Math Teaching on Properties of Circles

Thursday, July 23, 2009

More Tryouts on Properties of Circles

You may go to the Websites below to try out more Questions:
- http://www.justeducation.com/samples/Sample%20of%20EMaths%20Paper.pdf
- http://www.mathfair.com/puzzles.html

You can also go to the Websites below to look at the Videos:
- http://www.youtube.com/watch?v=91GjgQleAoA
- http://www.youtube.com/view_play_list?p=072C32B90D3B1388&search_query=Properties+of+Circles

# posted by Properties of Circles at 4:08:00 AM

Pictures of Circles

These are some Pictures of Circles:

- They can be use as Artefacts.

- They can be use as Floor Designs.

- They can be use in Nature.

- They can be use as Mathematical Instruments.

# posted by Properties of Circles at 3:47:00 AM

Tuesday, July 21, 2009

Real- Life Applications

Let us take a look at some of the Real- Life Applications where Circles are being used:
- Real- Life Application One

- This Picture shows a Ferris- Wheel.

- A ferris wheel is basically shaped like a wheel.
- Since most of these amusement rides are fairly large, the outer circle is not made from one piece of steel but rather several little arcs.
- The number of arcs is determines by the number of cars on the wheel.

Question:
- How long would the arc be between the cars if there were 12 cars on a 30-meter tall ferris wheel?

- Real- Life Application Two

- This is a Basketball Court Orientation.

- The geometry of a basketball court is a great study in areas and circles.

- There are three 12- foot circles in a Basketball Court:

- One lies at the center of the court and the other two circles lie at each of the free throw lines.

- The inner circle at center court has a 4- foot diameter.

- Here is a diagram.

- This is a Diagram of a 4- foot diameter centre court.

Question:

- How much paint would be needed to paint all of the areas shown in green colour if one gallon could only covers 130 square feets?

# posted by Properties of Circles at 7:09:00 AM

Radius, Diameter and Circumference

- The 1st Picture shows the radius of a circle.

- The 2nd Picture shows the Diameter of a circle.

- The 3rd Picture shows the Circumference of a circle.

# posted by Properties of Circles at 7:07:00 AM

Quiz Time

Quiz:

- In the diagram, the points A, B, C, D and E lie on the circumference of the two circles such that AEDC forms a cyclic quadrilateral and BCD is a straight line.

- The centre of the smaller circle is marked as O.

- Given that AB= BC, AE= CE, angle AED= 65° and angle ACE= 70°, find
i) angle ACB
ii) angle ABC
iii) angle OAC
iv) angle ADC

- In the diagram, line AB is the diameter of the circle with centre O.

- CE and BE are tangents to the circle.

- The tangent to the circle at point B meets OC produced at D.

- Given that angle CBE= 28° and that the radius of the circle is 5cm, find
i) angle BEC
ii) angle CAO
iii) length of BC
iv) angle CDE

# posted by Properties of Circles at 7:00:00 AM

Summary on Properties of Circles

Let us look at the Summary of the Properties of Circles:
~ Chords of a Circle ~
- The Perpendicular bisector of a chord of a circle passes through the centre of the circle.
- Equal chords of a circle are equildistant from the centre of the centre.
- The perpendicular from the centre of a circle to a chord bisects the chord.

~ Angles in a Circle ~
- The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at the circumference.
- An angle in a semicircle is a right- angle.
- The angles in the same segment of a circle are equal.

~ Angles in the Opposite Segments ~
The sum of the angles in the opposite segments of a circle is 180 degree.

~ Tangents to a Circle ~
- A tangent of a circle is perepndicular to the radius of the circle drawn from the point of Contact.

# posted by Properties of Circles at 6:50:00 AM

Teaching Concept Eleven

Let us look at the last Property of the Circle:
- The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

- How to proof if the angle between a tangent and a chord through the point of contact is always equal to the angle in the alternate segment.
Proof:
- angle SPT= 90° (right angle in a semicircle)
- angle x+ angle y = 90° (angle sum of triangle)
- angle y+ angle z= 90° (tangent perpendicular to radius)
- angle x= angle z
- angle PTB= angle PST
- angle PST= angle PQT ( angles in same segment)
- angle PTB= angle PQT

- Let us try some simple questions:

- In the figure shown, line PA and line PB are tangents to the circle. angle ABP= 60° and BAC= 40°. Find the value of angle x and angle y.

- Calculate the value of angle x and angle y.

# posted by Properties of Circles at 6:32:00 AM

Teaching Concept Ten

Let us take a look at the 2nd Property of Tangents to a Circle:
- From the previous Teaching Concept, you can see that if 2 tangents are drawn fom an external point to a circle, then

i) the lengths of the tangents are equal,

ii) and the line joining the external point and the centre bisects the angle between the tangents.

- Let us try a simple question:
- In the figure, line AB is a tangent to the circle, with centre O. Given that line AB= 8cm, BC= 5 cm and line OA= x cm, find
i) the value of x,

ii) angle AOB

iii) the area bounded by line AB, line BC and the arc AC.

# posted by Properties of Circles at 3:10:00 AM

Teaching Concept Nine

Let us take a look at the 1st Property of Tangents to a Circle:
- A tangent of a circle is perpendicular to the radius of the circle drawn from the point of contact.

- In the figure on the right, P is a point outside the circle, with centre O, line PA and line PB are two tangents drawn from P to touch the circle at A and B respectively.
- We can find that
i) line AP= line BP
ii) angle APO= angle BPO
iii) angle AOP= angle BOP

- Since, angle OAP= angle OBP= 90° (tangent perpendicular radius)
and △ AOP and △ BOP are congruent (R. H. S Property)
- Therefore, line AP= line BP,
angle APO= angle BPO,
and angle AOP= angle BOP

- We can also conclude that:
i) tangents drawn to a circle from an external point are equal.
ii) the tangents subtend equal angles at the centre.
iii) the line joining the external point to the centre of the circle bisects the angle between the tangents.

# posted by Properties of Circles at 3:08:00 AM

Teaching Concept Eight

Let us look at the 2nd Property of Angles in Opposite Segments:
- Exterior angle of Cyclic Quadrilaterals.
- If one side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the interior angle.

- How to proof if the exterior angle is always equal to the interior angle?
Proof:
angle b= 180° (opposite angles of cyclic quadrilaterals)
angle x+ angle d= 180° (adjacent angles on a straight line)
angle b+ angle d= angle x+ angle d
angle b= angle x
angle ABC= angle CDE

# posted by Properties of Circles at 3:04:00 AM

Teaching Concept Seven

Let us look at the 1st Property of Angles in Opposite Segments:
- The sum of the angles in the opposite segments of a circle is 180°.

- How to proof if the sums of the angles in the opposite segments of a circle is always 180° ?
Proof:
- Let angle b= 50°
- To find 2x angle d, 360°- 100°= 260°
- Therefore, angle d= 260°/ 2= 130°
- Thus, angle b+ angle d= 130°+ 50°= 180°

# posted by Properties of Circles at 3:03:00 AM

Teaching Concept Six

Let us take a look at the 3rd of Angles Properties of Circles:

- The angles in the same segment of a circle are equal.

- How to proof if the angles in the same segment are always equal?
Proof:

- angle AOB= 2x 1= 2x 2 (angle at the centre= 2x angle at the circumference)
Since x 1 = x 2
Therefore, angle APB= angle AQB

# posted by Properties of Circles at 2:50:00 AM

Monday, July 20, 2009

Teaching Concept Five

Let us take a look at the 2nd Angle Properties of Circles:
- Every angle at the circumference subtended by the diameter of a circle is a right angle triangle.

- How to proof if the the angle in the semicircle is always a right angle?
Proof:
- angle AOB = 2x AĈB ( angle at the centre= 2x angle at the circumference )
- Since angle AOB= 180°
- Therefore, AĈB= 90°

# posted by Properties of Circles at 4:51:00 AM

Teaching Concept Four

Let us take a look at the 1st Angle Properties of Circles:
- The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at the circumference.

- How to proof if the angle at the centre are always twice then the angle at the circumference when they are subtended by the same arc?

Proof:
- In the figure below, the angles are subtended by the minor arc AB.
- Since OA= OD [radii of circle], therefore, angle a= angle b [base angles of isosceles triangle]
- Not only that, angle AOE is the exterior angle of triangle AOD.
- Therefore, AOE= 2x angle a
- Similarly, angle c= angle d [base angles of isosceles triangle].
- Thus, angle BOE= 2x angle c
- Hence, angle AOB= 2 (angle a)+ 2 (angle c)= 2(angle a+ angle c)= 2 (angle ADB) [proven]

# posted by Properties of Circles at 3:46:00 AM

Teaching Concept Three

Let us look at the 2nd Symmetrical Property of Circles:
- SInce a circle is symmetrical about every diameter.
- Therefore any chord AB that is perpendicular to a diameter is bisected by the diameter.
- Also, any chord bisected by a diameter is perpendicular to the diameter.

- How to proof if the chord is always bisected to the diameter?
Proof:
- In a given a circle, centre O and chord AB, with a mid- point D, we are required to show that OĈB= 90°.
- 1st, you join OA and OB together.
- You can see that in triangle OAC and OBC,
OA= OB [radii of circle]
AC= BC [given]
OC is common.
- Triangle OCD is congruent to triangle OBC [SSS property].
OĈA = OĈB.
- Since these are adjacent angles on a straight line, therefore you can say that OĈA= OĈB= 90°. [proven]

# posted by Properties of Circles at 3:25:00 AM

Teaching Concept Two

Let us look at the 1st Symmetrical Property of Circles:
- The chord of a circle that passes through the centre of the circle is called the Perepndicular Bisector.
- Therefore, in equal circles or in the same circle, equal chords are equidistant from the centre.
- In another words, chords which are equidistant from the centre are equal.
- With this property, we are able to locate the centre of the circle.

- How to proof if the chords are always equidistant?
Proof:
- In the figure below, triangle OAB is rotated through an angle AOA' to triangle OA'B' about O.
- Since rotation preserves the shape and size, AB = A'B' and OG = OH. [Proven]

- Let us try a Simple Question.
If you are a designer for the construction of the Ferry Wheel in Genting Highland, using this property which you have just learnt, explain how would you find the centre of the Flyer?

# posted by Properties of Circles at 2:54:00 AM

Teaching Concept One

Before looking at the 1st Property..

There are some definitions that you might want to know.
- When any straight line is drawn through its centre, it divides the circle into 2 identical parts called the Semicircle. - When a line segment joined any 2 points that touches the circumference of a circle, it is called the chord.

[The Diameter is the longest chord.]

- A quadrilateral with its four vertices lying on the circumference of a circle is called a cyclic quadrilateral.

The Diameter cuts the circle into a Semicircle.

# posted by Properties of Circles at 2:11:00 AM

Saturday, July 18, 2009

A Welcome Speech

Welcome to Math Teaching Blog..! ^^
As you can see from my first sentence..
I am here to teach you Math but not Blogging..
So here..
There will be Teaching Concepts and Real- Life Applications on the Chapter of Properties of Circles.. =)

# posted by Properties of Circles at 9:36:00 PM