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 = 2.1cm, then the measure of EF is:
- ✓2.8cm.
- B4.2cm.
- C2.5cm.
- D4.1cm.
✓
Answer
Correct option: A.
2.8cm.
Given: $\text{Ar}(\triangle\text{ABC})=9\text{cm}^2,$ $\text{Ar}(\triangle\text{DEF})=16\text{cm}^2,\ \text{and BC}=2.1\text{cm}$
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.
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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