Loading presentation...

Present Remotely

Send the link below via email or IM


Present to your audience

Start remote presentation

  • Invited audience members will follow you as you navigate and present
  • People invited to a presentation do not need a Prezi account
  • This link expires 10 minutes after you close the presentation
  • A maximum of 30 users can follow your presentation
  • Learn more about this feature in our knowledge base article

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.



No description

Brianna Eggleton

on 22 May 2014

Comments (0)

Please log in to add your comment.

Report abuse

Transcript of Circles

P= 180-(80/2)
P= 180-40

<P= 140°
Theorem: Sum of opposite angles of a cyclic quadrilateral =180°.

A quadrilateral is cyclic if it can be inscribed in a circle, with each vertex touching the circumference.

If a quadrilateral has the property that the sums of opposite angles =180°, it is cyclic.

Central angle/Inscribed angle theorem
Angles in the same segment
Theorem of Thales
When the two endpoints of the diameter of a circle are connected to a third point on the circumference of the circle, the inscribed angle formed will always be a right angle
This means any triangle inscribed on the diameter of a semicircle will have a right angle
This can be used to find the centre of a circle
Chord and Angle Properties of Circles
Chord and Angle Properties of Circles
Cyclic quadrilaterals
Intersecting chords theorem
Proof 1
Proof 1 – using the inscribed angle theorem
Join two opposite corners to the centre (O).
The two angles at the centre = 360° (2D +2A =360°)
Therefore D+A =360°/2 =180°

An inscribed angle is formed when two chords of a circle intersect at a point on the circumference of the circle.
Measurement of the inscribed angle is half the measurement of the central angle.
Community.tes.co.uk, (2014).
Applications in "real" life
. [online] Available at: http://community.tes.co.uk/tes_mathematics/f/25/t/319382.aspx [Accessed 18 May. 2014].

Cut-the-knot.org, (2014).
Thales' Theorem
. [online] Available at: http://www.cut-the-knot.org/pythagoras/ThalesTheorem.shtml [Accessed 17 May. 2014].

Elwes, R. (2010).
Maths 1001
. 1st ed. London: Quercus Publishing, pp.92-96.

Jones, T. and Jackson, S. (2001).
Rugby and Mathematics
. 1st ed. [ebook] pp.1-6. Available at: http://wesclark.com/rrr/rugby_and_math.pdf [Accessed 18 May. 2014].

Mathopenref.com, (2014).
Central Angle Theorem.
[online] Available at: http://www.mathopenref.com/arccentralangletheorem.html [Accessed 17 May. 2014].

Mathopenref.com, (2014).
Thales theorem
. [online] Available at: http://www.mathopenref.com/thalestheorem.html [Accessed 17 May. 2014].

Mathsisfun.com, (2014).
Circle Theorems
. [online] Available at: http://www.mathsisfun.com/geometry/circle-theorems.html [Accessed 17 May. 2014].

Muljadi, P. (2014).
Regiomontanus' angle maximization problem
. [online] Academia.edu. Available at: http://www.academia.edu/3659994/Regiomontanus_angle_maximization_problem [Accessed 17 May. 2014].

Traxler, J. (2014).
Optimal Viewing Angle
. [online] Jwilson.coe.uga.edu. Available at: http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html [Accessed 18 May. 2014].

Wilson, J. (2014).
Angles in a Circle and Cyclic Quadrilateral
. [online] jwilson.coe.uga.edu. Available at: http://jwilson.coe.uga.edu/EMAT6680Su06/Byrd/Assignment%20Nine/CyclicQuad.pdf [Accessed 17 May. 2014].

Angles in Same Segment of Circle are Equal - ProofWiki. 2014. Angles in Same Segment of Circle are Equal - ProofWiki. [ONLINE] Available at: https://proofwiki.org/wiki/Angles_in_Same_Segment_of_Circle_are_Equal. [Accessed 21 May 2014].
A and C are the “end points” B is the “apex point”

Optimal Viewing Angle
Optimal Viewing Angle
-You are in a room with a TV screen 2.5m off the ground and 1m tall. You want to see the movie at the largest vertical viewing angle from eye level. How far away should your head be horizontally from the screen?

Step 3
Inscribed angles continued
Coordinates of the optimal viewing position are (√8.75, 0)
For optimal viewing, you should sit √8.75m, approximately 2.96m, away.
What is the size of angle POQ?
Step 1
Step 2
Figure 1.1
Figure 1. 2
Figure 1.3
Figure 1.4
Figure 1.0
Figure 2.1
References -figures
Figure 1.0: from http://www.webanswers.com/science/math/draw-the-circle-label-the-parts-of-a-circle-1d72cf
Figure 1.1: from http://www.mathsisfun.com/geometry/circle-theorems.html
Figure 1.2: from http://www.mathsisfun.com/geometry/circle-theorems.html
Figure 1.3: from http://www.mathopenref.com/arccentralangletheorem.html
Figure 1.4: from http://www.mathsisfun.com/geometry/circle-theorems.html
Figure 2.1: from http://www.mathopenref.com/thalestheorem.html
Figure 2.2: from http://www.mathopenref.com/thalestheorem.html
Figure 3.1: from http://www.regentsprep.org/Regents/math/geometry/GP14/CircleSegments.htm
Figure 3.2: from http://www.cut-the-knot.org/proofs/IntersectingChordsTheorem.shtml
Figure 3.3: from http://www.mathopenref.com/chordsintersecting.html
Figure 4.1: from https://proofwiki.org/w/images/4/47/Euclid-III-21.png
Figure 4.2: from https://proofwiki.org/w/images/thumb/7/73/IncsribedAngle.PNG/638px-IncsribedAngle.PNG
Figure 5.1: based off
Maths 1001
pg 95
Figure 5.2: based off
Maths 1001
pg 95
Figure 5.3: from http://jwilson.coe.uga.edu/EMAT6680Su06/Byrd/Assignment%20Nine/CyclicQuad.pdf
Figure 6.1: based off http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html
Figure 6.2: based off http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html
Figure 6.3: based off http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html
Figure 6.4: based off http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html
The central angle formed by the diameter line PQ is always 180°.
If you draw a line from P and Q to any other point on the circumference of the circle, the inscribed angle theorem tells us that it must create an angle of 90°.
Finding the Centre
The diameter always passes through the centre of the circle
Two straight lines that meet in a 90° point on the circumference of the circle also intersect the circumference at another point
The place where the lines intersect the circumference of the circle will always be the end points of the diameter
Midway between these points is the centre of the circle
Figure 2.2
= 3 because we want the radius to be 3
Substitute the coordinates into the equation
3 - 2.5 = 0.5
Square both sides of the equation to remove the square root
9 - 0.5^2 = 8.75
Square root both sides of the equation to remove the square
X - 0 = X
Figure 5.1
Figure 5.2
Figure 6.1
Figure 6.2
Figure 6.3
Figure 6.4
Theorem: When two chords intersect each other inside a circle, the products of their segments are equal.

When 2 chords intersect, 4 angles are created within the circle

Intersecting Chords rules:
AE x BE = CE x DE

4 angles created
Theorem Proof
Given: Chords AD & BC Intersecting at P

Prove: (AP)(DP) = (CP)(BP)

Triangle APB and Triangle CPD are similar triangles

Angle BAP = Angle DCP. Since both the angles are suspended by same arc

Angle ABP= Angle CDP. Since both the angles are suspended by same arc

Angle APB= Angle CPD. Since both are vertically opposite angles and vertically opposite angles are equal

Using the similar triangle property.

(AP)/(PC) = (BP)/(PD) = (AB)/(CD)

Using the first proportion,

(AP)/(PC) = (BP)/(PD)

On cross multiplying, we get,
(AP)(DP) = (BP)(CP)
Practical Use
In a circle, let angle BAD and angle BED be angles in the same segment, the major arc.
Let F be the center of the circle, and join BF and FD.
From the Inscribed Angle Theorem:
Angle BFD=2 x Angle BAD
Angle BFD=2 x Angle BED
So Angle BAD= Angle BED.

View ABC as a circle, now ∠BEC be an angle at the center. ∠BAC is an
angle at the circumference of the circle.
Now, these angles have the same arc BC at their base.
AE is joined and drawn straight through to F.
Since EA=EB, then from Isosceles Triangles have Two Equal Angles so we can see that Angle EBA= Angle EAB.
So Angle EBA+Angle EAB=2 x Angle EAB.
But from the because the sum of the angles of a triangle =180 = Two Right Angles, we have that Angle BEF=Angle EBA+Angle EAB.
That is, Angle BEF=2x Angle EAB.
For the same reason, Angle FEC=2 x Angle EAC.
So adding them together, we see that Angle BEC=2 x Angle BAC.
Inscribed Angle Theorem
Figure 4.1
Figure 4.2
Theorem: Two angles that are within the same segment (major/minor) are equal
Figure 3.3
Figure 3.1
Figure 3.2
Proof 2
Proof 2 -using angles in the same segment
Figure 5.3
Angle ACB = Angle ADB
Angle BAC = Angle BDC
Angle CAD = Angle CBD
Angle DBA= Angle DCA
Full transcript