Measuring Length and area Chapter 11 Measuring Length and area 11.1 areas of triangles and parallelograms 11.2 areas of trapezoids, rhombuses, and kites 11.3 perimeter and area of similar figures 11.4 Circumference and arc length 11.5 areas of circles and sectors 11.6 areas of regular polygons 11.7 use geometric probability

11.1 Area of Triangles and Parallelogram A = s2 s Postulate 24: Area of Square Postulate The area of the square is the square of the length of its sides.   Postulate 25: Area Congruence Postulate If two polygons are congruent, then they have the same area.   Postulate 26: Area Addition Postulate The area of the region is the sum of the areas of its non-overlapping parts.     Theorem 11.1: Area of the Rectangle The area of the rectangle is the product of its base and height. h b A = bh

11.1 Area of Triangles and Parallelogram Theorem 11.2: Area of a Parallelogram The area of a parallelogram is the product of the base and Its corresponding height. h b A = bh Theorem 11.3: Area of a Triangle The area of a triangle is one half the product of the base and its corresponding height. b h A = (1/2) bh Parallelograms Either pairs of the parallel sides can be used as the bases of the parallelogram. The height is the perpendicular distance between these bases. If you transform a rectangle to form other parallelograms with the same base and height, the area stays the same.

Chapter 11 Section 1 Practice Problems Find the area of the shaded region. Which postulate can be used to prove congruency? 4 8 12 4 A 4 B Area Congruence Postulate Area=(12)(8)-42 Area=96-16 Area=80 units2

Find the area of the shapes below. Chapter 11 Section 1 Practice Problems continued… Find the area of the shapes below. 5 9 3 6 A = bh A = 9 ∙ 3 = 27 The area of the parallelogram is 27 units 2. A = (1/2) bh H = √(52 – 32) = 4 A = (1/2) 6 ∙ 4 = 12 The area of the triangle is 12 units2.

11.2 Area of Trapezoids, Rhombuses, and Kites The height of the trapezoid is the perpendicular distance between its bases. h b1 b2 A = (0.5)h(b1 + b2) Theorem 11.4 Area of Trapezoid The area of Trapezoid is one half the product Of the height and the sum of the lengths of the bases. A = (0.5)d1d2 d2 Theorem 11.5 Area of a Rhombus The area of a Rhombus is one half the product of the Length of its diagonals. A = (0.5)d1d2 d2 Theorem 11.6 Area of a Kite The area of a kite is the product of the lengths of its diagonals.

Chapter 11 Section 2 Practice Problems The area of the trapezoid is 54ft2. Find the height of the trapezoid. Find the area of the rhombus. 4ft 20ft 16ft 8ft A = (0.5)h(b1 + b2) 54=(0.5)h(8+4) 54=(0.5)(12h) 54=(6h) h=9ft 202=162+x2 X=12 d1=32ft, d2=24ft A=(0.5)(32)(24) A=384ft2

11.3 Perimeter and Area of Similar Figures Recall… If two polygons are similar, then the ratio of their perimeters, or of any two corresponding lengths, is equal to the ratio of their corresponding side lengths. However, the areas of similar polygons have a different ratio. Theorem 11.7 Areas of Similar Polygons The area of two polygons are similar with the lengths of corresponding sides in the ratio of a : b, then the ratio of their areas is a2 : b2. Side length of Polygon I Side length of Polygon II Area of Polygon I a2 Area of Polygon II b2 a b = I II a b Polygon I ~ Polygon II Any two regular polygons with the same number of sides are similar. Also, any two circles are similar.

Chapter 11 Section 3 Practice Problems Find the height of figure B if it is similar to figure A (not drawn to scale obviously). Write the ratios of the corresponding side lengths of the 2 similar polygons from figure A to figure B given their areas. A B 8ft Area=40m2 Area of Figure A=108in2 Area of Figure B=432in2 108 432 = 1 4 Area=360m2 Ratio of Areas=1/9, so the ratios of corresponding side lengths in 1/3. 1/3=8/𝑥 X=24m. The ratio of the corresponding sides in 1 2 , as the square root of 1 4 is 1 2 .

11.4 Circumference and Arc Length The circumference of a circle is the distance around the circle. For all circles, the ratio of the circumference to the diameter is the same. This ratio is known as π, or pi. r d C C = πd = 2πr Theorem 11.8 Circumference of a Circle The circumference C of a circle is C = πd or C = 2πr, where d is the diameter of the circle and r is the radius of the circle AB Arc Length Corollary In a circle, the ratio of the length of a given arc to the circumference is equal to the ratio of the measure of the arc to 360 X 2πr = 2πr mAB 360 , or length of AB = B A

Chapter 11 Section 4 Practice Problems Find the circumference of the circle. Express your answer in terms of π. Find the length of AB. Express your answer to the nearest hundredth. 5in B A 30˚ 12mm Radius=5in, so diameter=10in Circumference=πd Circumference=10πin. Circumference=24π 30 360 = 𝑥 24π x=2π x=6.28mm

11.5 Area of Circles and Sectors Theorem 11.9 Area of a Circle The area of a circle is π times the square of the radius. Sector of a circle: the region bounded by two radii of the circle and their intercepted arc. Theorem 11.10 Area of a Sector The ratio of the area of the sector of a circle to the area of the whole circle (πr2) is equal to the ratio of the measure of the intercepted arc to 360◦ P B A r x πr2 πr2 360 ◦ = mAB Area of sector APB , or Area of sector APB =

Chapter 11 Section 5 Practice Problems The area of sector ABC is 24πcm2. Find the measure of angle ABC. Find the area of sector ABC to the nearest hundredth. A B C 45˚ 8m 12cm A C B 𝑥 64π = 45 360 X=8π Area of sector ABC=25.13m2 Area of circle B=144πcm2 24 144 = 𝑥 360 m ∠ABC=60˚

11.6 Areas of Regular Polygons M Q The center of the polygon (point P) and the radius of the polygon (MP) are the center and radius of the circle that circumscribes the regular polygon. The distance from the center to any side of the polygon is called the apothem of the polygon (QP). A central angle of a regular polygon is an angle formed by two radii drawn to consecutive vertices of the regular polygon (∠𝐌𝐏𝐍). s a = so A 1 2 Theorem 11.11 Area of a Regular Polygon The area of a regular n-gon with side length s is one half the product of the apothem a and the perimeter P, aP or A = a * ns

Chapter 11 Section 6 Practice Problems Find the area of the regular polygon. Express your answer to the nearest hundredth. (Hint=use trigonometric ratios to find the apothem.) Find the area of the shaded region if the triangle is equilateral. Express your answer to the nearest hundredth. 6in 4ft Apothem= 3 tan 36 15* 3 tan 36 Area=61.94in2 Area=29.48ft2

11.7 Use Geometric Probability The Probability of an event is the measure of the likelihood that the event will occur. A geometric probability is a ratio that involves a geometric measure such as length or area. Key concept Probability and Length Let AB be a segment that contains the segment CD. If a point K on AB is chosen at random, then the probability that it is on CD is the ratio of the length of CD to the length of AB. B C D A P(K is on CD) = Length of CD Length of AB Key concept Probability and Area Let J be a region that contains the region M. If a point K in J is chosen at random, then the probability that it is in region M is the ratio of the area of M to the area of J. M J P(K is in region M) = Area of M Area of J

Chapter 11 Section 7 Practice Problems What is the probability that a point randomly selected on AD is also on BC? Express your answer as a common fraction. What is the probability that a point randomly selected in region A is also in region B? Express your answer as a common fraction. (Region A is the entire figure). 13 182 A B C D A B 13 182 reduces to 1 14 The probability is 1 14 The probability is 1 4