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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 3 · Set 3 · Q35
In ABC, if AD ⊥ BC and AD^2= BD × DC, then prove that ∠ BAC =90^°. /CD1BA/22 This section consists of 3 case study based questions of 4 marks each.
In $\displaystyle \triangle \mathrm{ABC}$, if $\displaystyle \mathrm{AD} \perp \mathrm{BC}$ and $\displaystyle \mathrm{AD}^{2}=\mathrm{BD} \times \mathrm{DC}$, then prove that $\displaystyle \angle \mathrm{BAC}=90^{\circ}$. /CD1BA/$\displaystyle 22$ This section consists of $\displaystyle 3$ case study based questions of $\displaystyle 4$ marks each.
Marking-scheme solution
\[\begin{gathered}
\mathrm{AD}^{2}=\mathrm{BD} \times \mathrm{DC} \\
\frac{\mathrm{AD}}{\mathrm{DC}}=\frac{\mathrm{BD}}{\mathrm{AD}}
\end{gathered}
\]
\[\begin{gathered}
\triangle \mathrm{ADB} \sim \triangle \mathrm{CDA} \\
\angle \mathrm{BAD}=\angle \mathrm{ACD} \& \angle \mathrm{DAC}=\angle \mathrm{DBA} \\
\text { In } \triangle \mathrm{ABC} \\
\angle \mathrm{ABC}+\angle \mathrm{CAB}+\angle \mathrm{BCA}=180^{\circ} \\
\angle \mathrm{ABC}+\angle \mathrm{BAD}+\angle \mathrm{CAD}+\angle \mathrm{BCA}=180^{\circ} \\
\angle \mathrm{BAD}+\angle \mathrm{CAD}=90^{\circ} \\
\angle \mathrm{BAC}=90^{\circ}
\end{gathered}
\]CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.