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I'm Up and Down, and Round and Round - Chords and the Angles They Subtend

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A chord subtends an angle at the center of a circle. Equal chords of a circle subtend equal angles at the center.

Angle subtended by chord AB at center O
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The perpendicular from the center of a circle to a chord bisects the chord. Conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.

Perpendicular OM from center O bisects chord AB
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Equal chords of a circle (or of congruent circles) are equidistant from the center.

Two equal chords at equal distance from the center
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The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

📐Formulae

r2=d2+(l2)2r^2 = d^2 + \left(\frac{l}{2}\right)^2

l=2r2−d2l = 2\sqrt{r^2 - d^2}

d=r2−(l2)2d = \sqrt{r^2 - \left(\frac{l}{2}\right)^2}

💡Examples

Problem 1:

A chord of length 16 cm16\text{ cm} is drawn in a circle of radius 10 cm10\text{ cm}. Find the distance of the chord from the center of the circle.

Solution:

Given: Radius r=10 cmr = 10\text{ cm}, Chord length l=16 cml = 16\text{ cm}. Let the distance from the center be dd. The perpendicular from the center bisects the chord, so the half-length of the chord is: l2=162=8 cm\frac{l}{2} = \frac{16}{2} = 8\text{ cm} Using the Pythagoras theorem in the right-angled triangle formed by the radius, the distance, and the half-chord: r2=d2+(l2)2r^2 = d^2 + \left(\frac{l}{2}\right)^2 102=d2+8210^2 = d^2 + 8^2 100=d2+64100 = d^2 + 64 d2=100−64=36d^2 = 100 - 64 = 36 d=36=6 cmd = \sqrt{36} = 6\text{ cm}

Explanation:

We use the property that the perpendicular from the center to a chord bisects it, creating a right-angled triangle where the radius is the hypotenuse.

Problem 2:

In a circle with center OO, two chords ABAB and CDCD are such that AB=CDAB = CD. If ∠AOB=70∘\angle AOB = 70^\circ, find the value of ∠COD\angle COD.

Solution:

According to the theorem, equal chords of a circle subtend equal angles at the center. Given that chord AB=CDAB = CD. Therefore, ∠AOB=∠COD\angle AOB = \angle COD. Since ∠AOB=70∘\angle AOB = 70^\circ, then ∠COD=70∘\angle COD = 70^\circ.

Explanation:

This directly applies the theorem stating that equal chords subtend equal angles at the center of the circle.

Problem 3:

Find the length of a chord which is at a distance of 5 cm5\text{ cm} from the center of a circle of radius 13 cm13\text{ cm}.

Solution:

Given: r=13 cmr = 13\text{ cm} and d=5 cmd = 5\text{ cm}. We need to find the length of the chord ll. From the relation r2=d2+(l2)2r^2 = d^2 + (\frac{l}{2})^2: 132=52+(l2)213^2 = 5^2 + \left(\frac{l}{2}\right)^2 169=25+(l2)2169 = 25 + \left(\frac{l}{2}\right)^2 (l2)2=169−25=144\left(\frac{l}{2}\right)^2 = 169 - 25 = 144 l2=144=12 cm\frac{l}{2} = \sqrt{144} = 12\text{ cm} l=12×2=24 cml = 12 \times 2 = 24\text{ cm}

Explanation:

By calculating the half-chord length using Pythagoras theorem and then doubling it, we find the full length of the chord.

Problem 4:

Two chords ABAB and CDCD of a circle are 10 cm10\text{ cm} each. If the distance of chord ABAB from the center OO is 12 cm12\text{ cm}, what is the radius of the circle?

Geometry showing chord AB, center O, and radius r

Solution:

Let MM be the midpoint of chord ABAB. Since OM⊥ABOM \perp AB, AM=12AB=5 cmAM = \frac{1}{2} AB = 5\text{ cm}. In △OAM\triangle OAM, using Pythagoras theorem: OA2=OM2+AM2OA^2 = OM^2 + AM^2 r2=122+52=144+25=169r^2 = 12^2 + 5^2 = 144 + 25 = 169 r=169=13 cmr = \sqrt{169} = 13\text{ cm}.

Explanation:

Equal chords are equidistant from the center, so CDCD is also at 12 cm12\text{ cm}. We use the property that a perpendicular from the center bisects the chord to form a right-angled triangle with the radius.

Problem 5:

In a circle with center OO, chord PQPQ subtends an angle of 120∘120^\circ at the center. If the radius is 6 cm6\text{ cm}, find the length of the chord PQPQ.

Chord PQ subtending 120 degrees at center O

Solution:

Draw OM⊥PQOM \perp PQ. Since OMOM bisects ∠POQ\angle POQ, ∠POM=12×120∘=60∘\angle POM = \frac{1}{2} \times 120^\circ = 60^\circ. In △POM\triangle POM, sin⁡(60∘)=PMOP\sin(60^\circ) = \frac{PM}{OP} 32=PM6\frac{\sqrt{3}}{2} = \frac{PM}{6} PM=33 cmPM = 3\sqrt{3}\text{ cm} PQ=2×PM=63≈10.39 cmPQ = 2 \times PM = 6\sqrt{3} \approx 10.39\text{ cm}.

Explanation:

The perpendicular from the center to a chord bisects the angle at the center. By using trigonometry in the resulting right-angled triangle, we find half the chord length.