Chapter 17.4 - Length of Chords

In the previous section we saw properties related to 'equal chords'. In this section, we will see  'length of chords'.

Length of chords

Consider fig.17.18 below. AB is a chord. OP is the perpendicular from the centre. OA is the radius r.

Fig.17.18

• OAP is a right angled triangle. Applying Pythagoras theorem, we get: AP = √[OA² – OP²]. But AP is 'half the chord'. So this is an easy method to get the length of any chord. We will write it in the form of a theorem:

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Theorem 17.5:

■ Half of chord is the square root of:

    • the difference of two quantities:

        ♦ Square of the radius

        ♦ square of the perpendicular distance of the chord from the centre

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The steps can be combined to form a single equation:
Eq.17.1:

^c⁄₂ = √[r²- p²]

Where c is the length of chord, r is the radius of circle and p is the perpendicular distance of the chord from the centre

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Once we get 'the half', we will get 'the full', just by multiplying it with '2'. Let us see an example: 
A circle has a radius of 4 cm. One of it’s chords is situated at a perpendicular distance of 3 cm from the centre. What is the length of the chord?

Solution:

Applying theorem 17.5, we have:

Half of the chord is the square root of:

  • the difference of two quantities:

      ♦ square of radius = 4² = 16

      ♦ square of the perpendicular distance of the chord from the centre = 3² = 9

• So difference = 16-9 = 7

• So half of the chord is √7.

• Thus full chord = 2√7

Another method: We can avoid the detailed steps by using Eq.17.1:
• Using Eq.17.1, we have: ^c⁄₂ = √[r²- p²]
• Given r = 4 cm and p = 3 cm

• ^c⁄₂ = √[4²- 3²] = √[16- 9] = √7

• So c = 2√7 cm

Another type of problem:

The distance between the ends of a piece of bangle is 4 cm. It’s height is 1 cm. What is the radius of the full bangle?

Solution:

1. Fig.17.19(a) shows the piece of bangle. The distance between it’s ends A and B is 4 cm. It’s height is 1 cm.

Fig.17.19

2. Fig.b shows the full bangle. AB is taken as a chord. The height 1 cm is measured in a direction perpendicular to the chord AB. 
3. OP is the perpendicular from the centre to AB. OP is extended further until it meets the circle at Q. So OQ is the radius. O and B are joined by the line OB. 
4. OPB forms a right triangle. 
• It’s hypotenuse OB = r. 
• One leg OP = (r -1) cm. 
• Other leg PB = 2 cm. 
5. Applying Pythagoras theorem we get:

• r² – (r-1)² = 2² ⇒ r² – (r² - 2r + 1) = 4 ⇒ (r² – r² + 2r - 1) = 4 ⇒ 2r - 1 = 4 ⇒ 2r = 5
• Thus we get r = 2.5 cm

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Now we will see some solved examples

Solved example 17.6

In a circle, a chord 1 cm away from the centre is 6 cm long. What is the length of the chord which is 2 cm away from the centre?

Solution:

A rough sketch is shown in fig.17.20. Note that it is only a rough sketch. Actual construction using the given measurements is not required to solve the problem.

Fig.17.20

1. Given: length of chord c = 6 cm. So ^c⁄₂ = 3 cm

2. Also given: perpendicular distance p = 1 cm. These details relate to the lower ⊿OAP in fig.17.20

3. Using Eq.17.1, we have: ^c⁄₂ = √[r²- p²]

4. r is the unknown. So we must bring it to the left side.

• ^c⁄₂ = √[r²- p²] ⇒ (^c⁄₂)² = [r²- p²] (squaring both sides)

• ⇒ r² = [(^c⁄₂)² + p²] ⇒ r = √[(^c⁄₂)² + p²]

5. Thus we get r = √[3² + 1²] = √[9 + 1] = √10

• Now we use this r for the second part of the problem. This is related to the ⊿OCQ

6. We have ^c⁄₂ = √[r²- p²] = √[(√10)²- 2²] = √[10 - 4] = √6

7. So c = 2√6

Solved example 17.7

In a circle of radius 5 cm, two parallel chords of lengths 6 and 8 cm are drawn on either side of the diameter. What is the distance between them? If parallel chords of these lengths are drawn on the same side of a diameter, what would be the distance between them?

Solution:

Part I: Chords on either side of diameter

A rough sketch is shown in fig.17.21(a) below: 

Fig.17.21

Note that it is only a rough sketch. Actual construction using the given measurements is not required to solve the problem.

1. Consider the chord of 8 cm length. It is named as AB in the fig.17.21(a)

2. Given: length of chord c = 8 cm. So ^c⁄₂ = 4 cm

3. Also given: radius r = 5 cm. These details relate to ⊿OAP in fig.a

4. We have ^c⁄₂ = √[r²- p²]. Here p is the unknown. So we must bring it to the left side. 
5. ^c⁄₂ = √[r²- p²] ⇒ (^c⁄₂)² = [r²- p²] (squaring both sides)

⇒ p² = [ r² - (^c⁄₂)²] ⇒ p = √[ r² - (^c⁄₂)²]

6. Thus we get OP = p = √[5² - 4²] = √[25 - 16] = √9 = 3

7. Consider the chord of 6 cm length. It is named as CD in the fig.a

8. Given: length of chord c = 6 cm. So ^c⁄₂ = 3 cm

9. Also given: radius r = 5 cm. These details relate to ⊿OCQ in fig.b

10. We have p = √[ r² - (^c⁄₂)²]

11. Thus we get OQ = p = √[5² - 3²] = √[25 - 9] = √16 = 4

12. So the distance between the two chords = PQ = 3 + 4 = 7 cm

Part II: Chords on the same side of the diameter

• A rough sketch is shown in fig.c. Note that it is only a rough sketch. Actual construction using the given measurements is not required to solve the problem.

1. AB is the chord of 8 cm. We know that equal chords will be at equal distances from the centre. 
2. We have seen in part I of this problem that, the chord of 8 cm length is at a distance of 3 cm from the centre. 
3. So all chords of 8 cm length (in this particular circle of 5 cm radius) will be at 3 cm from the centre, what ever be the position. 
4. So we do not need to do the calculations. In fig.c, OP = 3 cm

5. Similarly, from part I, OQ = 4 cm. Thus, the distance between the two chords = PQ = OQ – OP = 4 – 3 = 1 cm

Solved example 17.8

The bottom side of the quadrilateral in the fig.17.22(a) is a diameter of the circle.

Fig.17.22

The top side is parallel to the bottom side. Calculate the area of the quadrilateral.

Solution:

1. The bottom side of the quadrilateral is a diameter. The top side is a chord.

2. Given that the top side is parallel to the bottom side. So it is a trapezium

3. Area of a trapezium = ¹⁄₂ × (top side + bottom side) × height

4. As the top and bottom sides are parallel, and the bottom side is the diameter, the 'height of the trapezium' is the perpendicular distance of the chord from the centre

5. In fig.b, the chord is named as AB, and the perpendicular distance is named as OP

6. We have ^c⁄₂ = √[r²- p²]. p is the unknown. So we must bring it to the left side. 
7. ^c⁄₂ = √[r²- p²] ⇒ (^c⁄₂)² = [r²- p²] (squaring both sides)

⇒ p² = [ r² - (^c⁄₂)²] ⇒ p = √[ r² - (^c⁄₂)²]

8. Thus we get OP = p = √[2.5² – 1.5²] = √[6.25 – 2.25] = √4 = 2

9. So from (3) we get: Area of a trapezium = ¹⁄₂ × (3 + 5) × 2= 8 sq.cm

Solved example 17.9

In a circle, two parallel chords of lengths 4 and 6 cm are 5 cm apart. What is the radius of the circle?

Solution:

A rough sketch is shown in fig.17.23. Note that it is only a rough sketch. Actual construction using the given measurements is not necessary to solve the problem.

Fig.17.23

1. In the fig., chord AB has a length of 6 cm, and chord CD has a length of 4 cm

2. The perpendicular distances are OP and OQ. We are given the total distance PQ as 5 cm

3. Let the distance OP be 'x' cm. Then OQ = (5-x) cm

4. In the upper ⊿OAP we have r² = 3² + x² ⇒ r² = 9 + x²

5. In the lower ⊿OCQ we have r² = 2² + (5-x)² ⇒ r² = 4 + 25 – 10x + x² ⇒ r² = 29 – 10x + x²

6. Both (4) and (5) gives r². So we can equate them:

7. 9 + x² = 29 – 10x + x² ⇒ 10x = 20. So x = 2

8. Substituting this value of x in (4) we get:
9. r² = 9 + 2² ⇒ r² = 9 + 4 ⇒ r² = 13. So r = √13

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In the next section, we will see 'Circles through given points'.

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Chapter 17.3 - Equal chords in a circle

In the previous section we saw some solved examples related to the centre of circles and chords. In this section, we will see 'equal chords'.

We know that the line joining two points on a circle is called a chord. When we follow this rule, a ‘diameter’ is also a chord. It is the chord which passes through the centre of the circle. In fact, it is the longest chord possible in a given circle. This is shown in the fig.11(a) below.

Fig.17.11 Diameter is the longest chord

All the other chords which do not pass through the centre, are shorter than the diameter. This is true for chords drawn in any direction, as shown in figs.17.11.a, b and c.

• From the figs., we get a feeling that chords which are situated at equal distances from the centre are equal in length. We will prove it mathematically:

• Consider the circle in fig.17.12(a) below.

Fig.17.12

1. We have two yellow lines OP and OQ. 

• They are of equal lengths

• They are drawn from the centre.

• They are drawn in any two convenient (nearly opposite) directions

2. A chord AB is drawn in such a way that it is perpendicular to OP. This is shown in fig.b [ We can draw a perpendicular to any given line using ruler and set square. Here we draw a perpendicular to OP through ‘P’. Then we extend the perpendicular both ways to meet the circle at A and B. Thus we get the chord AB]

3. Similarly, a chord CD is drawn in such a way that it is perpendicular to OQ. After completing fig.b, we can say these:

• AB is at a perpendicular distance of OP from the centre

• CD is at a perpendicular distance of OQ from the centre.

Since OP and OQ are equal, We can say:

■ The two chords AB and CD are at the same perpendicular distance from the centre.

• We do not know whether AB and CD are of the same length yet. But we will soon find out.

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It is important to note the significance of ‘perpendicular distance’. In the fig.c, OP’ and OP’’ are drawn. We could measure the distance from the centre along those lines also. But those measurements will not be ‘perpendicular distances’. Perpendicular distance is the shortest distance. For our present discussion we consider perpendicular lines OP and OQ only.

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4. Now draw OA and OC. (fig.17.12.d)

5. Consider ΔOAP. It is a right angled triangle. So applying Pythagoras theorem, we get:

AP = √[OA² - OP² ]

6. The right side of the above can be written as √[OC² - OQ² ] (∵ OA = OC, and OP = OQ)

7. But when we consider the right angled triangle OCQ, we find that √[OC² - OQ² ] = CQ

8. So we can write: AP = √[OA² - OP² ] = √[OC² - OQ² ] = CQ. ⇒ AP = CQ

9. Now, what is AP?

• We have the chord AB, and OP is the perpendicular drawn from the centre to that chord. So by theorem 17.1, AP is half of AB

• Similarly, CQ is half of CD. So from (8), we can write:

10. Half of AB = Half of CD (∵ AP = CQ)

11. So, 'full of AB' will be equal to 'full of CD'

12. That is., AB = CD

• So we proved that Chords at the same distance from the centre of the circle are equal in length. The converse can also be written:

• Chords of equal length are at equal perpendicular distances from the centre of the circle. Let us prove this also:

1. In the fig.17.13(a) below, two equal chords (in any two convenient directions) AB and CD are drawn.

Fig.17.13

• It is not difficult to draw two such chords. First, from any convenient point A, using any convenient radius, draw an arc, cutting the circle at B. Join AB. This is our first chord.

• Now, from any other convenient point ‘C’, draw an arc with the same radius AB, cutting the circle at ‘D’. Join CD. This is our second chord. AB and CD will be of the same length.

[Note that, in fig.17.12(a), we drew OP and OQ first. In fig.17.13(a), we are drawing AB and CD first.]

2. Now, draw a perpendicular from the centre ‘O’ to AB. We will call the foot of the perpendicular as ‘P’.

[Drawing a perpendicular to any given line can be done easily using ruler and set square. Here, we draw a perpendicular to AB in such a way that, the perpendicular passes through the centre ‘O’]

• Similarly draw the perpendicular OQ to CD.  This is shown in fig.17.13(b). We have to prove that OP = OQ

3. Now draw OA and OC. (fig.17.13.c)

5. Consider ΔOAP. It is a right angled triangle. So applying Pythagoras theorem, we get:

OP = √[OA² - AP² ]

6. The right side of the above can be written as √[OC² - CQ² ] (∵ OA = OC, and AP = CQ, according to theorem 17.1)

7. But when we consider the right angled triangle OCQ, we find that √[OC² - CQ² ] = OQ

8. So we can write: OP = √[OA² - AP² ] = √[OC² - CQ² ] = OQ. ⇒ OP = OQ

• So we proved that distances of 'equal chords' from the centre are the same.

We will write the above two findings as a theorem and it's converse

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Theorem 17.3:

■ Chords which are at equal distances from the centre, are of equal length

Converse of the theorem:

■ Chords which are equal in length, are at equal distances from the centre

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Based on the theorem 17.3 above, we can prove another property of chords.

• In fig.17.14(a) below, AB and CD are two equal chords. Then based on converse of theorem 17.3, the perpendicular distances OP and OQ will be equal.

Fig.17.14

• If we extend the chords AB and CD, they will meet at a point 'M' outside the circle. This is shown in fig.17.14(b). The extension beyond the circle is shown in 'dashed line' for clarity.

• In fig.(c), a diameter is drawn through 'O' in such a way that, when extended, it will pass through 'M'. Now we can begin our calculations.

1. Consider the triangle OMP. It is right angled. Applying Pythagoras theorem we get:

2. MP = √[MO² - OP² ]. In this we can put OQ instead of OP. The result will not change because OP = OQ. So we get:

3. MP = √[MO² - OQ² ]

4. Now consider triangle OMQ. It is also right angled. Applying Pythagoras theorem we get:

5. MQ = √[MO² - OQ² ]. The right side is same as that in (3). So we can write:

6. MP = √[MO² - OQ² ] = MQ. That is., MP = MQ

7. So we have an SSScongruence between ⊿OMP and ⊿OMQ (∵ OM=OM, OP=OQ, and MP = MQ)

8. Let us write the correspondence:

9. M↔M, O↔O and P↔Q. From M↔M, it is clear that the angles at M are equal. So we can write ∠OMP = ∠OMQ

• So we can write: OM bisects the angle at the point of intersection.

• In an earlier problem , we saw that that the diameter bisects the angle between 'equal chords'. In that problem, the two equal chords meet at a point on the circle.

• Now we find that the diameter will bisect the angle between two equal chords, even if the chords meet at a point outside the circle.

• In fact it is true even if the two chords meet at a point in the interior of the circle. This is shown in fig.17.15 below:

Fig.17.15

• In the fig.17.15, OP and OQ are equal. They are perpendicular to AB and CD. So AB and CD are two equal chords

• The equal chords AB and CD meet at 'M'

• The diameter (shown in red line) bisects ∠CMB and ∠AMD

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We can write a theorem based on the above discussion:

Theorem 17.4:

■ The angle at the meeting point of two equal chords is bisected by the diameter through the meeting point

The converse of this theorem can also be written:

■ If the diameter through the meeting point of two chords bisect the angle between those chords, then the chords are of equal length.

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Now we will show the proof for the converse:

• In the fig.17.16(a), any convenient chord AB is drawn. In the fig.b, a convenient diameter (shown in red colour) is drawn. Both the chord and the diameter are extended towards the left to meet at M

Fig.17.16

• Measure ∠OMB. Draw a line MD (fig.c) in such a way that MD makes the same ∠OMB with the diameter. So we have ∠DMO = ∠BMO

• Mark the point of intersection of MD with the circle as 'C'. Make the portion between M and C 'dashed'. Now CD is a chord.

• We have two chords AB and CD. We do not know whether they are equal in length. But we will soon find out.

• We can say this: The diameter bisects the angle between the two chords AB and CD (∵ ∠DMO = ∠BMO)

• Draw OP perpendicular to AB, and OQ perpendicular to CD. We can now begin the calculations

1. Consider the triangle OMP. It is right angled.

2. ∠MOP = [180 - (∠OMP + 90)] (∵ sum of interior angles in any triangle = 180^o)

⇒∠MOP = (180 -∠OMP – 90) = 90-∠OMP

3. Consider the triangle OMQ. It is right angled.

4. ∠MOQ = [180 - (∠OMQ + 90)] (∵ sum of interior angles in any triangle = 180^o)

⇒∠MOQ = (180 -∠OMQ – 90) = 90-∠OMQ = 90-∠OMP (∵ ∠OMP = ∠OMQ)

• So from (2) and (4) we get ∠MOP = ∠MOQ

• Thus, in ⊿OMP, we have:

    ♦ one side MO 

    ♦ two angles at it's ends: ∠OMP and ∠MOP.

• This same combination is present in ⊿OMQ also. It is a case of ASA congruence. Let us write the correspondence: M↔M, O↔O and P↔Q

• Using O↔O and P↔Q we can write: OP = OQ

• But OP and OQ are the perpendicular distances from the centre. If they are equal, according to theorem 17.3, the chords are also equal.So AB = CD
• Thus we have proved that, the chords are equal, and so this is the proof for the converse of theorem 17.4

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We will now see a solved example

Solved example 17.5

In the fig.17.17(a), the angle between the radii and the chords are equal. 

Fig.17.17

Prove that, the chords are equal in length.

Solution:

In fig.17.7(b), the chords are named. And also, two perpendicular lines OP and OQ are drawn. It is given that ∠OAP = ∠OCQ. We can now begin the calculations

1. Consider the triangle OAP. It is right angled.

2. ∠AOP = [180 - (∠OAP + 90)] (∵ sum of interior angles in any triangle = 180^o)

same as ∠AOP = (180 -∠OAP – 90) = 90-∠OAP

3. Consider the triangle OCQ. It is right angled.

4. ∠COQ = [180 - (∠OCQ + 90)] (∵ sum of interior angles in any triangle = 180^o)

⇒ ∠COQ = (180 -∠OCQ – 90) = 90-∠OCQ = 90-∠OAP (∵ ∠OAP = ∠OCQ)

5. So from (2) and (4) we get ∠AOP = ∠COQ

6. Thus, in ⊿OAP, we have:

    ♦ one side AO and

    ♦ two angles at it's ends: ∠OAP and ∠AOP.

7. This same combination is present in ⊿OCQ also:

    ♦ one side CO and

    ♦ two angles at it's ends: ∠OCQ and ∠COQ.

8. But the two combinations are the same as shown below:

    ♦ sides AO = CO (since radii of the same circle)

    ♦ ∠OAP = ∠OCQ (given)

    ♦ ∠AOP = ∠COQ (from (5))

9. It is a case of ASA congruence. Let us write the correspondence:

O↔O, A↔C and P↔Q

10. Using A↔C and P↔Q we can write: AP = CQ

11. But AP and CQ are ‘half chords’. Because, according to theorem 17.1, the perpendicular from the centre bisects the chord

12. So we can write: Half of AB = Half of CD

Then obviously, Full of AB = Full of CD. That is., AB = CD

Thus we have proved that, the chords are equal.

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In the next section, we will see 'Length of chords'.

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