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Unit 10 -Circles

Unit 10 -Circles. This unit addresses circles. It includes central angles, arcs (minor/major/semicircle), arc lengths, sectors, areas of sectors, segments, tangents to circles, circumscribed and inscribed circles, and chords. Standards. SPI’s taught in Unit 10:

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Unit 10 -Circles

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  1. Unit 10 -Circles • This unit addresses circles. • It includes central angles, arcs (minor/major/semicircle), arc lengths, sectors, areas of sectors, segments, tangents to circles, circumscribed and inscribed circles, and chords.

  2. Standards • SPI’s taught in Unit 10: • SPI 3108.1.1 Give precise mathematical descriptions or definitions of geometric shapes in the plane and space. • SPI 3108.4.8 Solve problems involving area, circumference, area of a sector, and/or arc length of a circle. • SPI 3108.4.13 Identify, analyze and/or use basic properties and theorems of circles to solve problems (including those relating right triangles and circles). • CLE (Course Level Expectations) found in Unit 10: • CLE 3108.1.7 Use technologies appropriately to develop understanding of abstract mathematical ideas, to facilitate problem solving, and to produce accurate and reliable models. • CLE3108.2.1 Establish the relationships between the real numbers and geometry; explore the importance of irrational numbers to geometry. • CLE 3108.3.1 Use analytic geometry tools to explore geometric problems involving parallel and perpendicular lines, circles, and special points of polygons. • CLE 3108.4.9 Develop the role of circles in geometry, including angle measurement, properties as a geometric figure, and aspects relating to the coordinate plane. • CLE 3108.5.1 Analyze, interpret, employ and construct accurate statistical graphs.

  3. Standards • CFU (Checks for Understanding) applied to Unit 10: • 3108.1.5 Use technology, hands-on activities, and manipulatives to develop the language and the concepts of geometry, including specialized vocabulary (e.g. graphing calculators, interactive geometry software such as Geometer’s Sketchpad and Cabri, algebra tiles, pattern blocks, tessellation tiles, MIRAs, mirrors, spinners, geoboards, conic section models, volume demonstration kits, Polyhedrons, measurement tools, compasses, PentaBlocks, pentominoes, cubes, tangrams). • 3108.1.7 Recognize the capabilities and the limitations of calculators and computers in solving problems. • 3108.2.1 Analyze properties and aspects of pi (e.g. classical methods of approximating pi, irrational numbers, Buffon’s needle, use of dynamic geometry software). • 3108.2.2 Approximate pi from a table of values for the circumference and diameter of circles using various methods (e.g. line of best fit). • 3108.3.3 Find the equation of a circle given its center and radius and vice versa. • 3108.4.13 Locate, describe, and draw a locus in a plane or space (e.g., fixed distance from a point on a plane, fixed distance from a point in space, fixed distance from a line, equidistant from two points, equidistant from two parallel lines, and equidistant from two intersecting lines). • 3108.4.40 Find angle measures, intercepted arc measures, and segment lengths formed by radii, chords, secants, and tangents intersecting inside and outside circles. • 3108.4.41 Use inscribed and circumscribed polygons to solve problems concerning segment length and angle measures. • 3108.5.1 Determine the area of each sector and the degree measure of each intercepted arc in a pie chart. • 3108.5.2 Translate from one representation of data to another (e.g., bar graph to pie graph, pie graph to bar graph, table to pie graph, pie graph to chart) accurately using the area of a sector.

  4. Review of Circles in a Plane • A circle is a set of points equally distant from a center point • A circle is named by it’s center point • A radius is a segment that has one endpoint in the center, and one on the circle. • Congruent circles have congruent radii (or diameters) • A diameter is a segment that contains the center of a circle and has both endpoints on the circle. • A central angle is an angle whose vertex is the center of the circle. . a m

  5. Find the Measure of the Central Angle • A study of 3600 people shows that this is how most people spend their time. The question is, what is the measure of each central angle used to make these pie slices? • Sleep = 31% of 360= • .31x360 = 111.6 • Food = 9% of 360= • .09x360 = 32.4 • Work = 20% of 360= • .20x360 = 72 • And so on… Other 15% Sleep 31% Entertainment 18% Must Do 7% Food 9% Work 20% How would I figure out how many people are in each category?

  6. Arcs • An arc is a part of a circle • One type of arc is a semicircle. A semicircle is half of a circle. • The measure of a semicircle is 180 degrees • A minor arcis smaller than a semicircle • The measure of a minor arc is the measure of its corresponding central angle • A major arcis greater than a semicircle • The measure of a major arc is 360 minus the measure of its related minor arc

  7. Identifying Arcs • Identify the following in circle O • Minor Arcs: • AD, CE, AC, DE • Semicircles: • ACE, CED, EDA, DAC • Major Arcs containing point A: • ACD, CEA, EDC, DAE A C O D E

  8. Arc Addition Postulate 7.1 • Adjacent Arcs are arcs of the same circle that have exactly ONE POINT in common. • The measure of the arc formed by two adjacent arcs is the sum of the measures of the two arcs. • Remembering that to measure an arc, you take the measure of the corresponding central angle, then the sum of the measures of two adjacent arcs is really the sum of the measures of two adjacent central angles. • And just like adjacent angles share one side, adjacent arcs share one point. • Congruent arcs are arcs that have the same measure AND are in the same circle or in congruent circles

  9. Find the Measure of the Arc • BC • 32 • BD • 32 + 58 = 90 • ABC • ABC is a semicircle so 1800 • AB • 180 – 32 = 148 • ADB • 180 + 32 = 212 C 580 D B 320 O A

  10. Circumference and Arc Length • Remember, circumference of a circle is π x the diameter • Theorem: The length of an arc of a circle is calculated like this: (measure of the arc) X π x D 360 • This is the same as saying the “fraction of the whole” i.e. ¾ x the circumference of a circle (which is π x D)

  11. Find the Arc Length • Find the length of xy • Length of XY = (mXY0/3600)x (π x D) = 90/360 x (π x 16) =.25 x (π x 16) = 4π inches X O 900 16 inches Y

  12. Find the length of the arc • Find the length of arc XPY • The radius here is 15 cm • = (mXPY0/3600)x (π x 2r) • = (240/360) x (π x 2(15)) • = 2/3 x 30 π • = 20π cm X O 15 CM P . Y 2400 Remember, Diameter = 2 Radii Of course... There is a way to do it on the calculator… 

  13. Assignment • Page 654 9-27 • Page 655 30-35,37-43 • Worksheet 7-6

  14. Unit 10 Quiz 1-round all answers to the nearest 10th • Using Circle O, find the following information: • Measure of arc AD in degrees • Measure of arc AB in degrees • Measure of arc ABD in degrees • Measure of arc ADC in degrees • Circumference of Circle O in pi • Length of arc BC in pi • Length of arc CD in pi • Length of arc DA in pi • Length of arc BDA in pi • If you rolled this circle 25 times, how far would you roll it? In inches (rounded to the nearest 10th) C 580 D B 10 inches 320 O A

  15. Areas of Circles and Sectors • Imagine taking a circle and cutting it into four quarters. • Cut each quarter into four wedge segments • Tape the wedges together to form a rough rectangle • Note that the areas are still the same • The base of this figure (b) is formed by the sum of the arcs of the circle  or ½ C • Remember, that the circumference of a circle is 2 π R. So one half of 2 π R would be π R. Therefore the base of this figure is π R (the base uses half, and the top of the figure uses the other half of the circumference. • The height of this figure is the radius, or R. • Therefore, the area of this figure is base times height, or π R times R, or π R2 which is also the formula for the area of a circle  R B (or π R)

  16. Area of a Circle • Theorem –As demonstrated with the rectangle in the previous slide, the area of a circle is Pi x R2 • Example: How much more pizza is in a 12 inch diameter pizza than a 10 inch pizza? • First find the radius of each 6 in. and 5 in. • Area of first pizza = π x (6)2 or 36 π • Area of second pizza = π x (5)2 or 25 π • The difference is about 11 π, or 34.55 square inches

  17. Sectors of Circles • A sector of a circle is a region bounded by an arc of the circle and the two radii to the arc’s end points. –In other words, it’s a slice of pie. • You name the sector by using one arc endpoint, the center of the circle, and the other arc endpoint. • A sector is a fractional part of the area of the circle. • Just as we measured the length of an arc by finding the ratio of the part of the circumference to the whole of the circumference (arc0/3600), we find the area of a sector by finding the same ratio of the part to the whole.

  18. Area of a Sector of a Circle • Theorem 7.16: The area of a sector of a circle is the product of the ratio (Arc0)/360 x the area of the circle (π R2) • Or Area of Sector AOB = m(arc AB)/360 x π R2 (Remember, the measure of the arc is the measure of degrees) A B r O

  19. Example –Find the area of a Sector of a Circle Z • Find the area of sector ZOM. Leave answer in terms of π. • Area of Sector ZOM = m(ZOM)/360 x πR2 • = 72/360 x π(20)2 • = 80 π • The area is 80 πcm2. 720 20 cm M O

  20. Check Understanding • A circle has a diameter of 20 cm. What is the area of a sector bounded by a 2080 major arc? Round your answer to the nearest tenth. • Use the equation m(arc)/360 x π x R2 • = 208/360 x πx (10)2 • = 181.5 cm2

  21. Segment of a Circle • A part of a circle bounded by an arc and a line segment joining the arc’s endpoints is called a Segment of a Circle. • To find the area of a segment for a minor arc, draw radii to form a sector. • The area of the segment equals the area of the sector minus the area of the triangle formed. Area of Sector Area of Triangle Area of Segment - = Make a sector Make a Triangle Calculate area Given a Segment

  22. Example Find the Area of a Segment of a Circle • Find the area of the shaded segment of the circle. Round to the nearest tenth. • Area of sector = m(AB)/360 x π(R )2 • = 90/360 x Pi(10)2 • = 25 π in2 • Area of Triangle AOB = ½ B x H • = ½ x 10 x 10 • = 50 in2 • Area of Segment = 25 π – 50 • = 28.5 in2 A O 10 in. B

  23. Assignment • Page 663/64 7-31 • Worksheet 7-7

  24. Unit 10 Quiz 2 Hint: Convert Percent to a decimal • Use the following information to calculate the area of each slice of pie (round to the nearest 10th ): Hint: What is the area of the circle? Radius = 15 inches

  25. Circles in the Coordinate Plane • You can use the Distance Formula to find an equation of a circle with center point (h,k) and radiusr. Chose (x,y)as any point on a circle, then you can express the radius r as the distance from point (h,k) to point (x,y). • Original distance formula: • d = √ (x2-x1)2 + (y2-y1)2 • Substitute “h” for x1 and “k” for y1 (x,y) -same as x2, y2 y r r = √ (x-h)2 + (y-k)2 -Revised Distance Formula r2= (x-h)2 + (y-k)2 -Square both sides (x-h)2 + (y-k)2 = r2 - Re-write in the “equation of a circle” format (h,k) -same as x1, y1 x 0

  26. Equation of a Circle • An equation of a circle with center (h,k) and radius r is • (x-h)2 + (y-k)2=r2 • This equation is in standard form. It is also known as the standard equationof a circle • Normally you are given a point (h,k) and a radius “r”

  27. Example of a Standard Equation of a Circle • Write the standard equation of a circle with center (5, -2) and radius 7 • Use standard form (x-h)2 + (y-k)2=r2 • (x-5)2 + (y-(-2))2 = 72 Substitute variables • (x-5)2 + (y+2)2 = 49 Simplify • Write the standard equation of the circle with center (3,5) and radius 6 • (x-3)2 + (y-5)2 = 36 • Write the standard equation of the circle with center (-2,-1) and radius √2 • (x+2)2 + (y+1)2 = 2

  28. Using the Center, and another point on the Circle • Suppose you have a circle with center (1,-3) and the circle passes through point (2,2). Write the standard equation for this circle • First solve for r • r = √(x-h)2 + (y-k)2 or √(2-1)2+(2-(-3))2 • = √1 + 25 = √26 • Now use standard form (x-h)2 + (y-k)2=r2 • (x-1)2 + (y-(-3))2 = (√26)2 Substitute • (x-1)2 + (y+3)2 = 26 Simplify

  29. Another Look • Using the standard form, determine a circle’s center and it’s radius. • (x-7)2 + (y+2)2 = 64 • (x-7)2 + (y-(-2))2 = 82 Put in Standard Form • h k r • Therefore the center is (7, -2) and the radius is 8

  30. Assignment • Page 800 8-36 • Worksheet 11-5

  31. Unit 10 Quiz 3 • Find the area of a circle with a radius of 12 inches • Find the area of 45% of that circle • Find the area of 67% of that circle • Find the area of 4% of that circle • Find the area bounded by a 88 degree arc of that circle • Find the area bounded by a 129 degree arc of that circle • Find the area of a circle with a diameter of 6 feet • Find the area of 34% of that circle • Find the area bounded by a 230 degree arc of that circle • Find the area bounded by a 310 degree arc of that circle

  32. Tangent Lines In Relation to a Circle • What if you drew a circle and labeled it O • Then you drew a line which intersected the circle in only one point, and labeled it Point A • Then you drew radius OA • What seems to be true about the two angles created by your line, and radius OA? • They are Right Angles, therefore the line is perpendicular to the radius. O A

  33. Tangent to a Circle • Previously we learned about the tangent ratio in Right Triangles. Here we will earn about tangents in relation to circles. • A tangent to a circle is a line (in the same plane) that intersects the circle in exactly one point. • This point (where they intersect, or share one point) is called the point of tangency. • You can have tangent lines, tangent rays, or tangent segments, but they all intersect in exactly one point (the point of tangency).

  34. Theorem 11.1 • If a line is a tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency. • Here line AB is perpendicular to segment OP P A B O

  35. Example Finding Angle Measures • Segment MN and Segment ML are tangent to Circle O. Find the value of X • Since the segments are tangent, Angle L and Angle N are right angles. • LMNO is a quadrilateral whose interior angle measures is 3600 • Therefore, 360 – 90 – 90 – 117 = X0 • X = 630 O N L 1170 X0 M

  36. Check Understanding • ED is tangent to Circle O. Find the value of X. • Because ED is perpendicular to the radius OD, angle D is a right angle • The sum of the interior angles in a triangle is 180 degrees. • Therefore, 180 – 90 – 38 = 52 • X = 520 E D X0 . O 380

  37. Finding the Tangent • Is ML tangent to Circle N at point L? Explain • Determine whether triangle LMN is a right triangle -because we need to know if angle L is a right angle • Does 72 + 242 = 252? • 625 = 625 YES • We can conclude that angle L is a right angle • Therefore we can conclude that radius NL is perpendicular to segment LM • Therefore the segment is tangent to Circle N at point L N 25 M 7 24 L

  38. Circumscribed and Inscribed Circles • Previously, we learned that a circle is circumscribed about a triangle if all the vertices (corners) of the triangle lie on points of the circle. In this case, the triangle is inscribed in the circle. • Similarly, when a circle is inscribed inside a triangle, then we can say the triangle is circumscribed about the circle. Each side of the triangle would then be considered tangent to the circle.

  39. Theorem 11.3 • Look at the inscribed circle below. • What conclusions could you draw about segment AD and Segment AF? • Or segment BD and Segment BE? • Or segment CF and Segment EC? • Each pair of segments is congruent • Theorem 11.3: The two segments tangent to a circle from one point outside the circle are congruent. Or: • AD = AF • BD = BE • CE = CF B D E A C F

  40. Find the Perimeter B 8 cm • Circle O is inscribed in triangle ABC • Find the perimeter of triangle ABC • Knowing that inscribed circles are tangent, we can conclude that AF = AD, BE = BD, and CE = CF • Therefore, the perimeter of triangle ABC = 2(10) + 2(15) + 2(8) = 66 cm D E O C A 15 cm F 10 cm

  41. Find a Segment Length B • Circle O is inscribed in Triangle ABC • Triangle ABC has a perimeter of 88 cm • Find the length of segment BE • Perimeter (88) = 2(17) + 2(15) + 2(BE) • 2(BE) = 88 – 34 – 30 • 2(BE) = 24 • BE = 12 cm D E 17 cm A C F 15 cm

  42. Assignment • Page 767 6-19 • Worksheet 11-1

  43. Unit 10 Quiz 4 • Write a standard equation of a circle with a center point of (3,5) and a radius of 6 • Write a standard equation of a circle with a center point of (7,2) and a radius of 9 • Write a standard equation of a circle with a center point of (3,5) and goes through point (6,9) • Write a standard equation of a circle with a center point of (7,5) and goes through point (16,9) • Write a standard equation of a circle with a center point of (0,0) and goes through point (6,1) • What is the center point and radius of a circle with this standard equation: x2 + y2 = 169 • What is the center point and radius of a circle with this standard equation: (x-3)2 + (y+2)2 = 256 • Find the area of a circle with a 15 foot diameter (leave answer in pi) • Find the area bounded by a 111 degree arc of that circle (in pi) • Find the area bounded by a 300 degree arc of that circle (in pi)

  44. Chords • A segment whose endpoints are on a circle is called a chord • A diameter is technically a chord, but normally when talking about chords, we are talking about all chords but the diameter • Segment PQ is a chord P O Q

  45. Some Basic Ideas • Within a circle (or in congruent circles): • Congruent Central Angles have congruent Chords • Congruent Chords have congruent Arcs • Congruent Arcs have Congruent Central Angles • Chords which are equally distant from the center are congruent P A X0 X0 B Q

  46. Example 9 a 9 • Find the value of a • Find the value of x 9 12.5 18 16 X 36 18

  47. More Theorems • In a circle, a diameter that is perpendicular to a chord bisects the chord, and its arc • Vice-Versa is true too: If a diameter bisects a chord, it is perpendicular • Any perpendicular bisector of a chord goes through the center of the circle (thus it must be a diameter)

  48. Example • Find the missing length • We are looking for the length of r • One length of the right triangle is 3 • One length is 7 –because the line that goes through the center of the circle (Segment KN) is perpendicular to the chord, therefore it bisects the 14 cm chord to 7 cm) • Therefore, we use the Pythagorean theorem • R2 = 32+72 • R = 7.6 K r 3 L M 14 N

  49. Example • Find the value of X • Here, we have the standard right triangle, so we can use the Pythagorean Theorem • 42+ (X/2)2 = 6.82 • (X/2)2 = 6.82-42 • X/2 = √(6.82-42) • X = 2√(6.82-42) • X = 2 x 5.5 • X = 11 6.8 4 X

  50. Assignment • Page 776-777-778 8-10,13-18, 30-32 • Worksheet 11-2

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