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:
Option 1: 169 cm2
Option 2: 80 cm2
Option 3: 144 cm2
Option 4: 125 cm2
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Correct Answer: 125 cm 2
Solution : $\triangle$ ABC ~ $\triangle$ DEF, hence the ratio of areas of two similar triangles is equal to the ratio of the square of their corresponding sides. $\frac{\text{area of }(\triangle ABC)}{\text{area}(\triangle DEF)}=\frac{BC^{2}}{EF^{2}}$ $⇒\frac{80}{\text{area of}(\triangle DEF)}=\frac{4^{2}}{5^{2}}$ $⇒\frac{80}{\text{area of}(\triangle DEF)}=\frac{16}{25}$ $⇒\text{area of}(\triangle DEF) = \frac{80 \times 25}{16}$ $⇒\text{area of}(\triangle DEF) = 125$ cm 2 Hence, the correct answer is 125 cm 2 .
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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:
Question : $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are similar triangles and their areas are 49 cm2 and 144 cm2 respectively, If $EF$ = 16.80 cm, then find $BC$.
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 : $\triangle BAC\sim \triangle PQR$. The area of $\triangle BAC$ and $\triangle PQR$ is 25 cm2 and 36 cm2 respectively. If BA = 4 cm, then what is the length of PQ?
Question : If in $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}, \frac{A B}{D E}=\frac{B C}{F D}$, then they will be similar when:
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