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Triangles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTriangles1 Mark
MCQ
If $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are two triangles such that $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}=\frac{2}{5},$ then area $\text{Area }(\triangle\text{ABC}):\text{Area}(\triangle\text{DEF})=$
  • A
    $2 : 5$
  • ✓
    $4 : 25$
  • C
    $4 : 15$
  • D
    $8 : 125$
✓

Answer

Correct option: B.
$4 : 25$
Given: $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are two triangles such that $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}=\frac{2}{5}$
To find: $\text{Ar}(\triangle\text{ABC}):\text{Ar}(\triangle\text{DEF})$
We know that if the sides of two triangles are proportional, then the two triangles are similar.
Since $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{CA}}{\text{FD}}=\frac{2}{5},$ therefore, $\triangle\text{ABC}$ and $\triangle\text{DEF}$ are similar.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}$
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{2^2}{5^2}$
$\frac{\text{Ar}(\triangle\text{ABC})}{\text{Ar}(\triangle\text{DEF})}=\frac{4}{25}$
Hence the correct answer is $b.$

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