A Chord On A Circle

Chord of a Circle Definition

The chord of a circumvolve can be divers as the line segment joining any 2 points on the circumference of the circumvolve. It should be noted that the diameter is the longest chord of a circle which passes through the center of the circle. The figure below depicts a circle and its chord.

Chord Of Circle

In the given circle with 'O' as the heart, AB represents the diameter of the circumvolve (longest chord), 'OE' denotes the radius of the circumvolve and CD represents a chord of the circle.

Allow us consider the chord CD of the circle and ii points P and Q anywhere on the circumference of the circumvolve except the chord every bit shown in the effigy below. If the endpoints of the chord CD are joined to the point P, then the bending ∠CPD is known equally the angle subtended past the chord CD at signal P. The angle ∠CQD is the bending subtended by chord CD at Q. The angle ∠COD is the bending subtended past chord CD at the center O.

Angle Subtended by Chord

Chord Length Formula

There are 2 basic formulas to discover the length of the chord of a circumvolve which are:

Formula to Calculate Length of a Chord
Chord Length Using Perpendicular Distance from the Middle	Chord Length = two × √(r²− d²)
Chord Length Using Trigonometry	Chord Length = ii × r × sin(c/2)

Chord Length of a Circle Formula

Where,

r is the radius of the circumvolve
c is the angle subtended at the centre by the chord
d is the perpendicular altitude from the chord to the circle center

Instance Question Using Chord Length Formula

Question: Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance from the chord to the center is 4 cm.

Solution:

Given radius, r = 7 cm

and altitude, d = four cm

Chord length = 2√(r²−dⁱⁱ)

⇒ Chord length = 2√(7²−4²)

⇒ Chord length = 2√(49−xvi)

⇒ Chord length = 2√33

⇒ Chord length = 2×5.744

Or , chord length = 11.48 cm

Video Related to Chords

Chord of a Circle Theorems

If we try to institute a relationship between dissimilar chords and the angle subtended past them in the heart of the circle, nosotros run across that the longer chord subtends a greater angle at the center. Similarly, two chords of equal length subtend equal angle at the center. Let u.s. attempt to prove this statement.

Theorem 1 : Equal Chords Equal Angles Theorem

Statement: Chords which are equal in length subtend equal angles at the center of the circle.

Chords which are equal in length subtend equal angles at the center of the circle.

Proof:

From fig. iii, In ∆AOB and ∆POQ

S.No.	Statement	Reason
one.	AB=PQ	Chords of equal length (Given)
2.	OA = OB = OP = OQ	Radius of the same circle
3.	△AOB ≅ △POQ	SSS precept of Congruence
iv.	∠AOB = ∠POQ	By CPCT from argument three

Note: CPCT stands for congruent parts of congruent triangles.

The converse of theorem 1 besides holds true, which states that if two angles subtended by two chords at the center are equal so the chords are of equal length. From fig. 3, if ∠AOB =∠POQ, then AB=PQ. Let us try to prove this statement.

Theorem 2: Equal Angles Equal Chords Theorem (Converse of Theorem i)

Statement: If the angles subtended by the chords of a circumvolve are equal in measure, so the length of the chords is equal.

If the angles subtended by the chords of a circle are equal in measure then the length of the chords is equal.

Proof:

From fig. 4, In ∆AOB and ∆POQ

S.No.	Statement	Reason
1.	∠AOB = ∠POQ	Equal angle subtended at centre O (Given)
ii.	OA = OB = OP = OQ	Radii of the same circle
3.	△AOB ≅ △POQ	SAS axiom of Congruence
4.	AB = PQ	From Statement iii (CPCT)

Theorem three: Equal Chords Equidistant from Center Theorem

Statement: Equal chords of a circle are equidistant from the centre of the circle.

Proof:

Given: Chords AB and CD are equal in length.

Structure: Join A and C with centre O and drop perpendiculars from O to the chords AB and CD.

Equal chords of a circle are equidistant from the center of the circle.

S.No.	Statement	Reason
one	AP = AB/2, CQ = CD/2	The perpendicular from centre bisects the chord
In △OAP and △OCQ
ii	∠1 = ∠ii = 90°	OP⊥AB and OQ⊥CD
3	OA = OC	Radii of the same circle
iv	OP = OQ	Given
5	△OPB ≅ △OQD	R.H.S. Axiom of Congruency
6	AP = CQ	Respective parts of congruent triangle
7	AB = CD	From argument (i) and (6)

Solved Examples

Example 1:

A chord of a circumvolve is equal to its radius. Find the angle subtended by this chord at a point in the major segment.

Solution:

Let O exist the center, and AB exist the chord of the circle.

So, OA and OB be the radii.

Given that chord of a circle is equal to the radius.

AB = OA = OB

Chord of a circle Example 1

Thus, ΔOAB is an equilateral triangle.

That means ∠AOB = ∠OBA = ∠OAB = 60°

Also, we know that the angle subtended by an arc at the centre of the circle is twice the angle subtended by information technology at any other indicate in the remaining office of the circle.

So, ∠AOB = 2∠ACB

⇒ ∠ACB = 1/2 (∠AOB)

⇒ ∠ACB = i/2 (60°)

= 30°

Hence, the bending subtended by the given chord at a point in the major segment is 30°.

Example 2:

Two chords AB and Ac of a circumvolve subtend angles equal to 90º and 150º, respectively at the centre. Discover ∠BAC, if AB and AC lie on the opposite sides of the centre.

Solution:

Given,

Two chords AB and AC of a circle subtend angles equal to 90º and 150º.

And so, ∠AOC = 90º and∠AOB = 150º

Chord of a circle Example 2

In ΔAOB,

OA = OB (radius of the circle)

Equally we know, angles opposite to equal sides are equal.

So, ∠OBA = ∠OAB

According to the angle sum property of triangle theorem, the sum of all angles of a triangle = 180°

In ΔAOB,

∠OAB + ∠AOB +∠OBA = 180°

∠OAB + 90° + ∠OAB = 180°

2∠OAB = 180° – 90°

2∠OAB = ninety°

⇒ ∠OAB = 45°

Now, in ΔAOC,

OA = OC (radius of the circumvolve)

Equally mentioned above, angles contrary to equal sides are equal.

∴ ∠OCA = ∠OAC

Using the angle sum property in ΔAOB, we take;

∠OAC + ∠AOC +∠OCA = 180°

∠OAC + 150° + ∠OAC = 180°

2∠OAC = 180° – 150°

2∠OAC = 30°

⇒ ∠OAC = fifteen°

At present, ∠BAC = ∠OAB + ∠OAC

= 45° + xv°

= 60°

Therefore, ∠BAC = 60°

Practice Issues

Two equal chords AB and CD of a circle, when produced, intersect at a signal P. Prove that PB = PD.
Two circles of radii five cm and iii cm intersect at ii points, and the altitude between their centres is 4 cm. Find the length of the common chord.
The lengths of two parallel chords of a circle are half-dozen cm and viii cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre?

Frequently Asked Questions

What is a Circle?

A circle is defined equally a airtight two-dimensional figure whose all the points in the boundary are equidistant from a single indicate (called heart).

What is the Chord of a Circle?

The chord is a line segment that joins 2 points on the circumference of the circle. A chord but covers the office inside the circle.

What is the Formula of Chord Length?

The length of whatever chord can be calculated using the following formula:

Chord Length = ii × √(r²− d²)

Is Bore a Chord of a Circle?

Yes, the bore is also considered equally a chord of the circle. The diameter is the longest chord possible in a circle and it divides the circle into two equal parts.