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: $6\frac{3}{4} \mathrm{~cm}$
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Correct Answer: $6\frac{3}{4} \mathrm{~cm}$
Solution : Given: $A B=9 \mathrm{~cm}, A C=11 \mathrm{~cm}, D F=16 \mathrm{~cm}$ and $D E=12 \mathrm{~cm}$ $\triangle ABC \sim \triangle FDE$ Now, $\frac{AB}{DF}= \frac{BC}{DE}$ Substituting the given values: ⇒ $\frac{9}{16}= \frac{BC}{12}$ $⇒BC = 12 \times \frac{9}{16} = 6\frac{3}{4} \mathrm{~cm}$ Hence, the correct answer is $6\frac{3}{4} \mathrm{~cm}$.
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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:
Question : If $\triangle A B C \sim \triangle E D F$ such that $AB=6$ cm, $DF=16$ cm, and $DE=8$ cm, then the length of $BC$ is:
Question : If $\triangle A B C \sim \triangle D E F$, and $B C=4 \mathrm{~cm}, E F=5 \mathrm{~cm}$ and the area of triangle $A B C=80 \mathrm{~cm}^2$, then the area of the $\triangle DEF$ is:
Question : In $\triangle \mathrm{ABC}$, $AB=20$ cm, $BC=7$ cm and $CA=15$ cm. Side $BC$ is produced to $D$ such that $\triangle \mathrm{DAB} \sim \triangle \mathrm{DCA}$. $DC$ is equal to:
Question : $ABC$ is a triangle and $D$ is a point on the side $BC$. If $BC = 16\mathrm{~cm}$, $BD = 11 \mathrm{~cm}$ and $\angle ADC = \angle BAC$, then the length of $AC$ is equal to:
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