MCQ
$\triangle \text{ABC}\sim\triangle\text{DEF}, \text{ar}(\triangle\text{ABC})=9\text{ cm}^2,\ \text{ar} (\triangle\text{DEF})=16\text{ cm}^2.$ If $BC = 2.1\ cm,$ then the measure of $EF$ is :
- ✓$2.8\ cm.$
- B$4.2\ cm.$
- C$2.5\ cm.$
- D$4.1\ cm.$
✓
Answer
Correct option: A.
$2.8\ cm.$
Given : $\text{Ar}(\triangle\text{ABC})=9\text{ cm}^2,$
$\text{Ar}(\triangle\text{DEF})=16\text{ cm}^2,$ and ${BC}=2.1\text{ cm}$
To find : measure of $EF$
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\frac{9}{16}=\frac{2.1^2}{\text{EF}^2}$
$\frac{3}{4}=\frac{2.1}{\text{EF}}$
$\text{EF}=2.8\text{ cm}$
Hence the correct answer is $A$.
$\text{Ar}(\triangle\text{DEF})=16\text{ cm}^2,$ and ${BC}=2.1\text{ cm}$
To find : measure of $EF$
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\frac{9}{16}=\frac{2.1^2}{\text{EF}^2}$
$\frac{3}{4}=\frac{2.1}{\text{EF}}$
$\text{EF}=2.8\text{ cm}$
Hence the correct answer is $A$.
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