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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 2 · Set 2 · Q31
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.If a regular hexagon ABCDEF circumscribes a circle, then prove that $\displaystyle \mathrm{AB}+\mathrm{CD}+\mathrm{EF}=\mathrm{BC}+\mathrm{DE}+\mathrm{FA}$.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
If a regular hexagon ABCDEF circumscribes a circle, then prove that $\displaystyle \mathrm{AB}+\mathrm{CD}+\mathrm{EF}=\mathrm{BC}+\mathrm{DE}+\mathrm{FA}$.
Marking-scheme solution
for correct Given, To Prove
for correct Proof
\[\left.\begin{array}{l}
\mathrm{AM}=\mathrm{AR} \\
\mathrm{BM}=\mathrm{BN} \\
\mathrm{CN}=\mathrm{CO} \\
\mathrm{DO}=\mathrm{DP} \\
\mathrm{EQ}=\mathrm{EP} \\
\mathrm{FQ}=\mathrm{FR}
\end{array}\right]
\]
\(\displaystyle \mathrm{LHS}=\mathrm{AB}+\mathrm{CD}+\mathrm{EF}=(\mathrm{AM}+\mathrm{BM})+(\mathrm{CO}+\mathrm{OD})+(\mathrm{EQ}+\mathrm{QF})\)
\(\displaystyle =\mathrm{AR}+\mathrm{BN}+\mathrm{CN}+\mathrm{DP}+\mathrm{EP}+\mathrm{EP}+\mathrm{FR}\)
\(\displaystyle =(\mathrm{AR}+\mathrm{FR})+(\mathrm{BN}+\mathrm{CN})+(\mathrm{DP}+\mathrm{EP})\)
\(\displaystyle =\mathrm{BC}+\mathrm{DE}+\mathrm{FA}=\mathrm{RHS}\)
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.