Question : $\triangle \mathrm{XYZ} \sim \triangle \mathrm{GST}$ and $\mathrm{XY}: \mathrm{GS}=2: 3, \mathrm{XV}$ is the median to the side $\mathrm{YZ}$, and $\mathrm{GD}$ is the median to the side ST. The value of $\left(\frac{\mathrm{YV}}{\mathrm{SD}}\right)^2$ is:
Option 1: $\frac{4}{9}$
Option 2: $\frac{3}{5}$
Option 3: $\frac{1}{4}$
Option 4: $\frac{2}{3}$
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Correct Answer: $\frac{4}{9}$
Solution : Given: $\Delta \mathrm{XYZ} \sim \Delta \mathrm{GST}$ $\left(\frac{\mathrm{YV}}{\mathrm{SD}}\right)=\left(\frac{\mathrm{XY}}{\mathrm{GS}}\right)$ $\therefore\left(\frac{\mathrm{YV}}{\mathrm{SD}}\right)^2=\left(\frac{\mathrm{XY}}{\mathrm{GS}}\right)^2=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^2=\frac{4}{9}$ Hence, the correct answer is $\frac{4}{9}$.
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Question : In a $\triangle \mathrm{XYZ}, \mathrm{XO}$ is the median and $\mathrm{XO}=\frac{1}{2} \mathrm{YZ}$. If $\angle \mathrm{YXO}=30^{\circ}$, then what is the value of $\angle \mathrm{XYZ}$?
Option 1: 15°
Option 2: 90°
Option 3: 30°
Option 4: 60°
Question : $\triangle$ABC is similar to $\triangle$PQR and AB : PQ = 2 : 3. AD is the median to the side BC in $\triangle$ABC and PS is the median to the side QR in $\triangle$PQR. What is the value of $(\frac{BD}{QS})^2$?
Option 1: $\frac{3}{5}$
Option 2: $\frac{4}{9}$
Option 3: $\frac{2}{3}$
Option 4: $\frac{4}{7}$
Question : If $\triangle A B C \sim \triangle F D E$ such that $A B=9 \mathrm{~cm}, A C=11 \mathrm{~cm}, D F=16 \mathrm{~cm}$ and $D E=12 \mathrm{~cm}$, then the length of $BC$ is:
Option 1: $5 \frac{3}{4} \mathrm{~cm}$
Option 2: $4\frac{3}{5} \mathrm{~cm}$
Option 3: $3\frac{5}{7} \mathrm{~cm}$
Option 4: $6\frac{3}{4} \mathrm{~cm}$
Question : $3\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right]+\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right]^3=?$
Option 1: $a^2-\frac{1}{a^3}$
Option 2: $a^3-\frac{1}{a^3}$
Option 3: $a^3+\frac{1}{a^3}$
Option 4: $a^2-\frac{1}{a^2}$
Question : What is the value of $ \left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)? $
Option 1: $\frac{\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}$
Option 2: $\frac{\mathrm{k}^{32}-\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}-\frac{1}{\mathrm{k}}}\\$
Option 3: $\frac{\mathrm{k}^{32}-\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}\\$
Option 4: $\frac{\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}$
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