}=\frac{(1)^2}{(3)^2}=\frac{1}{9}$
$\therefore9\text{ area}(\triangle\text{APQ})=\text{area}(\triangle\text{ABC})$
$\Rightarrow9\text{ area}(\triangle\text{APQ})=\text{area}(\triangle\text{APQ})+\text{area (trap BPQC})$
$\Rightarrow9\text{ area}(\triangle\text{APQ})-\text{are}(\triangle\text{APQ})=\text{are trap BPQC}$
$\Rightarrow8\text{ and}(\triangle\text{APQ})=\text{area (trap BPQC)}$
$\frac{\text{area}(\triangle\text{APQ)}}{\text{area}(\text{trap. }\text{BPQC})}=\frac{1}{8}$
$\therefore\text{Ratio} = 1 : 8$
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