In the previous section we saw properties related to 'equal chords'. In this section, we will see 'length of chords'.
Theorem 17.5:
Now we will see some solved
examples
Length of chords
Consider fig.17.18 below. AB is a
chord. OP is the perpendicular from the centre. OA is the radius r.
• 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:
 |
| Fig.17.18 |
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
The steps can be combined to form a single equation:
Eq.17.1:
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?
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 centreA 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
• 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.
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:
 |
| Fig.17.19 |
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
• Thus we get r = 2.5 cm
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:
Note that it is only a rough sketch. Actual construction using
the given measurements is not required to solve the problem.
 |
| Fig.17.21 |
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)
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
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.
The top side
is parallel to the bottom side. Calculate the area of the
quadrilateral.
 |
| Fig.17.22 |
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)
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.
1. In the fig., chord AB has a
length of 6 cm, and chord CD has a length of 4 cm
 |
| Fig.17.23 |
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
9. r2 = 9 + 22 ⇒ r2 = 9 + 4 ⇒ r2 = 13. So r = √13
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