Lesson 30 35


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Lesson 30 35

  1. 1. Lessons 30 -35 Information on Circles
  2. 2. Lesson 30 & 31 Notes Terminology Circle with centre k diameter semicircles chord Radius from centre to outside edge Perpendicular bisector central angle Inscribed angle tangent The properties that should be discovered and used are as follows: k point of tangency
  3. 3. EC = 3.19 cm FE = 3.19 cm m AEC = 90.00 E A F C A) The perpendicular (line) from the centre of a circle to a chord bisects the chord This also works in the reverse. If you find the middle point of any chord and draw a perpendicular line from the middle point it will pass through the center of the circle. This property can help you find the center because, if you apply this property more than once inside one circle, the point where the perpendicular lines intersect is where the center of the circle is. B) The measure of the central angle is equal to twice the measure of the inscribed angle subtended (shared) by the same arc. This property becomes very useful for finding the center when we start to create central angles of 1800. This means that the inscribed angle will be 900. Center F This angle will be half the central This Angle will be twice as much as the inscribed The Arc that is shared is called the Subtended Arc
  4. 4. If students create a 900 angle on the circumference of the circle the end of each ray will be the diameter. If you apply this property more than once, the point where the diameters insect will be the center of the circle. C) The inscribed angles subtended by the same arc are congruent. Again, this is a property that only works when the inscribed angles share an arc. If you are using computer software, ask the students to drag one of the vertices into the subtended arc and see what happens to the measurements. They will no longer be congruent because they are no longer sharing the same arc. D) A tangent to a circle is perpendicular to the radius at the point of tangency. This becomes a useful tool for finding the center of a circle. If we extend the radius to a diameter, and apply the property more than once, the intersection point will be the center. m JML = 42.94 m JKL = 42.94 L J K M
  5. 5. I H G J E) An angle inscribed in a semi-circle is ALWAYS 900. This is directly related item B in the list. Do this twice and the point where the hypotenuses intersect will be the center of the circle. F) The opposite angle of a cyclic quadrilateral (are quadrilaterals inside a circle with its vertices on the circumference of the circle) are supplementary (add up to 1800). G) An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. 40HJG and 110HGK Then we know that angle JGH = 700 because it is in a straight line with HGK (180 – 110 = 70) From this we can then figure out that 70GHJ because interior angles of 180GHJ If we know that 70GHJ we also know that 70GIJ because both are inscribed angles subtended by the minor arc JG. 180I G K I H G J
  6. 6. O E D C F Examples 1) Determine the measures of angles ECB BDC BAD DBE where E is the center of the circle. 2) Determine the length of a chord AC. 3) Given a circle with the center O, and OF CD and DE=20 and OF=DF find the length of OF EO DF CD 30 100 E C A B D 610 B A C
  7. 7. O A BC Lesson 32 – Properties of Tangents 1) Tangent segments to a circle’s circumference from any external point are congruent in length. 2) The angle between the tangent and the chord is half the measure of the intercepted arc. 3) The angle between the tangent and the chord is equal to the inscribed angle on the opposite side of the chord. Examples 1) The circle has a radius of 5 and AB is tangent to the circle at point C. 50B and OA = OB Find the Area of AOB
  8. 8. 2) Find the length of AC 3) Given that ED is tangent at c 60 (2 ) (3 5) BCA ACD x BCE x Find B 5 30 50 A D B C 60
  9. 9. 15 AD BC ACB Find DBC 4) Lesson 33 – No Notes – Just Review Lesson 34 – Circle and Polygon Properties Example 1 – If diameter CD is perpendicular to a chord AB at E, verify that ABC is isosceles. This means we want to prove AC=CB
  10. 10. 8 6 4 2 -2 -4 -6 -8 -10 -15 -10 -5 5 10 15 Example 2) Determine the measure of BAC if 60 70DEF and EFC Example 3) Find the center of the circle that passes through the points (0,0), (0,6) and (4,0).
  11. 11. Example 4) With a circle centered at (0,0) and tangent to the circle at (3,4) find the equation of AB 10 8 6 4 2 -2 -4 -6 -8 -15 -10 -5 5 10 15