Question : In triangle RST, M and N are two points on RS and RT such that MN is parallel to the base ST of the triangle RST. If $\text{RM}=\frac{1}{3} \text{MS}$ and $\text{ST}=5.6 \text{ cm}$, what is the ratio of $\frac{\text { Area of triangle RMN}}{\text {Area of trapezium MNST}}$?
Option 1: $\frac{14}{15}$
Option 2: $\frac{15}{16}$
Option 3: $\frac{1}{15}$
Option 4: $\frac{1}{16}$
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Correct Answer: $\frac{1}{15}$
Solution :
Given: $RM=\frac{1}{3}MS$
⇒ $\frac{RM}{MS}=\frac{1}{3}$
$ST=5.6$ cm
Since $\triangle RMN$ and $\triangle RST$ are similar triangles.
Then, $\frac{RM}{RS}=\frac{MN}{ST}$
⇒ $\frac{1}{1+3}=\frac{MN}{5.6}$
⇒ $MN=\frac{5.6}{4}$
⇒ $MN=1.4$ cm
So, $\frac{\text { Area of triangle RMN }}{\text {Area of trapezium MNST}}=\frac{\text{Area of triangle RMN}}{\text{Area of RST–Area of RMN}}$
Putting the values, we have,
= $\frac{\frac{1}{2}×1.4×1}{\frac{1}{2}×4×5.6-\frac{1}{2}×1.4×1}$
= $\frac{0.7}{11.2-0.7}$
= $\frac{0.7}{10.5}$
= $\frac{1}{15}$
Hence, the correct answer is $\frac{1}{15}$.
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