Confidential 2 WARM UP 1.The following regular polygons tessellate. Determine how many of each polygon you need at each vertex. Squares and Octagons Determine.

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Confidential 2 WARM UP 1.The following regular polygons tessellate. Determine how many of each polygon you need at each vertex. Squares and Octagons Determine whether each polygon can be used by itself to make a tessellation. Verify your results by finding the measures of the angles at a vertex. The sum of the measures of the angles of each polygon is given. 2. Triangle 180°

Confidential 3 3. Octagon 1080° 4. Pentagon 540° 5. If an equilateral triangle, a square, and a regular hexagon are used in a tessellation, how many of each do you need at a vertex?

Confidential 4 Lets Review what we have learnt in the last lesson A tessellation is created when a shape is repeated over and over again covering a plane without any gaps or overlaps.

Confidential 5 Only three regular polygons tessellate in the Euclidean plane: Triangles. Squares or hexagons. We can't show the entire plane, but imagine that these are pieces taken from planes that have been tiled Regular polygons tessellate

Confidential 6 Examples a tessellation of triangles a tessellation of squares a tessellation of hexagons

Confidential 7 ShapesAngles Triangle60 Square90 Pentagon108 Hexagon120 <6 Sides<120 Interior Measure of angles for the Polygon

Confidential 8 The regular polygons in a tessellation must fill the plane at each vertex, the interior angle must be an exact divisor of 360 degrees. This works for the triangle, square, and hexagon, and you can show working tessellations for these figures. For all the others, the interior angles are not exact divisors of 360 degrees, and therefore those figures cannot tile the plane.

Confidential 9 Semi-regular Tessellations You can also use a variety of regular polygons to make semi-regular tessellations. A semiregular tessellation has two properties which are:  It is formed by regular polygons.  The arrangement of polygons at every vertex point is identical.

Confidential 10 Examples of semi-regular tessellations

Confidential 11 What is a circle?- A circle is a closed plane figure where all the points have the same distance from the center. It has no beginning or end point.

Confidential 12 Radius and Diameter Radius: A line segment with one point at the center and the other endpoint on the circle is called a Radius. AB is the radius of circle A. Diameter: A line segment that passes through the center of the circle and has its endpoints on the circle is called a Diameter. CD is the Diameter of circle A. Diameter is double of radius in size.

Confidential 13 Example Find the diameter of the circle, if radius = 4 cm. Diameter = 2 × radius Diameter = 2 × 4 Diameter = 8 cm. To find the diameter, multiply the radius by 2.

Confidential 14 Example Find the radius of the circle, if diameter = 14 cm. Radius = diameter ÷ 2 Radius = 14 ÷ 2 Radius = 7 cm. To find the radius, divide the diameter by 2.

Confidential 15 Chords Chord: A line segment with its endpoints on the circle is called a chord. EF is a chord of circle A. Diameter is the longest chord of a circle.

Confidential 16 Example Example: In the given circle name a chord, radius and diameter of the circle. GH is a chord. CD is a radius. AB is the diameter of the given circle.

Confidential 17 Central Angle The sum of all the angles in any circle is 360°. Example: Find the unknown measure of an angle in the given circle. Sum of angles = 90°+ 90° + 90° +25° = 295° Unknown angle = 360° - 295° = 65° First find the sum of the angles that are given. Then subtract that sum from 360°.

Confidential 18 Circumference The distance around a circle is called circumference. Example: Distance around a coke can The ratio of the circumference of a circle to the diameter of a circle, C:d = π. So, we can say that C= π × d circumferencediameter π=3.14

Confidential 19 Circumference (Example) Find the circumference of the circle that has a diameter of 15 cm. C= π × d C≈ 3.14 × 15 C≈ 47.10

Confidential 20 Area of a Circle Area of a circle = π × r × r = π × r 2 Where π = 3.14 and r is the radius of a circle. If a circle has radius 5cm, find its area. Area of a circle = π × 5 × 5 = 3.14 × 5 × 5 = 78.5 cm 2

Confidential 21 Area of a Circle (Example) Find the area of the circle that has a diameter of 22 cm. First find the radius of the circle. r = Diameter/2 = 22/2 = 11cm Area of a circle = π × r 2 = 3.14 × 11 × 11 = cm 2

Confidential 22 Your Turn Find the diameter if 1. Radius = 3 cm. 2. Radius = 1.4 ft 3. Radius = 32 cm. Find the radius if 1. Diameter = 9 cm 2. Diameter = 11 cm. 3. Diameter = 70 cm

Confidential 23 Your Turn Find the Area if 1. Radius = 10 cm. 2. Radius = 12 ft 3. Radius = 6 in. 4. Find the Circumference if 5. Diameter = 12.5 ft 6. Radius = 5 cm 7. Diameter = 20 cm

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Confidential 26 Q1. John is putting shiny lace around his birthday cap. The diameter of his cap is 6.4 cm. How many centimeters of lace does he need?

Confidential 27 Q2. The radius of a pizza is 4 in. Find the diameter and area of the pizza.

Confidential 28 Q3. What is the degree measure of the angle between the 5 and the 6 on a clock?

Confidential 29 Let Us Review A circle is a plane figure where all the points have the same distance from the center. Diameter is double of radius in size. Radius is half of the diameter in size. Diameter is the longest chord of a circle.

Confidential 30 Let Us Review The sum of all the angles in any circle is 360°. The distance around a circle is called circumference and C= π × d Area of a circle = π × r 2

Confidential 31 You had a Great Lesson! Do Practice some more!!