In the previous section we have seen the central angle of an arc for each 1 cm length. In this section we will see some solved examples.
Solved example 21.16
In a circle, the length of an
arc is 3π cm. The central angle of this arc is 40o. What is the
perimeter of the circle? What is it's radius?
Solution:
Solution:
1. Both values of the arc is given to us:
• Length of arc = 3π cm
• Central angle of arc = 40o
2. We need an equation which gives the relation between the two. We can use theorem 21.1
According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm
3. So for 40o, the length of arc will be (πr⁄180) × 40 = (πr⁄4.5) cm
4. We can equate it to the given length. So we get: (πr⁄4.5) = 3π cm ⇒ r = 3 × 4.5 = 13.5 cm
5. Perimeter = 2πr = 2 × 13.5 × π = 27π cm
Solved example 21.17
In a circle, the length of an arc is 4 cm. It's central angle is 25o
(i) In the same circle, what is the length of an arc whose central angle is 75o?
(ii) In a circle of radius one and a half times the radius of this circle, what is the length of an arc whose central angle is 75o?
Solution:
1. Both values of the arc is given to us:
• Length of arc = 4 cm
• Central angle of arc = 25o
2. We need an equation which gives the relation between the two. We can use theorem 21.1
According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm
3. So for 25o, the length of arc will be (πr⁄180) × 25 = (πr⁄7.2) cm
4. We can equate it to the given length. So we get: (πr⁄7.2) = 4 cm ⇒ πr = 4 × 7.2 = 28.8 cm
• We need not divide 28.8 by π to get the actual radius because, we will be using 'πr' as a whole in our calculations.
Part (i)
1. We use theorem 21.1 again:
For 75o, the length of arc = (πr⁄180) × 75 = (28.8⁄180) × 75 = 12 cm
Part (ii)
1. The radius is one and a half times. So new πr = 28.8 ×1.5 = 43.2 cm
For 75o, the length of arc = (πr⁄180) × 75 = (43.2⁄180) × 75 = 18 cm
Solved example 21.18
From a bangle of radius 3 cm, a piece is to be cut out to make a ring of radius 1⁄2 cm.
(i) What is the central angle of the piece to be cut out?
(ii) The remaining part of the bangle was bent to make a smaller bangle. What is it's radius?
Solution:
1. When a piece is cut out from a bangle, it would be an arc. What is the length of this arc?
Ans: Enough to make a ring of radius 1⁄2 cm
2. So, if r is the radius of the ring, it's perimeter 2πr must be equal to the length of the arc cut out
3. But radius of the ring is given as 1⁄2 cm. So length of cut out arc = 2π × 1⁄2 = π cm
4. So we have the length of arc. From that, we need to find the central angle. We can use theorem 21.2.
5. For every 1 cm of an arc on a circle of radius r, the central angle will be (180⁄πr)o
6. So for π cm, the angle will be (180⁄πr) × π = (180⁄r)o .
7. But r is given as 3 cm. So angle = (180⁄3) = 60o. This is the answer for part (i)
8. π cm is cut out. So remaining arc length = 2πr - π = (2r-1)π = (2×3 - 1)π = 5π
9. So the smaller bangle is made using an arc of length 5π. That means, perimeter of the smaller bangle = 5π
10. Let r1 be the radius of the smaller bangle. Then it's perimeter = 2πr1
11. So we get: 2πr1= 5π ⇒ r1 = 2.5 cm. This is the answer for part (ii)
7. So for 60o, the length of arc will be (π×4⁄180) × 60 = (4⁄3)π cm
8. Thus the total perimeter = 3×(4⁄3)π = 4π cm
Solved example 21.20
• Length of arc = 3π cm
• Central angle of arc = 40o
2. We need an equation which gives the relation between the two. We can use theorem 21.1
According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm
3. So for 40o, the length of arc will be (πr⁄180) × 40 = (πr⁄4.5) cm
4. We can equate it to the given length. So we get: (πr⁄4.5) = 3π cm ⇒ r = 3 × 4.5 = 13.5 cm
5. Perimeter = 2πr = 2 × 13.5 × π = 27π cm
Solved example 21.17
In a circle, the length of an arc is 4 cm. It's central angle is 25o
(i) In the same circle, what is the length of an arc whose central angle is 75o?
(ii) In a circle of radius one and a half times the radius of this circle, what is the length of an arc whose central angle is 75o?
Solution:
1. Both values of the arc is given to us:
• Length of arc = 4 cm
• Central angle of arc = 25o
2. We need an equation which gives the relation between the two. We can use theorem 21.1
According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm
3. So for 25o, the length of arc will be (πr⁄180) × 25 = (πr⁄7.2) cm
4. We can equate it to the given length. So we get: (πr⁄7.2) = 4 cm ⇒ πr = 4 × 7.2 = 28.8 cm
• We need not divide 28.8 by π to get the actual radius because, we will be using 'πr' as a whole in our calculations.
Part (i)
1. We use theorem 21.1 again:
For 75o, the length of arc = (πr⁄180) × 75 = (28.8⁄180) × 75 = 12 cm
Part (ii)
1. The radius is one and a half times. So new πr = 28.8 ×1.5 = 43.2 cm
For 75o, the length of arc = (πr⁄180) × 75 = (43.2⁄180) × 75 = 18 cm
Solved example 21.18
From a bangle of radius 3 cm, a piece is to be cut out to make a ring of radius 1⁄2 cm.
(i) What is the central angle of the piece to be cut out?
(ii) The remaining part of the bangle was bent to make a smaller bangle. What is it's radius?
Solution:
1. When a piece is cut out from a bangle, it would be an arc. What is the length of this arc?
Ans: Enough to make a ring of radius 1⁄2 cm
2. So, if r is the radius of the ring, it's perimeter 2πr must be equal to the length of the arc cut out
3. But radius of the ring is given as 1⁄2 cm. So length of cut out arc = 2π × 1⁄2 = π cm
4. So we have the length of arc. From that, we need to find the central angle. We can use theorem 21.2.
5. For every 1 cm of an arc on a circle of radius r, the central angle will be (180⁄πr)o
6. So for π cm, the angle will be (180⁄πr) × π = (180⁄r)o .
7. But r is given as 3 cm. So angle = (180⁄3) = 60o. This is the answer for part (i)
8. π cm is cut out. So remaining arc length = 2πr - π = (2r-1)π = (2×3 - 1)π = 5π
9. So the smaller bangle is made using an arc of length 5π. That means, perimeter of the smaller bangle = 5π
10. Let r1 be the radius of the smaller bangle. Then it's perimeter = 2πr1
11. So we get: 2πr1= 5π ⇒ r1 = 2.5 cm. This is the answer for part (ii)
Solved example 21.19
In fig.21.33(a) below, parts
of a circle centred at each vertex of an equilateral triangle, and
passing through the other two vertices is shown.
What is the
perimeter of this fig.?
 |
| Fig.21.33 |
Solution:
1. In fig.21.33(b), more details are added. ABC is the equilateral triangle.
2. A circle is drawn centred at each vertex. What is the radius of those circles?
Ans: Any of the above three circles centred on a vertex, passes through the other two vertices. For example, in the fig.b, the circle centred at A passes through B and C.
Also, it is an equilateral triangle. So radius of any circle is the side of the triangle, and is equal to 4 cm.
3. An arc is taken out between the 'other two vertices', from each circle.
4. For an equilateral triangle, all three angles are 60o
5. So we have 3 equal arcs
• Each of them have a radius of 4 cm. (∵ they are part of a circle with 4 cm radius)
• Each of them have central angle 60o.
According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm 7. So for 60o, the length of arc will be (π×4⁄180) × 60 = (4⁄3)π cm
8. Thus the total perimeter = 3×(4⁄3)π = 4π cm
Solved example 21.20
Parts of a circle are drawn,
centred at each vertex of a regular octagon, and a fig. is cut out as
in fig.21.34(b) below. Calculate the perimeter of fig.b
Solution:
1. The measure of each interior angle of a regular polygon is ([180(n-2)]⁄n). Where n is the number of sides.
2. We are given a regular octagon. It has 8 sides. So each interior angle is equal to:
([180(8-2)]⁄8) = (180×6⁄8) = 135o.
3. We want to find the perimeter in fig.b. It consists of equal arcs. How are these arcs formed?
We get the answer from fig.c
Equal circles are centred at each vertex. Then the portions outside the octagon are removed.
Radius of each circle is 1 cm
6. So for 135o, the length of arc will be (π×1⁄180) × 135 = (3⁄4)π cm
7. Thus the total perimeter = 8×(3⁄4)π = 6π cm
 |
| Fig.21.34 |
1. The measure of each interior angle of a regular polygon is ([180(n-2)]⁄n). Where n is the number of sides.
2. We are given a regular octagon. It has 8 sides. So each interior angle is equal to:
([180(8-2)]⁄8) = (180×6⁄8) = 135o.
3. We want to find the perimeter in fig.b. It consists of equal arcs. How are these arcs formed?
We get the answer from fig.c
Equal circles are centred at each vertex. Then the portions outside the octagon are removed.
Radius of each circle is 1 cm
4. So we have 8 equal arcs
• Each of them have a radius of 1 cm. (∵ they are part of a circle with 1 cm radius)
• Each of them have central angle 135o.
According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm 6. So for 135o, the length of arc will be (π×1⁄180) × 135 = (3⁄4)π cm
7. Thus the total perimeter = 8×(3⁄4)π = 6π cm
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