In the previous section we completed the discussion on Arithmetic progressions. In this chapter we will learn about Circles.
We have learned about circles in a previous chapter 21.
• A circle is a collection of points. All those points are at equal distances from a fixed point.
♦ The 'equal distances' is the radius of the circle.
♦ The fixed point is the centre of the circle.
• We have also learned about chord, segment, sector, arc etc.,
In this chapter, we will learn about angles related to circles.
1. Consider a line AB shown in red colour in fig.27.1(a) below:
2. Length of AB is not important for our present discussion. What we want is a triangle with AB as hypotenuse.
3. It is very easy to draw such a triangle. We know that, the angle at the corner C, which is opposite the hypotenuse AB will be always equal to 90o.
4. Let the angle at A be xo. And let the angle at B be yo. Since ABC is right angled, y will always be equal to (90-x). For example, if x = 30o, y will be equal to 60o.
5. So, to draw the required triangle:
• choose any convenient value for x. Draw a line at this angle at A.
• Calculate y. Draw a line at this angle at B.
• The two lines will intersect at C.
■ Thus we obtained the required triangle with AB as hypotenuse.
2. For each value of x, we calculate the corresponding y.
3. For each set of (x,y) we can draw a triangle. All of those triangles will have AB as the hypotenuse. 4. Fig(b) shows 7 such triangles. Note that, in the fig(b), only 7 triangles are shown. A larger number is avoided for the purpose of clarity. The reader may draw a larger number of triangles in the note book.
5. Now we draw a smooth curve joining all the top corners and also the ends A and B. It is shown in magenta colour in fig.27.1(c). The smooth curve resembles a semi circle.
[We can check this using a compass. First draw the perpendicular bisector of AB and mark it's midpoint. With this midpoint as centre and half of AB as radius, draw a semi circle. It will pass through all the top corners of the triangles]
6. We obtained the semi circle above the line AB. The triangles can be drawn below AB also. This is shown in fig 27.2(b) below:
7. Connect all the points as in the previous case, and a semi circle can be obtained below AB also. This is shown in fig.27.2(c)
8. Thus we get a full circle as shown in fig.27.3(a) below:
1. AB is a diameter of a circle.
2. Mark any point C on the circle
3. Draw lines AC and BC
4. Then the angle between AC and BC will be 90o
This is shown in fig.27.3(b) above.
2. A line OC is drawn in yellow colour. OC splits the original ⊿ABC into two: ΔAOC and ΔBOC
3. Consider ΔAOC.
• OA = OC (∵ radii of the same circle)
• So ΔOAC is an isosceles triangle. It's base angles will be equal
• Thus ∠CAO = ∠OCA = xo
4. Consider ΔBOC.
• OB = OC (∵ radii of the same circle)
• So ΔBOC is an isosceles triangle. It's base angles will be equal
• Thus ∠CBO = ∠OCB = yo
5. ∠ACB = ∠ACO + ∠BCO = x + y
6. Now consider ⊿ABC as a whole.
• We have: sum of interior angles = x + y + y + x = 2x + 2y = 2(x+y)o
• But sum of the interior angles of any triangle is 180o
• So we can write: 2(x+y) = 180 ⇒ (x+y) = 90o
7. But from (5), (x+y) = ∠ACB.
8. Thus it is proved that angle at C is 90o. Theorem 27.1 is also proved.
Let us write a summary of the above discussion:
• IF the ends A and B of a diameter of the circle becomes two corners of a triangle AND
• The third corner lies any where on the circle, THEN
• The angle at the third corner will be 90o.
■ Now we are going to consider another case:
• The ends A and B of a diameter of the circle becomes two corners of a triangle (same as above) AND
• The third corner lies any where on the interior of the circle
Such a case is shown in fig.27.4(a) below:
• We want to know the peculiarity of the angle at P. For finding that, we do the following steps:
1. Extend AP until it meets the circle at Q. This is shown in fig(b)
2. Draw BQ.
3. Now consider the new ⊿PBQ. Based on theorem 27.1, we have: ∠AQB = 90
4. ∠APB is an exterior angle of triangle PBQ
• Exterior angle = sum of remote interior angles
5. So we get: ∠APB = ∠PQB + ∠PBQ
⇒ ∠APB = 90o + ∠PBQ
6. Consider the right side of the above equation.
• A quantity '∠PBQ' is added to 90o. The sum will obviously be greater than 90
• That means, the left side ∠APB will be greater than 90o
This is a very use full result. We will write it as a theorem. We will write it in steps:
Theorem 27.2:
1. AB is a diameter of a circle.
2. Mark any point P on the interior of the circle
3. Draw lines AP and BP
4. Then the angle between AP and BP, that is., ∠APB will be greater than 90o
■ Next we will consider the case when P is outside the circle. Let us write the specifications:
• The ends A and B of a diameter of the circle becomes two corners of a triangle (same as above) AND
• The third corner lies any where on the exterior of the circle
Such a case is shown in fig.27.5(a) below:
• We want to know the peculiarity of the angle at P. For finding that, we do the following steps:
1. Mark the point of intersection of AP and the circle. Let this point be Q. This is shown in fig(b)
2. We know that, based on theorem 27.1, ∠AQB = 90o
3. This ∠AQB is an exterior angle of BPQ.
• Exterior angle = sum of remote interior angles
4. So we can write:
∠AQB = ∠QPB + ∠QBP
⇒ 90 = ∠QPB + ∠QBP
5. Consider the right side of the above equation:
• A quantity '∠QBP' is added to the angle at P
• Only when such an addition takes place, we get 90o
• That means, angle at P is less than 90o
This is also a very useful result. We will write it as a theorem. We will write it in steps:
Theorem 27.3:
1. AB is a diameter of a circle.
2. Mark any point P on the exterior of the circle
3. Draw lines AP and BP
4. Then the angle between AP and BP, that is., ∠APB will be less than 90o.
We have learned about circles in a previous chapter 21.
• A circle is a collection of points. All those points are at equal distances from a fixed point.
♦ The 'equal distances' is the radius of the circle.
♦ The fixed point is the centre of the circle.
• We have also learned about chord, segment, sector, arc etc.,
In this chapter, we will learn about angles related to circles.
1. Consider a line AB shown in red colour in fig.27.1(a) below:
 |
| Fig.27.1 |
3. It is very easy to draw such a triangle. We know that, the angle at the corner C, which is opposite the hypotenuse AB will be always equal to 90o.
4. Let the angle at A be xo. And let the angle at B be yo. Since ABC is right angled, y will always be equal to (90-x). For example, if x = 30o, y will be equal to 60o.
5. So, to draw the required triangle:
• choose any convenient value for x. Draw a line at this angle at A.
• Calculate y. Draw a line at this angle at B.
• The two lines will intersect at C.
■ Thus we obtained the required triangle with AB as hypotenuse.
Now we want a large number of such triangles. That is., we want a large number of triangles and all of them must have the same AB as the hypotenuse.
1. This task is also easy. We choose different values of x. 2. For each value of x, we calculate the corresponding y.
3. For each set of (x,y) we can draw a triangle. All of those triangles will have AB as the hypotenuse. 4. Fig(b) shows 7 such triangles. Note that, in the fig(b), only 7 triangles are shown. A larger number is avoided for the purpose of clarity. The reader may draw a larger number of triangles in the note book.
5. Now we draw a smooth curve joining all the top corners and also the ends A and B. It is shown in magenta colour in fig.27.1(c). The smooth curve resembles a semi circle.
[We can check this using a compass. First draw the perpendicular bisector of AB and mark it's midpoint. With this midpoint as centre and half of AB as radius, draw a semi circle. It will pass through all the top corners of the triangles]
6. We obtained the semi circle above the line AB. The triangles can be drawn below AB also. This is shown in fig 27.2(b) below:
 |
| Fig.27.2 |
8. Thus we get a full circle as shown in fig.27.3(a) below:
 |
| Fig.27.3 |
Considering the full circle in fig.27.3(a), we can write a theorem. We will write it in steps:
Theorem 27.1:1. AB is a diameter of a circle.
2. Mark any point C on the circle
3. Draw lines AC and BC
4. Then the angle between AC and BC will be 90o
This is shown in fig.27.3(b) above.
Now we have to write the proof:
1. Consider fig.27.3(c) above. We have to prove that ∠ACB = 90o2. A line OC is drawn in yellow colour. OC splits the original ⊿ABC into two: ΔAOC and ΔBOC
3. Consider ΔAOC.
• OA = OC (∵ radii of the same circle)
• So ΔOAC is an isosceles triangle. It's base angles will be equal
• Thus ∠CAO = ∠OCA = xo
4. Consider ΔBOC.
• OB = OC (∵ radii of the same circle)
• So ΔBOC is an isosceles triangle. It's base angles will be equal
• Thus ∠CBO = ∠OCB = yo
5. ∠ACB = ∠ACO + ∠BCO = x + y
6. Now consider ⊿ABC as a whole.
• We have: sum of interior angles = x + y + y + x = 2x + 2y = 2(x+y)o
• But sum of the interior angles of any triangle is 180o
• So we can write: 2(x+y) = 180 ⇒ (x+y) = 90o
7. But from (5), (x+y) = ∠ACB.
8. Thus it is proved that angle at C is 90o. Theorem 27.1 is also proved.
Let us write a summary of the above discussion:
• IF the ends A and B of a diameter of the circle becomes two corners of a triangle AND
• The third corner lies any where on the circle, THEN
• The angle at the third corner will be 90o.
■ Now we are going to consider another case:
• The ends A and B of a diameter of the circle becomes two corners of a triangle (same as above) AND
• The third corner lies any where on the interior of the circle
Such a case is shown in fig.27.4(a) below:
 |
| Fig.27.4 |
1. Extend AP until it meets the circle at Q. This is shown in fig(b)
2. Draw BQ.
3. Now consider the new ⊿PBQ. Based on theorem 27.1, we have: ∠AQB = 90
4. ∠APB is an exterior angle of triangle PBQ
• Exterior angle = sum of remote interior angles
5. So we get: ∠APB = ∠PQB + ∠PBQ
⇒ ∠APB = 90o + ∠PBQ
6. Consider the right side of the above equation.
• A quantity '∠PBQ' is added to 90o. The sum will obviously be greater than 90
• That means, the left side ∠APB will be greater than 90o
This is a very use full result. We will write it as a theorem. We will write it in steps:
Theorem 27.2:
1. AB is a diameter of a circle.
2. Mark any point P on the interior of the circle
3. Draw lines AP and BP
4. Then the angle between AP and BP, that is., ∠APB will be greater than 90o
■ Next we will consider the case when P is outside the circle. Let us write the specifications:
• The ends A and B of a diameter of the circle becomes two corners of a triangle (same as above) AND
• The third corner lies any where on the exterior of the circle
Such a case is shown in fig.27.5(a) below:
 |
| Fig.27.5 |
1. Mark the point of intersection of AP and the circle. Let this point be Q. This is shown in fig(b)
2. We know that, based on theorem 27.1, ∠AQB = 90o
3. This ∠AQB is an exterior angle of BPQ.
• Exterior angle = sum of remote interior angles
4. So we can write:
∠AQB = ∠QPB + ∠QBP
⇒ 90 = ∠QPB + ∠QBP
5. Consider the right side of the above equation:
• A quantity '∠QBP' is added to the angle at P
• Only when such an addition takes place, we get 90o
• That means, angle at P is less than 90o
This is also a very useful result. We will write it as a theorem. We will write it in steps:
Theorem 27.3:
1. AB is a diameter of a circle.
2. Mark any point P on the exterior of the circle
3. Draw lines AP and BP
4. Then the angle between AP and BP, that is., ∠APB will be less than 90o.
Now we will see a solved example.
Solved example 27.1
In ΔABC, ∠A = 60o and ∠B = 70o. Is the vertex C inside or outside the circle with AB as diameter?
Solution:
We will first draw a rough sketch as shown in the fig.27.6(a) below:
1. AB is the diameter. A triangle ABC is formed with ∠A = 60o and ∠B = 70o.
2. Then ∠C = 180 -(60 + 70) = 180 - 130 = 50o
3. 50o < 90o. So, based on theorem 27.3, the vertex C will lie outside the circle.
Note: In this problem, actual construction is not required. However, an actual construction is shown in fig.b. It shows that our calculations are correct.
In the next section, we will see the cases when AB is not a diameter.
Solved example 27.1
In ΔABC, ∠A = 60o and ∠B = 70o. Is the vertex C inside or outside the circle with AB as diameter?
Solution:
We will first draw a rough sketch as shown in the fig.27.6(a) below:
 |
| Fig.27.6 |
2. Then ∠C = 180 -(60 + 70) = 180 - 130 = 50o
3. 50o < 90o. So, based on theorem 27.3, the vertex C will lie outside the circle.
Note: In this problem, actual construction is not required. However, an actual construction is shown in fig.b. It shows that our calculations are correct.
In the next section, we will see the cases when AB is not a diameter.
No comments:
Post a Comment