In the previous section we derived the formula for the perimeter of any circle. In this section we will see some solved examples.
Solved example 21.4
A wire was bent into a circle of 4 cm diameter. What would be the diameter of a circle made by bending a wire of half the length?
Solution:
1. Perimeter of a circle of 4 cm diameter = πd = π × 4 = 4π
2. So length of the wire = 4π
3. Half length of wire = 1⁄2 × 4π = 2π
4. Let d1 be the diameter of the new circle.
5. Then perimeter of the new circle = πd1= 2π
From this we get d1 = 2 cm
Solved example 21.5
1. We are given two circles. One with diameter 2 m, and the other with diameter 3 m
2. Imagine that they are placed in such a way that, their centres are at the same point (see fig.below)
3. Then, we have:
• Perimeter of outer circle = k × perimeter of inner circle
• Where k = outer diameter⁄inner diameter.
4. So we get: k = 3⁄2.
5. Perimeter of the inner circle with 2 m radius is given to be 6.28 m
6. So Perimeter of outer circle with 3 m radius = 3⁄2 × 6.28 = 9.42 m
Solved example 21.6
Solved example 21.7
2. The diagonals will interset at the centre O of the circle
3. The diagonals will give 6 equilateral triangles
4. Consider any one of those 6 triangles. Say OAB
5. OB is half of diameter.
6. But OB = length of one side of the hexagon = 4 cm
7. So total diameter = 4×2 = 8 cm
Part (i):
1. Now, in fig.b, a square is drawn inside the same circle
Solved example 21.4
A wire was bent into a circle of 4 cm diameter. What would be the diameter of a circle made by bending a wire of half the length?
Solution:
1. Perimeter of a circle of 4 cm diameter = πd = π × 4 = 4π
2. So length of the wire = 4π
3. Half length of wire = 1⁄2 × 4π = 2π
4. Let d1 be the diameter of the new circle.
5. Then perimeter of the new circle = πd1= 2π
From this we get d1 = 2 cm
Solved example 21.5
The perimeter of a circle of diameter 2 m was measured and found out to be 6.28 m. How do we compute the perimeter of a circle of diameter 3 m, with out measuring?
Solution:1. We are given two circles. One with diameter 2 m, and the other with diameter 3 m
2. Imagine that they are placed in such a way that, their centres are at the same point (see fig.below)
3. Then, we have:
• Perimeter of outer circle = k × perimeter of inner circle
• Where k = outer diameter⁄inner diameter.
4. So we get: k = 3⁄2.
5. Perimeter of the inner circle with 2 m radius is given to be 6.28 m
6. So Perimeter of outer circle with 3 m radius = 3⁄2 × 6.28 = 9.42 m
Solved example 21.6
In the fig.21.12 below, the
circles have the same centre and the line drawn is the diameter of
the large circle.
 |
| Fig.21.12 |
What is the difference in the perimeter of the
large and small circles?
Solution:
1. It is given that the line is
the diameter of the large circle. So the line passes through the
centre of the large circle.
2. Also it is given that, the
circles have the same centre. So the line is the inner portion of the
line is the diameter of the inner circle
3. Let di and do be the diameters
of the inner and outer circles respectively.
4. Then do = (di +2) cm
5. Perimeter of the inner circle
= πd = πdi
6. Perimeter of the outer circle
= πd = πdo = π(di + 2)
7. So difference = π(di + 2) - πdi = πdi + 2π - πdi = 2π
Solved example 21.7
The perimeter of a regular
hexagon, with vertices on a circle is 24 cm
(i) What is the perimeter of a
square with vertices on this circle?
(ii) What is the perimeter of a
square with vertices on a circle with double the diameter?
(iii) What is the perimeter of an
equilateral triangle with vertices on a circle with half the diameter
of the first circle?
Solution:
Fig.21.13(a) below shows the regular
hexagon drawn inside a circle. It's perimeter is 24 cm.
So length of
one side will be = 24⁄6 = 4 cm
1. Draw the three diagonals of the hexagon |
| Fig.21.13 |
2. The diagonals will interset at the centre O of the circle
3. The diagonals will give 6 equilateral triangles
4. Consider any one of those 6 triangles. Say OAB
5. OB is half of diameter.
6. But OB = length of one side of the hexagon = 4 cm
7. So total diameter = 4×2 = 8 cm
Part (i):
1. Now, in fig.b, a square is drawn inside the same circle
2. Draw the two diagonals of the square.
3. Length of each of these diagonals are same as the diameter of the circle = 8 cm [from (7)]
4. The diagonals bisect each other at right angles. So OP = OQ = 4 cm
5. As the diagonals are perpendicular to each other, angle at O = 90o
6. Consider any of the right triangles. Say ⊿POQ. Applying Pythagoras theorem we get:
PQ2 = OP2 + OQ2
⇒ PQ2 = (4)2 + (4)2 = 16 + 16 = 32
⇒ PQ = √32 = √[16×2] = √16 × √2 = 4√2
Thus we get length of one side
7. Perimeter of the square = 4 × 4√2 = 16√2
Part (ii):
1. Now, the circle in fig.b, has a diameter of 16 cm
PQ2 = OP2 + OQ2
⇒ PQ2 = (4)2 + (4)2 = 16 + 16 = 32
⇒ PQ = √32 = √[16×2] = √16 × √2 = 4√2
Thus we get length of one side
7. Perimeter of the square = 4 × 4√2 = 16√2
Part (ii):
1. Now, the circle in fig.b, has a diameter of 16 cm
2. Draw the two diagonals of the square.
3. Length of each of these diagonals are same as the diameter of the circle = 16 cm
4. The diagonals bisect each other at right angles. So OP = OQ = 8 cm
5. As the diagonals are perpendicular to each other, angle at O = 90o
6. Consider any of the right triangles. Say ⊿POQ. Applying Pythagoras theorem we get:
PQ2 = OP2 + OQ2
⇒ PQ2 = (8)2 + (8)2 = 64 + 64 = 128
⇒ PQ = √128 = √[64×2] = √64 × √2 = 8√2
Thus we get length of one side
7. Perimeter of the square = 4 × 8√2 = 32√2
Part (iii)
In this part, we are placing an equilateral triangle inside a circle with a diameter half that of the first circle. Half of 8 is 4. So the diameter is 4 cm. This is shown in the fig.21.14 below:
We have only two information:
PQ2 = OP2 + OQ2
⇒ PQ2 = (8)2 + (8)2 = 64 + 64 = 128
⇒ PQ = √128 = √[64×2] = √64 × √2 = 8√2
Thus we get length of one side
7. Perimeter of the square = 4 × 8√2 = 32√2
Part (iii)
In this part, we are placing an equilateral triangle inside a circle with a diameter half that of the first circle. Half of 8 is 4. So the diameter is 4 cm. This is shown in the fig.21.14 below:
 |
| Fig.21.14 |
• The triangle is equilateral
• The diameter of the circle is 4 cm
With these, we have to calculate the perimeter of the triangle.
1. Let us draw two medians as in fig.b
• Median AE (E is the midpoint of BC)
• Median CD (D is the midpoint of AB)
Now look at the two medians carefully.
• The median CD is drawn from the vertex C to the midpoint D of the opposite side AB. This median will be perpendicular to the side AB. This is because ABC is an equilateral triangle. Also, CD passes through the midpoint D. So CD is the perpendicular bisector of AB
• Similarly, AE is the perpendicular bisector of BC.
So we can write:
The 'medians' in any equilateral triangle will serve another purpose also:
They will bisect the side perpendicularly. In other words, they are perpendicular bisectors also.
• We know that the point of intersection of any two perpendicular bisectors of a triangle is it’s circumcentre. (details here)
2. So O is the circumcentre of the given equilateral triangle. Let us mark the radius. OB is the radius marked with a dashed green arrow. OB = half of diameter = 1⁄2 of 4 cm = 2 cm
3. Now we use one important property of medians: The point of intersection splits the medians in the ratio 2:1, measured from the vertex. (Theorem 18.6)
4. So we can write: OC:OD = 2:1
5. That means: If we divide CD into 3 equal parts, OC will constitute 2 such parts, and OD will constitute 1 such part
6. When we divide CD into 3 equal parts, each part will be CD⁄3
7. Two such parts = 2CD⁄3
8. So OC = 2CD⁄3
9. But OC = radius = 2 cm
10. Substituting this in (8) we get: 2 cm = 2CD⁄3. So we get CD = 3 cm
11. Now, AD = AB⁄2. But AB = AC (∵ ABC is an equilateral triangle)
12. So AD = AC⁄2
13. Now consider the right triangle ADC
14. Applying Pythagoras theorem, we get: AC2 = AD2 + CD2.
⇒ AC2 = (AC⁄2)2 + (3)2. ⇒ AC2 = (AC2⁄4) + 9
⇒ AC2 - (AC2⁄4) = 9 ⇒ (3AC2⁄4) = 9
⇒ AC2 = 12 ⇒ AC = √[4 × 3] = √4 × √3 = 2√3
15. Thus we got one side of the equilateral triangle.
So perimeter = 3 × 2√3 = 6√3
⇒ AC2 - (AC2⁄4) = 9 ⇒ (3AC2⁄4) = 9
⇒ AC2 = 12 ⇒ AC = √[4 × 3] = √4 × √3 = 2√3
15. Thus we got one side of the equilateral triangle.
So perimeter = 3 × 2√3 = 6√3
Solved example 21.8
In the figs.21.15(a), (b) and (c) below, a regular
hexagon, square and a rectangle are drawn with their vertices on a
circle.
Calculate the perimeter of each circle
Solution:
Part (i): Regular hexagon
1. In fig.21.16(a) below, it's three diagonals can be drawn, giving 6 inner triangles.
3. OB is half of diameter.
4. But OB = length of one side of the hexagon = 2 cm
5. So total diameter = 2×2 = 4 cm
6. So perimeter of the circle = πd = π × 4 = 4π cm
Part (ii): Square
1. In fig.21.16(b), it's two diagonals are drawn
 |
| Fig.21.15 |
Solution:
Part (i): Regular hexagon
1. In fig.21.16(a) below, it's three diagonals can be drawn, giving 6 inner triangles.
 |
| Fig.21.16 |
)
2. All the 6 inner triangles are equilateral, and are also equal. So we can consider any one of them. Say OAB3. OB is half of diameter.
4. But OB = length of one side of the hexagon = 2 cm
5. So total diameter = 2×2 = 4 cm
6. So perimeter of the circle = πd = π × 4 = 4π cm
Part (ii): Square
1. In fig.21.16(b), it's two diagonals are drawn
2. Length of each of these diagonals are same as the diameter of the circle. Let this be equal to 'd' cm
3. The diagonals bisect each other at right angles. So OP = OQ = d⁄2 cm
4. As the diagonals are perpendicular to each other, angle at O = 90o
5. Consider any of the right triangles. Say ⊿POQ. Applying Pythagoras theorem we get:
PQ2 = OP2 + OQ2
⇒ PQ2 = (d⁄2)2 + (d⁄2)2 = d2⁄4 + d2⁄4 = 2d2⁄4 = d2⁄2.
⇒ 22 = d2⁄2. ⇒ 4 = d2⁄2 ⇒ d2 = 8
⇒ d = √8 = √[4×2] = √4 × √2 = 2√2
Thus we get diameter of the circle as 2√2 cm
6. So perimeter of the circle = πd = π × 2√2 = 2π√2 cm
Part(iii): Rectangle
1. In fig.21.16(c), One of it's diagonal is drawn. This diagonal VW is a diameter of the circle.
2. From the right triangle UVW, we get: VW2 = UV2 + UW2
⇒ VW2 = 22 + 1.52 = 4 + 2.25 = 6.25
⇒ VW = √6.25 = 2.5
3. Thus we get the diameter as 2.5 cm
4. So perimeter of the circle = πd = π × 2.5 = 2.5π cm
Solved example 21.9
An isosceles triangle with it's vertices on a circle is shown in the fig.21.17(a) below.
What is the
perimeter of the circle?
Solution:
1. In fig.b, the triangle is named as ABC. AC and BC are the equal sides.
2. Let 'O' be the centre of the circle, and let 'x' be the radius
3. Then OB = OC = x
4. From the right triangle OBD, we get: OB2 = BD2 + OD2.
⇒ x2 = 22 + (4-x)2 ⇒ x2 = 22 + 16 -8x + x2
⇒ 8x = 16 + 4 ⇒ 8x = 20 ⇒ x = 20⁄8 = 2.5 cm
5. Thus we get the radius as 2.5 cm
6. So perimeter of the circle = 2πr = 2 × π × 2.5 = 5π cm
PQ2 = OP2 + OQ2
⇒ PQ2 = (d⁄2)2 + (d⁄2)2 = d2⁄4 + d2⁄4 = 2d2⁄4 = d2⁄2.
⇒ 22 = d2⁄2. ⇒ 4 = d2⁄2 ⇒ d2 = 8
⇒ d = √8 = √[4×2] = √4 × √2 = 2√2
Thus we get diameter of the circle as 2√2 cm
6. So perimeter of the circle = πd = π × 2√2 = 2π√2 cm
Part(iii): Rectangle
1. In fig.21.16(c), One of it's diagonal is drawn. This diagonal VW is a diameter of the circle.
2. From the right triangle UVW, we get: VW2 = UV2 + UW2
⇒ VW2 = 22 + 1.52 = 4 + 2.25 = 6.25
⇒ VW = √6.25 = 2.5
3. Thus we get the diameter as 2.5 cm
4. So perimeter of the circle = πd = π × 2.5 = 2.5π cm
Solved example 21.9
An isosceles triangle with it's vertices on a circle is shown in the fig.21.17(a) below.
 |
| Fig.21.17 |
Solution:
1. In fig.b, the triangle is named as ABC. AC and BC are the equal sides.
2. Let 'O' be the centre of the circle, and let 'x' be the radius
3. Then OB = OC = x
4. From the right triangle OBD, we get: OB2 = BD2 + OD2.
⇒ x2 = 22 + (4-x)2 ⇒ x2 = 22 + 16 -8x + x2
⇒ 8x = 16 + 4 ⇒ 8x = 20 ⇒ x = 20⁄8 = 2.5 cm
5. Thus we get the radius as 2.5 cm
6. So perimeter of the circle = 2πr = 2 × π × 2.5 = 5π cm
Solved example 21.10
In all the figs.21.18(a), (b) and (c) below, the
centres of the circles are on the same line.
In the figs (a) and (b),
the small circles are of the same diameter. Prove that, in all the
figs, the perimeter of the large circle is the sum of the perimeters
of the small circles
Solution:
Case 1:
 |
| Fig.21.18 |
Solution:
Case 1:
1. Let the diameters of the two
inner circles be di
2. Let the diameter of the outer
circle be do
3. Then we get do = 2di
4. Perimeter of each of the inner
circles = πdi
5. Sum of the perimeters of the
two inner circles = πdi + πdi = 2πdi
6. Perimeter of the outer circle
= πdo.
7. But from (3), we have do = 2di .
8. So from (6) we get: Perimeter of the outer circle = πdo. = π × 2di = 2πdi.
9. (5) and (8) are equal. So we can write: The perimeter of the large circle is the sum of the perimeters of the two small circles
Case 2:
1. Let the diameters of the three inner circles be di
2. Let the diameter of the outer circle be do
3. Then we get do = 3di
4. Perimeter of each of the inner circles = πdi
5. Sum of the perimeters of the two inner circles = πdi + πdi + πdi = 3πdi
6. Perimeter of the outer circle = πdo.
7. But from (3), we have do = 3di .
8. So from (6) we get: Perimeter of the outer circle = πdo. = π × 3di = 3πdi.
9. (5) and (8) are equal. So we can write: The perimeter of the large circle is the sum of the perimeters of the three small circles
Case 3:
1. Let the diameters of the three inner circles be d1, d2 and d3
2. Let the diameter of the outer circle be do
2. Let the diameter of the outer circle be do
3. Then we get do = d1 + d2 + d3
4. Perimeter of the inner circles will be πd1, πd2 and πd3
5. Sum of the perimeters of the three inner circles = πd1 + πd2 + πd3
6. Perimeter of the outer circle = πdo.
7. But from (3), we have do = d1 + d2 + d3
8. So from (6) we get: Perimeter of the outer circle = πdo. = π × (d1 + d2 + d3 ) = πd1 + πd2 + πd3
9. (5) and (8) are equal. So we can write: The perimeter of the large circle is the sum of the perimeters of the three small circles
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