✓ Board-verified
Mathematics · 2026 · 1 mark
CBSE 2026 · Region 2 · Set 1 · Q15
Area of a segment of a circle of radius ' r ' and central angle 60^° is:
Area of a segment of a circle of radius ' $\displaystyle \mathrm{r}$ ' and central angle $\displaystyle 60^{\circ}$ is :
Official answer
(C)
$\displaystyle \frac{\pi \mathrm{r}^{2}}{6}-\frac{\sqrt{3}}{4} \mathrm{r}^{2}$
All Introduction to Trigonometry questions2026 question paperNCERT solutions: Introduction to TrigonometryPractice Introduction to Trigonometry →
More from Introduction to Trigonometry
- Prove that: (1+ cot θ-cosec θ)(1+ tan θ+ sec θ)=22026
- Evaluate: (5 cos^2 60^°+4 sec^2 30^°- tan^2 45^°)/(sin^2 30^°+ cos^2 30^°) OR Prove that: 1+(cot^2 α)/(1+cosec α)=cosec α2026
- [5/8 sec^2 60^°- tan^2 60^°+ cos^2 45^°] is equal to2023
- If sin θ+ cos θ=√3, then prove that tan θ+ cot θ=1 OR Prove that: ( sin A+ sec A)^2+( cos A+cosec A)^2=(1+ sec A cosec A)^22026
- The hour-hand of a clock is 6 cm long. The angle swept by it between 7:20 a.m. and 7:55 a.m. is:2023
- If x=a cos θ and y=b sin θ, then the value of b^2 x^2+a^2 y^2 is:2024
- Assertion: For an acute angle θ, value of cosec θ cannot be 1/(√2). Reason (R): cosec θ ≥ 1 for 0^° ≤ θ ≤ 90^°2025
- Prove that (cosec A- sin A)( sec A- cos A)=1/(cot A+ tan A).2023
CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.