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# Unit 10 -Circles This unit addresses circles.

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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:
SPI Give precise mathematical descriptions or definitions of geometric shapes in the plane and space. SPI Solve problems involving area, circumference, area of a sector, and/or arc length of a circle. SPI 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 Use technologies appropriately to develop understanding of abstract mathematical ideas, to facilitate problem solving, and to produce accurate and reliable models. CLE Establish the relationships between the real numbers and geometry; explore the importance of irrational numbers to geometry. CLE Use analytic geometry tools to explore geometric problems involving parallel and perpendicular lines, circles, and special points of polygons. CLE Develop the role of circles in geometry, including angle measurement, properties as a geometric figure, and aspects relating to the coordinate plane. CLE Analyze, interpret, employ and construct accurate statistical graphs.

Standards CFU (Checks for Understanding) applied to Unit 10:
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). Recognize the capabilities and the limitations of calculators and computers in solving problems. Analyze properties and aspects of pi (e.g. classical methods of approximating pi, irrational numbers, Buffon’s needle, use of dynamic geometry software). Approximate pi from a table of values for the circumference and diameter of circles using various methods (e.g. line of best fit). Find the equation of a circle given its center and radius and vice versa. 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). Find angle measures, intercepted arc measures, and segment lengths formed by radii, chords, secants, and tangents intersecting inside and outside circles. Use inscribed and circumscribed polygons to solve problems concerning segment length and angle measures. Determine the area of each sector and the degree measure of each intercepted arc in a pie chart. 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.

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

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?

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 arc is smaller than a semicircle The measure of a minor arc is the measure of its corresponding central angle A major arc is greater than a semicircle The measure of a major arc is 360 minus the measure of its related minor arc

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

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

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

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 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)

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

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… 

Assignment Page Page ,37-43 Worksheet 7-6

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

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)

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 square inches

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.

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

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 π R2 = 72/360 x π(20)2 = 80 π The area is 80 π cm2. 720 20 cm M O

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 π x (10)2 = cm2

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 - = Given a Segment Make a sector Make a Triangle Calculate area

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

Assignment Page 663/ Worksheet 7-7

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 ): Car Owner Mustang 41% Corvette 10% Camaro 25% G8 8% BMW 5 5% Jetta 7% Viper 4% Hint: What is the area of the circle? Radius = 15 inches

Circles in the Coordinate Plane
You can use the Distance Formula to find an equation of a circle with center point (h,k) and radius r. 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 y (x,y) -same as x2, y2 r r = √ (x-h)2 + (y-k)2 -Revised Distance Formula r2= (x-h)2 + (y-k) Square both sides (x-h)2 + (y-k)2 = r Re-write in the “equation of a circle” format (h,k) -same as x1, y1 x

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 equation of a circle Normally you are given a point (h,k) and a radius “r”

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 = 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

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 = √ = √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 = Simplify

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

Assignment Page Worksheet 11-5

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

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

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).

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

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

. 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

Chains and Gears A dirt bike chain fits tightly around two gears. The chain and gears form a figure like the one shown. Find the distance between the center of the gears. Label your center points and points of Tangency Create rectangle ABCE Now create Right Triangle AED Side AE is congruent to BC, or 26.5 In. Side DE = 9.3 – 2.4, or 6.9 In Hypotenuse AD can be calculated using the Pythagorean Theorem, or = AD2 Therefore AD2 = Therefore AD = 27.38 Or about 27.4 Inches . B 2.4 In A 26.5 In 26.5 In . C Hypotenuse E 9.3 In. D = 6.9 In

Check with Belts and Pulleys
A belt is wrapped tightly around two pulleys. What is the distance between the center of the pulleys? Again, create a right triangle The sides of the triangle are: 35 Inches 14 – 8 Inches (or 6 inches) Hypotenuse2 = Hypotenuse2= 1261 Hypotenuse = 35.5 Inches It is about 35.5 inches between the centers of the pulleys 6 Inches 35 Inches 14 In 8 In Hypotenuse

Theorem 11.2 If a line in the plane of a circle is perpendicular to a radius at the endpoint of the circle, then the line is tangent to the circle In other words, if you are given, or determine, that a line touching a circle creates a right angle, then you can conclude the line is tangent to the circle

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 = 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

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.

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

Find the Perimeter Circle O is inscribed in triangle ABC
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

Find a Segment Length 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

Assignment Page Worksheet 11-1

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)

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

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

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

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)

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

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 = X/2 = √( ) X = 2√( ) X = 2 x 5.5 X = 11 6.8 4 X

Assignment Page ,13-18, 30-32 Worksheet 11-2

Warning track (180 degree arc)
Unit 10 Quiz 5 Warning track (180 degree arc) The distance from home plate to the wall at a baseball field is 320 feet. The warning track is 5 feet wide all the way around The warning track arc is 180 degrees The ground crew needs to replace the warning track gravel How much area will they have to replace (How much area is the warning track)? 5 ft 320 Feet Home Plate

Inscribed Angles An inscribed angle is an angle that is on one side of a circle, and the rays of the angle extend to the other side of the circle The arc that is between each ray that forms the angle is called the intercepted arc The measure of an inscribed angle is ½ the measure of the intercepted arc The measure of angle B = ½ the measure of arc ac Remember, if it is not given, then the measure of the arc is found by measuring the central angle If arc AC = 50 degrees, what is the measure of angle B? A B C

Example Find the measure of arc a and angle b
The measure of the inscribed angle q is 60 degrees The intercepted arc (which is labeled a) is twice the measure of the angle, so arc a is 120 degrees The measure of the inscribed angle b is the measure of arc a, plus 300 Therefore angle b = = 1500/2=750 a 600 n q 300 b p

Theorems 2 inscribed angles that intercept the same arc are congruent
If an angle is inscribed inside a semicircle (half of a circle) then it is a right angle (because a semicircle is 180 degrees, and it is half of that) The opposite angles of a quadrilateral that are inscribed inside a circle are supplementary

Example Find the measure of the numbered angle
Angle 1 is inscribed in a semicircle –it is 90 degrees Angle 2 captures the same arc as the 25 degree angle –it is 25 degrees Angle 3 and 5 are both inscribed in a semicircle, they are both 90 degrees Angle 6 is ½ the sum of 600 and 800, so it is 70 degrees Angle 4 is ½ the other part of the circle –there are 360 degrees, minus 60 minus 80= 220, therefore Angle 4 = 220/2 = 110 Or, you could say 360 degrees in a quadrilateral, minus 2x90, minus 70 = 110 1 2 250 3 600 4 6 800 5

Theorem The measure of an angle formed by a tangent line and a chord is ½ the measure of the intercepted arc The measure of angle C = ½ the measure of arc BDC Notice: Angle c does not equal 90 degrees, because the line is not tangent to the diameter (or radius). B D C

. Unit 10 Quiz 6 How many degrees is angle c?
How many degrees is angle a? How many degrees is angle b? How many degrees is angle d? How many degrees is arc mn? How many degrees is arc man? How many degrees is arc pn? If arc bw is 10 degrees, how many degrees is arc nb? How many degrees is angle p? How many degrees is arc pd? 300 n m 300 . p b c w a 600 d

Assignment Text P ,20,21,23-25 Worksheet 11-3

Angle Measures and Segment Lengths
Angle 1 is (60+50)/2 = 550 A secant is a line that intersects a circle at two points A part of a secant forms a chord inside a circle The measure of an angle formed by two lines that intersect INSIDE a circle is ½ the sum of the measures of the two intercepted arcs in other words, you take the average of 2 angles What is the measure of angle 1? 600 1 500

Another Theorem The measure of an angle formed by two lines that intersect OUTSIDE the circle is ½ the difference of the measures of the intersected arcs In other words, instead of adding the 2 angles, you subtract them and then divide by 2 –what is angle 1? 1 300 The measure of angle 1 is ( )/2 = 350 1000

Examples Find the value of the variable X = (46+90)/2 = 68
950 Z0 460 200 X 900 X = (46+90)/2 = 68 (95-Z)/2 = 20 95 – Z = 2 x 20 = 40 -Z = 40 – 95 = -55 Z = 55

The Ice Cream Cone Problem
Find the value of X Here, all we are given is that the measure of the angle is 40 degrees We know that part of the circle is X degrees The other part of the circle would then be 360-X degrees Therefore, we can say that (360-X –X)/2 = 40 360 – 2X = 80 -2X = 80 – 360 = -280 X = -280/-2 = 1400 What is the measure of the other side of the cone? X 400

Theorem For a given point and a circle, the product (we multiply) of the lengths of the two segments from the point to the circle is constant along the line through the point and the circle a x b = c x d w(w+x) = y(y+z) t2 = y(y+z) a x c w t d b y z y z

Example Find the value of a Find the value of y Find the value of z
6.5 x a = 7 x 3 6.5a = 21 a = 3.2 6 8 7 y z 8 16 6.5 3 7 a

Assignment Text P ,15-20,24-26 Worksheet 11-4

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