In a circle with centre O, the central angle AOB = 60^. If the radius of the circle is 12 ~cm, what is the length of the chord AB? [Page no. 110]

In ΔOAB, OA = OB = 12 cm (radii), so the triangle is isosceles. The angle at the vertex ∠AOB = 60°. The base angles are OAB = OBA = 180^ - 60^/2 = 60^. So, all angles are 60° i.e., ΔOAB is equilateral. Therefore, AB = OA = OB = 12 cm. Length of chord AB = 12 cm.

Question:

In a circle with centre $O$ , the central angle $\angle AOB = 60^{\circ}$ . If the radius of the circle is $12 \mathrm{~cm}$ , what is the length of the chord $AB$ ? [Page no. 110]

Answer: Verified

In ΔOAB, OA = OB = 12 cm (radii), so the triangle is isosceles. The angle at the vertex ∠AOB = 60°.

The base angles are $\angle OAB = \angle OBA = \frac{180^\circ - 60^\circ}{2} = 60^\circ$ .

So, all angles are 60° i.e., ΔOAB is equilateral.

Therefore, AB = OA = OB = 12 cm.

Length of chord AB = 12 cm.

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In a circle with centre OOO, the central angle ∠AOB=60∘\angle AOB = 60^{\circ}∠AOB=60∘. If the radius of the circle is 12 cm12 \mathrm{~cm}12 cm, what is the length of the chord ABABAB? [Page no. 110]

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In a circle with centre $O$ , the central angle $\angle AOB = 60^{\circ}$ . If the radius of the circle is $12 \mathrm{~cm}$ , what is the length of the chord $AB$ ? [Page no. 110]