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Triangles — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTriangles1 Mark
MCQ
If $\triangle\text{ABC}\sim\triangle\text{QRP},\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{9}{4},\text{AB}=18\text{cm}$ and $\text{BC}=15\text{cm}$ then $\text{PR}=?$
  • A
    8cm
  • ✓
    10cm
  • C
    12cm
  • D
    $\frac{20}{3}\text{cm}$
✓

Answer

Correct option: B.
10cm
$\triangle\text{ABC}\sim\triangle\text{QRP}$
$\Rightarrow\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}=\frac{\text{AC}}{\text{PQ}}$
Also, $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{AB}^2}{\text{BC}^2}$
$\Rightarrow\frac{9}{4}=\frac{\text{AB}^2}{\text{BC}^2}$
$\Rightarrow\frac{\text{AB}}{\text{QR}}=\frac{3}{2}$
So, $\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}\Rightarrow\frac{\text{BC}}{\text{PR}}=\frac{3}{2}$
$\Rightarrow\frac{15}{\text{PR}}=\frac{3}{2}$
$\Rightarrow\text{PR}=10\text{cm}.$

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