In the previous section we completed the discussion on some of the properties of circles. We also saw some solved examples. In this section, we will see some solved examples that are related to the topics that we discussed so far in this chapter in general.
Solved example 17.13
Solved example 17.13
Find the radius of the circle
shown in fig.17.30(a). AB is 24 cm long. CD is the perpendicular bisector of
AB, and it is 8 cm long.
 |
| Fig.17.30 |
Solution:
1. CD is the perpendicular
bisector of the chord AB. So if it is extended inwards, it will meet
the centre ‘O’ of the circle. (See theorem 17.1)
2. Let the radius OB = r cm
3. Consider ⊿OBC. Applying
Pythagoras theorem, we get: r2-122 = x2
4. Also we have OD = r = 8+x
5. Substituting this value of
r in (3), we get: (8+x)2-122 = x2 ⇒ 64 + 16x + x2 -122 = x2
⇒ 64 + 16x - 144 = 0 ⇒ 16x = 144 – 64 = 80. So x = 80/16 = 5
6. Substituting this value of
x in (4), we get: r = 8 + 5 = 13 cm
Solved example 17.14
In the fig.17.31(a), AB is the
diameter of the circle. PQ is parallel to AB.
 |
| Fig.17.31 |
If AB = 50 cm and PQ =
14 cm, find BQ
Solution:
1. AB is the diameter. So
radius r = 25 cm. PQ is a chord. It is shown in magenta colour in
fig.b.
2. Perpendicular distance of
PQ from O can be obtained by rearranging Eq.17.1: p = √[r2-(c⁄2)2]
3. So we get: OD = p =
√[252-72] = √[625-49] = √576 = 24
4. QR is drawn perpendicular
to AB.
5. QR = OD = p = 24 cm (∵ AB
and PQ are parallel)
6. Consider ⊿QRB.
• QR = p [from (5)]
• RB = OB – OR = 25 – 7
= 18
7. Applying Pythagoras theorem
we get: BQ = √[QR2 + RB2] = √[242 +
182] = √[576 + 324] = √[900] = 30
Solved example 17.15
In the fig.17.32(a), OE = 6 cm and ED
= 15 cm. Find the radius of the circle
 |
| Fig.17.32 |
Solution:
1. BD is a chord, and OE is a
perpendicular drawn from the centre to this chord
3. Thus BE = DE = 15 cm (∵
it is given that BE = 15 cm)
4. Consider ⊿OBE. Applying
Pythagoras theorem we get: OB = √[OE2 + BE2]
= √[62 + 152]
= √[36 + 225] =
√[261] = 16.2 cm
Solved example 17.16
From a circular disc of radius
17 cm, a portion in the shape of a trapezium is cutoff. Parallel
sides are 30 cm and 16 cm. Both the parallel sides are on the same
side of the centre of the circle. Find the area of the trapezium.
Solution:
1. Fig.17.33(a) shows the given data.
 |
| Fig.17.33 |
2. In
fig.b, the trapezium is named as ABCD. The parallel sides AB and CD
become two chords of the circle. Radius r of the circle = 17 cm
3. OP and OQ are the
perpendicular distances of the two chords from the centre. They can
be obtained rearranging Eq.17.1:
p = √[r2-(c⁄2)2]
4. So we get OP = √[172-152]
= √[289-225] = √64 = 8
5. OQ = √[172-82]
= √[289-64] = √225 = 15
7. Here height of trapezium = PQ = OQ
– OP = 15 – 8 = 7
8. So area = 1⁄2 × (30 + 16) × 7 = 161 sq.cm
Solved example 17.17
Three points A, B and C are situated at equal distances from each other, along the periphery of a circle. The radius of this circle is 20 m. What is the distance between the points?
Solution:
1. Given that the distances between the points are equal. So they form the vertices of an equilateral triangle.
2. Also, the three points are on the periphery of a circle. So the circle is the circumcircle.
3. We have seen the relation between the side of an equilateral triangle and the radius of it's circumcircle. The relation is S = r√3. Where S is the side and r is the circumradius
4. So we get: Distance between the points = S = 20√3 = 20 × 1.732 = 34.64 m
5. The actual construction is shown in fig.17.34 below:
Solved example 17.17
Three points A, B and C are situated at equal distances from each other, along the periphery of a circle. The radius of this circle is 20 m. What is the distance between the points?
Solution:
1. Given that the distances between the points are equal. So they form the vertices of an equilateral triangle.
2. Also, the three points are on the periphery of a circle. So the circle is the circumcircle.
3. We have seen the relation between the side of an equilateral triangle and the radius of it's circumcircle. The relation is S = r√3. Where S is the side and r is the circumradius
4. So we get: Distance between the points = S = 20√3 = 20 × 1.732 = 34.64 m
5. The actual construction is shown in fig.17.34 below:
 |
| Fig.17.34 |
Solved example 17.18 Solved example 17.19 Solved example 17.20
In the next section, we will see 'Parallel lines'.
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