Chapter 11-Areas of Plane Figures

Chapter 11-Areas of Plane Figures By Lilli Leight, Zoey Kambour, and Claudio Miro

11.1-Areas of Rectangles • Postulate 17-The area of a square is the square of the length of a side. • A=S2 Length: 1 unit Area: 1 square unit By counting, Area=9 square units By using the formula, Area=32=9

11.1-Areas of Rectangles • Postulate 18 (Area Congruence Postulate)-If two figures are congruent, then they have the same area. • Postulate 19 (Area Addition Postulate)-The area of a region is the sum of the areas of its non-overlapping parts.

11.1-Areas of Rectangles • Theorem 11.1-The area of a rectangle equals the product of its base and height. • A=bh Given: A rectangle with base b and height h Prove: A=bh Proof: Draw-the given rectangle with area A, a congruent rectangle with area A, a square with area b2, a square with area h2 Area of big square= 2A+b2+h2 Area of big square= (b+h)2= b2+2bh+h2 2A+b2+h2 = b2+2bh+h62 2A = 2bh A = bh b h A H2 B2 A h b h b b h

11.1 Practice Problems • What is the area of a rectangle with a base of 5 and a height of 7? • What is the area of a square with a base of 4 and a height of 4?

11.2-Areas of Parallelograms, Triangles, and Rhombuses • Theorem 11.2-The area of a parallelogram equals the product of a base and the height to that base • A=bh S V R T Given: PQRS Prove: A=bh Key steps of proof: 1.) Draw altitudes PV and QT, forming two rt, triangles 2.)Area I=Area III 3.) Area of PQRS= Area II + Area I = Area II + Area III = Area of rect. PQTV = bh I II III h h P b Q

11.2 Areas of Parallelograms, Triangles, and Rhombuses • Theorem 11.3-The area of a triangle equals half the product of a base and the height to that base • A=1/2bh W Given: XYZ Prove: A=1/2bh Key steps of proof: 1.) Draw XW parallel to YZ and ZW parallel to YX forming WXYZ 2.) XYZ congruent to ZWX (SAS or SSS) 3.) Area of XYZ = ½ x Area of WXYZ =1/2bh X h Y Z b

11.2-Areas of Parallelograms, Triangles, and Rhombuses • 11.4-The area of a rhombus equals half the product of its diagonals. • A=1/2d1d2 D C 1/2d2 Given: Rhombus ABCD with diagonals d1 and d2 Prove: A=1/2d1d2 Key steps of proof: 1.) ADC congruent ABC (SSS) 2.)Since DB is perpendicular to AC, the area of ADC = ½ bh = ½ x d1 x 1/2d2=1/4d1d2 3.)Area of rhombus ABCD=2 x 1/4d1d2= 1/2d1d2 1/2d2 d1 A B

11. 2 Examples 1.) Find the area of a parallelogram with sides 8 and 15, and the acute angle equal to 35 degrees. Sin(35)=X/8 X=4.588 x 15 = 68.83 8 2.) The area of a triangle is 410 with a base of 41. Find its height. A=410 A=1/2bh 410=1/2(41)(h) 410=(20 x 5)(h) h=20 35 15

11.2 Practice Problems • Find the area of a parallelogram with sides 6 cm and 8 cm, and a 135 degree angle. • A rhombus has a perimeter of 60 and one diagonal of 24. Find its area. • Find the area of: • 1.) An equilateral triangle with a perimeter of 24 • 2.) An isosceles triangle with sides 13, 13, and 10. • 3.) 30-60-90 triangle with a hypotenuse of 12 inches.

11.3 Area of a Trapezoid • Theorem 11.5-The area of a trapezoid equals half the product of the height and the sum of the bases. • (A=1/2h(b1+b2)) • Also A=h(median) D b2 C II h h I Key steps of proof: 1.) Draw a diagonal BD of trap. ABCD, forming two triangular regions, I and II, each with a height of h. 2.) Area of a trapezoid = Area I+ Area II = 1/2b1h+1/2b2h = 1/2h(b1+b2) A B b1

11.3 Example • Find the area of a trapezoid with a height of 7 and a median of 15. • 15 x 7=105 15 7

11.3 Practice Problem • A trapezoid has an area of 75 cm2 and a height of 5 cm. How long is the median? 5 cm

11.4 Regular Polygon • A regular Polygon is any convex shape whose sides are all the same length and angles are all the same measure.

11.4 Regular Polygons • Regular Polygons can be inscribed inside of a circle. Using this relationship, some definitions were derived. • Center of a regular polygon= Center of circumscribed circle. • Radius of a regular polygon= distance from a center to a vertex. • Central Angle of a regular polygon= an angle formed by 2 radii drawn to consecutive vertices. • Apothem of a regular polygon= Perpendicular distance from the center to one of the sides of the polygon.

11.4 Practice Problems • Find the perimeter and the area of each figure. • A regular octagon with sides 4 and apothem a. • A regular pentagon side s and apothem of 3. • A regular decagon with side s and apothem a. 5 9

11.4 Theorem 11-6 • The area of a regular polygon is one half of the product of the apothem and the perimeter. • A=1/2*P*a • P=Perimeter • a= Apothem • Ex: • P=10*6=60 • a=8.66 • A=(1/2)*60*8.66=259.8 sq. units 10 8.66

11.5 Circumference • Circumference: Perimeter of a circle. • It is found by the product of twice the radius (diameter) and pi. • C= 2πr or C= πd • C= circumference r= radius d= diameter • Ex: Find the circumference of a circle with a radius of 12. • C=? r= 12 • C= (2)π(12)= 24π units

11.5 Area • Area= As the area of the inscribed regular polygons get closer and closer to a limiting number defined to be the area of a circle. • It is found by the product of the radius square and pi. • A= πr2 r = radius • Ex: Find the area of a circle with a radius of 27 • A=? r= 27 • A= (272)π = 729π sq. units

11.5 Practice Problems • A circle has an area of 18 in. Find the circumference of the rim. • Find the area of a circle with a radius of 7 • Find the radius and area of a circle with a circumference of 20π. • Find the radius and circumference of a circle with an area of 25π

11.6 Arc Lengths and Areas of Sectors • Sector- region bounded by two radii and an arc of the circle.

11.6 Finding the length of a sector • Length of sector: x/360(2πr) • x = number of degrees in sector • r = radius Ex. Find the length of the sector 120/360(2x9π) 1/3(18 π) = 6 π 9 120

11.6 Finding the area of a sector • Area of sector: x/360 = πr2 • x = degree of sector • r = radius Ex. Find the area of the sector 45/360(42 π) 5/72(16 π)= 10 π/9 4 45

11.6Finding the length of the radii • If you are only given the area or length of a sector and the sector degree measurement, you can still find the length of the radius. • Practice problems • Find the length of the radius with length of the sector is 2π and the degree measurement is 90. • Find the length of the radius with length of sector being 4π and area of sector 120.

11.6 Practice Problems • Find the area and the length of each sector 1 1

11.7 Ratios of Areas • Theorem 11-7: If the scale factor of two similar figures a:b, then • The ratio of the perimeters is a:b • The ratio of the areas is a2:b2 Ex. Find the ratio of the perimeters and the areas of the two similar figures. The scale factor is 8:12 or 2:3. Therefore, the ratio of the perimeter is 2:3. The ratio of the areas is 23:32 or 4:9. 8 12

11.7 Practice Problems • If the scale factor is 1:4, what is the ratio of the perimeter and the ratio of the areas? • If the ratio of the areas I 25:1, what is the ratio of the perimeter and the scale factor? • A quadrilateral with sides 8cm, 9cm, 6cm and 5cm has area 45 cm2. Find the area of a similar quadrilateral who's longest side is 15 cm.

Chapter 11-Areas of Plane Figures