Q190 / 270 · NDA 2026 (Phase 1) · Mathematics
What is the length of the chord of a unit circle which subtends at the centre of the circle an angle of $45^\circ$ ?
- A$2\sqrt{2+\sqrt{2}}$ units
- B$2\sqrt{2-\sqrt{2}}$ units
- C$\sqrt{2+\sqrt{2}}$ units
- Correct answerD$\sqrt{2-\sqrt{2}}$ units
Solution
Let the chord be $AB$ and the centre of the circle be $O$.
In $\triangle OAB$, the lengths of the radii are $OA = OB = 1$ and the angle subtended at the centre is $\angle AOB = 45^\circ$.
Using the cosine rule in $\triangle OAB$:
$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(45^\circ)$
Substituting the given values:
$AB^2 = 1^2 + 1^2 - 2(1)(1)\left(\dfrac{1}{\sqrt{2}}\right)$
$AB^2 = 2 - \sqrt{2}$
$AB = \sqrt{2 - \sqrt{2}}$ units
Answer: $\sqrt{2-\sqrt{2}}$ units
In $\triangle OAB$, the lengths of the radii are $OA = OB = 1$ and the angle subtended at the centre is $\angle AOB = 45^\circ$.
Using the cosine rule in $\triangle OAB$:
$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(45^\circ)$
Substituting the given values:
$AB^2 = 1^2 + 1^2 - 2(1)(1)\left(\dfrac{1}{\sqrt{2}}\right)$
$AB^2 = 2 - \sqrt{2}$
$AB = \sqrt{2 - \sqrt{2}}$ units
Answer: $\sqrt{2-\sqrt{2}}$ units
Hence, the correct answer is option: 'D'.