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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Question : In $\triangle$ABC, D and E are two points on the sides AB and AC, respectively, so that DE $\parallel$ BC and $\frac{AD}{BD}=\frac{2}{3}$. Then $\frac{\text{Area of trapezium DECB}}{\text{Area of $\triangle$ABC}}$ is equal to:
Option 1: $\frac{5}{9}$
Option 2: $\frac{21}{25}$
Option 3: $1\frac{4}{5}$
Option 4: $5\frac{1}{4}$
Question : In $\triangle ABC$, $D$ and $E$ are the points of sides $AB$ and $BC$ respectively such that $DE \parallel AC$ and $AD : DB = 3 : 2$. The ratio of the area of trapezium $ACED$ to that of $\triangle DBE$ is:
Option 1: $4:15$
Option 2: $15:4$
Option 3: $4:21$
Option 4: $21:4$
Question : The sides of similar triangle $\triangle ABC$ and $\triangle DEF$ are in the ratio of $\frac{\sqrt{3}}{\sqrt{5}}$. If the area of $\triangle ABC$ is $90 \text{ cm}^2$, then the area of $\triangle DFF\left(\right.$ in $\left.\text{cm}^2\right)$ is:
Option 1: 150
Option 2: 152
Option 3: 154
Option 4: 156
Question : A can finish a work in 7 days. B can finish the same work in 9 days. The number of days required to finish the same work by both of them together is:
Option 1: $1\frac{15}{16}$
Option 2: $2\frac{15}{16}$
Option 3: $3\frac{15}{16}$
Option 4: $4\frac{15}{16}$
Question : ABCD is a square. Draw a triangle QBC on side BC considering BC as a base and draw a triangle PAC on AC as its base such that $\Delta$QBC$\sim\Delta$PAC. Then, $\frac{\text{Area of $\Delta$QBC}}{\text{Area of $\Delta$PAC}}$ is equal to:
Option 1: $\frac{1}{2}$
Option 2: $\frac{2}{1}$
Option 3: $\frac{1}{3}$
Option 4: $\frac{2}{3}$
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