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Triangles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTriangles1 Mark
MCQ
In the given figure, $\angle\text{BAC}=90^\circ$ and $\text{AD}\perp\text{BC}.$ Then:
  • A
    $\text{BC}\cdot\text{CD}=\text{BC}^2$
  • B
    $\text{AB}\cdot\text{AC}=\text{BC}^2$
  • ✓
    $\text{BD}\cdot\text{CD}=\text{AD}^2$
  • D
    $\text{AB}\cdot\text{AC}=\text{AD}^2$
✓

Answer

Correct option: C.
$\text{BD}\cdot\text{CD}=\text{AD}^2$
In $\triangle\text{ABC},$
$\angle\text{ABD}=90^\circ-\angle\text{C}$
Similarly, in $\triangle\text{ACD},$
$\angle\text{CAD}=90^\circ-\angle\text{C}$
In $\triangle\text{DBA}$ and $\triangle\text{DAC}$
$\angle\text{ADB}=\angle\text{CDA}=90^\circ$
$\angle\text{ABD}=\angle\text{CAD}=90^\circ-\angle\text{C}$
So, $\triangle\text{DBA}\sim\triangle\text{DAC} .....(AA$ criterion of similarity$)$
$\frac{\text{BD}}{\text{AD}}=\frac{\text{AD}}{\text{CD}}$
$\Rightarrow\text{BD}\cdot\text{CD}=\text{AD}^2$

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