Question : In $\triangle \text{ABC}, \mathrm{DE} \| \mathrm{BC}$ and $\frac{\text{AD}}{\text{DB}}=\frac{4}{5}$. If $\mathrm{DE}=12 \mathrm{~cm}$, find the length of $\mathrm{BC}$.
Option 1: 48 cm
Option 2: 12 cm
Option 3: 30 cm
Option 4: 27 cm
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Correct Answer: 27 cm
Solution : According to the question, $\mathrm{DE} \| \mathrm{BC}$ $\frac{\text{AD}}{\text{DB}}=\frac{4}{5}$ DE = 12 cm By the SAS criterion of similarity, we have △ABC $\sim$ △ADE $\therefore$ $\frac{\text{AB}}{\text{AD}} = \frac{\text{BC}}{\text{DE}}$ ⇒ $\text{BC} = \frac{\text{AB}}{\text{AD}}\times \text{DE}$ ⇒ $\text{BC} = \frac{\text{AD+DB}}{\text{AD}}\times 12$ ⇒ $\text{BC} = (1+\frac{\text{DB}}{\text{AD}})\times 12$ ⇒ $\text{BC}=12\times(1+\frac{5}{4}) = 3\times 9 = 27$ cm Hence, the correct answer is 27 cm.
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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 : If $\triangle ABC \sim \triangle QRP, \frac{\operatorname{area}(\triangle A B C)}{\operatorname{area}(\triangle Q R P)}=\frac{9}{4}, A B=18 \mathrm{~cm}, \mathrm{BC}=15 \mathrm{~cm}$, then the length of $\mathrm{PR}$ is:
Option 1: 16 cm
Option 2: 14 cm
Option 3: 10 cm
Option 4: 12 cm
Question : In $\mathrm{\Delta ABC, \angle BAC = 90^{\circ}}$ and $\mathrm{AD}$ is drawn perpendicular to $\mathrm{BC}$. If $\mathrm{BD} = 7\;\mathrm{cm}$ and $\mathrm{CD }= 28\;\mathrm{cm}$, what is the length of $\mathrm{AD}$?
Option 1: $3.5 \text{ cm}$
Option 2: $7 \text{ cm}$
Option 3: $10.5 \text{ cm}$
Option 4: $14 \text{ cm}$
Question : $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$ and the perimeters of these triangles are 32 cm and 12 cm, respectively. If $\mathrm{DE}=6 \mathrm{~cm}$, then what will be the length of AB?
Option 3: 12 cm
Option 4: 18 cm
Question : If $\mathrm{D}$ is the midpoint of $\mathrm{BC}$ in $\triangle \mathrm{ABC}$ and $\angle A=90^{\circ}$, then $AD=$ ______.
Option 1: $\frac{\mathrm{BC}}{4}$
Option 2: $2 \mathrm{BC}$
Option 3: $\frac{\mathrm{BC}}{2}$
Option 4: $\mathrm{BC}$
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