Chapter 10 Section 10.1 - Areas of Parallelograms and Triangles Objectives: To find the area of a parallelogram To find the area of a triangle

h b A = bh Theorem 10.1 – Area of a Rectangle The area of a rectangle is the product of its base and height. h b A = bh

Base of a Parallelogram  any of its sides Altitude  a segment perpendicular to the line containing that base, drawn from the side opposite the base. Height  the length of an altitude. Altitude Base

A = bh h b Theorem 10.2 – Area of a Parallelogram The area of a parallelogram is the product of a base and the corresponding height. A = bh h b

Base of a Triangle  any of its sides Height  the length of the altitude to the line containing that base

A = 1 2 bh h b Theorem 10.3 – Area of a Triangle The area of a triangle is half the product of a base and the corresponding height. A = 1 2 bh h b

Ex: Find the area of each Parallelogram 4.6 cm 3.5 cm 4 in 5 in 2 cm

Ex: Find the area of each Triangle. 13 cm 6.4 ft 5 cm 10 ft 4 ft 12 cm

When designing a building, you must be sure that the building can withstand hurricane-force winds, which have a velocity of 73 mi/h or more. The formula F = 0.004A 𝑣 2 gives the force F in pounds exerted by a wind blowing against a flat surface. A is the area of the surface in square feet, and v is the wind velocity in miles per hour. How much force is exerted by a 73 mi/h wind blowing against the side of the building shown below? 6 ft 12 ft 20 ft

Homework #18 Due Tuesday (March 12) Page 536 – 538 # 1 – 27 odd

Section 10.2 – Areas of Trapezoids, Rhombuses, and Kites Objectives: To find the area of a trapezoid To find the area of a rhombus or a kite

A = 1 2 h( 𝑏 1 + 𝑏 2 ) Theorem 10.4 – Area of a Trapezoid 𝑏 1 h 𝑏 2 The area of a trapezoid is half the product of the height and the sum of the bases. A = 1 2 h( 𝑏 1 + 𝑏 2 ) 𝑏 1 h 𝑏 2

A = 1 2 𝑑 1 𝑑 2 Theorem 10.5 – Area of a Rhombus or a Kite 𝑑 1 𝑑 2 The area of a rhombus or a kite is half the product of the lengths of its diagonals. 𝑑 1 𝑑 2 A = 1 2 𝑑 1 𝑑 2

Ex: Find the area of the trapezoid 12 cm 7 cm 15 cm

What is the area of trapezoid PQRS What is the area of trapezoid PQRS? What would the area be if <P was changed to 45°? 5 m S R 60° P Q 7 m

Ex: Find the area of the kite. 3 m 5 m 3 m 2 m

Find the area of the rhombus. C D

Homework #19 Due Wednesday (March 13) Page 542 – 543 # 1 – 29 odd

Section 10.3 – Areas of Regular Polygons Objectives: To find the area of a regular polygon

You can circumscribe a circle about any regular polygon. The center of a regular polygon is the center or the circumscribed circle. The radius is the distance from the center to a vertex. The apothem is the perpendicular distance from the center to a side. Center Radius Apothem

Ex: Finding Angle Measures The figure below is a regular pentagon with radii and apothem drawn. Find the measure of each numbered angle. m<1 = 360 5 = 72 (Divide 360 by the number of sides) m<2 = 1 2 m<1 = 36 (apothem bisects the vertex angle) m<3 + 90 + 36 = 180 m<3 = 54 3 2 1

Find the measure of each angle of the half of an octagon. 1 2 3

Suppose you have a regular n-gon with side s Suppose you have a regular n-gon with side s. The radii divide the figure into n congruent isosceles triangles. Each isosceles triangle has area equal to 1 2 as (a being apothem/s being side). Since there are n congruent triangles, the area of the n-gon is A = n · 1 2 as. The perimeter p of the n-gon is ns. Substituting p for ns results in a formula for the area of the polygon.

A = 1 2 ap p a Theorem 10.6 – Area of a Regular Polygon The area of a regular polygon is half the product of the apothem and the perimeter. A = 1 2 ap p a

Ex: Find the area of each regular polygon. A regular decagon with a 12.3 in. apothem and 8 in. sides A regular pentagon with 11.6 cm sides and an 8 cm apothem

Ex: Find the area of the hexagon. 10 mm 5 mm

Homework #20 Due Thurs/Fri (March 14/15) Page 548 # 1 – 23 all

Section 10.4 – Perimeters and Areas of Similar Figures Objectives: To find the perimeters and areas of similar figures

Theorem 10.7 – Perimeters and Areas of Similar Figures If the similarity ratio of two similar figures is 𝑎 𝑏 , then 1. the ratio of their perimeters is 𝑎 𝑏 2. the ratio of their areas is 𝑎 2 𝑏 2

Ex: Finding Ratios in Similar Figures The trapezoids below are similar. Find the ratio of their perimeters and ratio of their areas. 9 m 6 m

Ex: Two similar polygons have corresponding sides in the ratio 5 : 7. a. Find the ratio of their perimeters b. Find the ratio of their areas

Ex: Finding Areas Using Similar Figures The area of the smaller regular pentagon is 27.5 𝑐𝑚 2 . What is the area of the larger pentagon? 4 cm 10 cm

Ex: The corresponding sides of two similar parallelograms are in the ratio 3 : 4. The area of the larger parallelogram is 96 𝒊𝒏 𝟐 . Find the area of the smaller parallelogram.

Ex: Finding Similarity and Perimeter Ratios The areas of two similar triangles are 50 𝑐𝑚 2 and 98 𝑐𝑚 2 . What is the similarity ratio? What is the ratio of their perimeters? The areas of two similar rectangles are 1875 𝑓𝑡 2 and 135 𝑓𝑡 2 . Find the ratio of their perimeters.

Homework #21 Due Monday (March 18) Page 555 – 556 # 1 – 23 all

Section 10.5 – Trigonometry and Area Objectives: To find the area of a regular polygon using trigonometry To find the area of a triangle using trigonometry

In the last lesson, we learned how to find the area of a regular polygon by using the formula A = 1 2 ap. By using this formula and trigonometric ratios, you can solve other types of problems.

Trigonometry Review Sine = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 SOH Cosine = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 CAH Tangent = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 TOA

Ex: Find the area of the regular pentagon with 8cm sides.

Ex: Finding the Area and Perimeter Find the area and perimeter of a regular octagon with radius 16m.

Suppose we want to find the area of triangle ABC (below), but you are only given m<A and lengths b and c. To use the formula A = 1 2 bh, you need to find the height. This can be found by using the sine ratio: sin A = ℎ 𝑐 therefore h = c(sin A) B Area = 1 2 bc(sin A) c a h A C b

Area ΔABC = 1 2 bc(sin A) Theorem 10.8 – Area of a Triangle Given SAS The area of a triangle is one half the product of the lengths of two sides and the sine of the included angle. B Area ΔABC = 1 2 bc(sin A) a c A C b

Ex: Finding the area of a Triangle Two sides of a triangular building plot are 120 ft and 85 ft long. They include an angle of 85°. Find the area of the building plot to the nearest square foot.

Quiz Tuesday (10.1 – 10.5) Homework #22 Due Monday (March 18) Page 561 – 562 # 1 – 27 odd Quiz Tuesday (10.1 – 10.5)

Section 10.6 – Circles and Arcs Objectives: To find the measures of central angles and arcs To find circumference and arc length

Circle  the set of all points equidistant from a given point called the center. Radius  a segment that has one endpoint at the center and the other endpoint on the circle. Congruent Circles  have congruent radii Diameter  a segment that contains the center of a circle and has both endpoints on the circle Central Angle  an angle whose vertex is the center of the circle.

Circle = ΘP Central Angle = <CPA Radius = CP Diameter = AB

Semicircle  half of a circle (180°) Minor Arc  smaller than a semicircle (< 180°). Its measure is the measure of its corresponding angle. Major Arc  greater than a semicircle (> 180°). Its measure is 360 minus the measure of its related minor arc. Adjacent Arcs  arcs of the same circle that have exactly one point in common.

Ex: Identify the following in Θ O: The minor arcs (4) The semicircles (4) The major arcs that contain point A (4) A C O D E

mABC = mAB + mBC Postulate 10.1 – Arc Addition Postulate B C A The measure of the arc formed by two adjacent arcs is the sum of the measures of the two arcs. B C A mABC = mAB + mBC

C = Πd C = 2Πr Circumference  the distance around a circle Pi (Π)  the ratio or the circumference of a circle to its diameter Theorem 10.9 – Circumference of a Circle The circumference of a circle is Π (pi) times the diameter d C = Πd C = 2Πr r C O

Arc Length  a fraction of a circle’s circumference Theorem 10.10  Arc Length The length of an arc of a circle is the product of the ratio 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑐 360 and the circumference of the circle. A r Length of AB = 𝑚𝐴𝐵 360 · 2Πr O B

Ex: Find the measure of each arc. a. BC b. BD c. ABC d. AB 58° D O 32° A

Ex: The diameter of a bicycle wheel is 22in. To the nearest whole number, how many revolutions does the wheel make when the bicycle travels 100 feet?

Homework #23 Due Thurs/Fri (March 21/22) Page 569 – 570 # 1 – 39 odd

Section 10.7 – Areas of Circles and Sectors Objectives: To find the areas of circles, sectors, and segments of circles.

A = Π 𝑟 2 Theorem 10.11 – Area of a Circle r O The area of a circle is the product of Π (pi) and the square of the radius. A = Π 𝑟 2 r O

Sector of a Circle  a region bounded by an arc of the circle and the two radii to the arc’s endpoints. A sector is named using one arc endpoint, the center of the circle, and the other arc endpoint.

Theorem 10-12 – Area of a Sector of a Circle The area of a sector of a circle is the product of the ratio 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑐 360 and the area of the circle. A Area Sector AOB = 𝑚𝐴𝐵 360 · Π 𝑟 2 O r B

Segment of a circle  a part of a circle bounded by an arc and the segment joining its endpoints. To find the area of a segment, draw radii to form a sector. The area of the segment equals the area of the sector minus the area of the triangle formed. T

Ex: You’re hungry one day and decide to go to Warehouse Pizza to get some food. When you arrive, you check the menu and notice they are having deals on 14-in and 12-in pizzas. If a 14-in pizza costs $20.00 and a 12-in pizza costs $16.00, which price gives you the most pizza for your dollar?

Ex: Find the Area of a Segment of a Circle Find the area of the circle segment if the radius is 10-in and central angle ATB forms a right angle. B T

Section 10.8 – Geometric Probability Objectives: To use segment and area models to find the probabilities of events.

P(event) = 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 You may remember that the probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes P(event) = 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

P(event) = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑛𝑡𝑖𝑟𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 Geometric Probability  a model in which points represent outcomes. We find probabilities by comparing measurements of sets of points. P(event) = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑛𝑡𝑖𝑟𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

Chapter 10 Test Tuesday/Wednesday (Notebooks also due) Homework #24 Due Monday (March 25) Page 577 – 578 # 1 – 27 odd Chapter 10 Test Tuesday/Wednesday (Notebooks also due)