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Triangles — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTriangles1 Mark
MCQ
$\triangle\text{ABC}$ is a right triangle right-angled at A and $\text{AD}\perp\text{BC}.$ Then, $\frac{\text{BD}}{\text{DC}}=$
  • ✓
    $\Big(\frac{\text{AB}}{\text{AC}}\Big)^2$
  • B
    $\frac{\text{AB}}{\text{AC}}$
  • C
    $\Big(\frac{\text{AB}}{\text{AD}}\Big)^2$
  • D
    $\frac{\text{AB}}{\text{AD}}$
✓

Answer

Correct option: A.
$\Big(\frac{\text{AB}}{\text{AC}}\Big)^2$

In right angled $\triangle\text{ABC},\ \angle\text{A}=90^\circ$
$\text{AD}\perp\text{BC}$
$\therefore\triangle\text{ABD}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AB}}{\text{BC}}=\frac{\text{BD}}{\text{AB}}\Rightarrow\text{AB}^2=\text{BD}\times\text{BC}\ \ ...(\text{i})$
Similarly $\triangle\text{ACD}\sim\triangle\text{ABC}$
$\therefore\frac{\text{AC}}{\text{BC}}=\frac{\text{DC}}{\text{AC}}\Rightarrow\text{AC}^2=\text{DC}\times\text{BC}\ \ ...(\text{ii})$
Dividing (ii) by (i)
$\frac{\text{BD}\times\text{BC}}{\text{DC}\times{\text{BC}}}=\frac{\text{AB}^2}{\text{AC}^2}\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}^2}{\text{AC}^2}$
Hence $\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}^2}{\text{AC}^2}$

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