Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Basic College Mathematics (ALEKS) Section 8 – 1 Chapter Eight GEOMETRY Lines and Angles Point is a specific location in space. . Line is a straight path (infinite number of points). …A……B….. Line Segment is part of a line between TWO points. A…………B Ray is part of the line that includes an ENDPOINT. A………B…..> ∙ Point A Line AB (no Endpoints) Ray with Endpoint A Line Segments with Endpoints A and B Angle is formed by TWO Rays with a common Endpoint. ( Rays: and Vertex is the common Endpoint. Vertex = B Angle ABC, Angle CBA, Angle B, ABC, Circle = 360° 1 Chapter 8 CBA ) Basic College Mathematics (ALEKS) Right Angle = 90° Box at vertex shows 90° Straight Angle = 180° Acute Angle is > 0° and < 90° (small) Obtuse Angle is > 90° and < 180° (Fat) Congruent Angles has the same measure. Complementary Angles measures add to 90°. Supplementary Angles measures add to 180°. What is the supplement of 35°? 180° − 35° = 145° What is the complement of 52°? 90° − 52° = 38° Parallel Lines never intersect. 2 Chapter Eight GEOMETRY Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY Two intersecting Lines for four angles. A and C are vertical angles. m A=m C B and D are vertical angles. m B=m D Adjacent Angles share side (Ray). A is adjacent to B and D Perpendicular Lines intersect at 90°. When a Line intersects two Parallel Lines, Eight Angles are formed. 1= 4= Section 8 – 2 5= 8 and 2= 3= 7 Triangles and the Pythagorean Theorem Triangle is a three sided figure. The sum of the angles of a triangle = 180°. 3 6= Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY What is measure of angle a? 90 + 43 + X = 180 133 + X = 180 X = 180 – 133 = 47° What are the measures of angles a and b? 100 + a = 180 a = 80 39 + 80 + b = 180 119 + b = 180 b = 180 – 119 = 61 Names of Triangles: Acute – all angles acute. Right – has a right angle. Obtuse – has an obtuse angle. Equilateral – all three sides equal. Isosceles – two sides equal. Scalene – NO sides equal. SQUARE ROOTS Square of a number = that number times itself. N2 = N x N Reverse is the Square Root. 4 Chapter 8 Basic College Mathematics (ALEKS) √ = 02 = 0 √0 = 0 12 = 1 √1 = 1 22 = 4 √4 = 2 33 = 9 √9 = 3 44 = 16 √16 = 4 52 = 25 √25 = 5 62 = 36 √36 = 6 72 = 49 √49 = 7 Chapter Eight GEOMETRY √ x √ = N 82 = 64 √64 = 8 92 = 81 √81 = 9 102 = 100 √100 = 10 112 = 121 √121 = 11 122 = 144 √144 = 12 132 = 169 √169 = 13 142 = 196 √196 = 14 152 = 225 √225 = 15 PYTHAGOREAN THEOREM AND APPLICATIONS Right Triangles have a special property. Side a and side b are called legs and are on either side of the 90° angle. Side c is the Hypotenuse and is opposite the 90° angle. (Leg)2 + (Leg)2 = (Hypotenuse)2 a2 + b 2 = c2 5 Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY 4 2 + 3 2 = c2 19 + 9 = c2 25 = c2 √25 = √ 5=c ? = a or b 26 = c a2 + 242 = 262 a2 + 576 = 676 a2 = 100 = √100 a = 10 The bottom of a 17 foot ladder is placed 8 feet from the building. How far up is the ladder on the building? c = 17, a = 8, b = building 82 + b2 = 172 64 + b2 = 289 b2 = 225 √ = √225 b = 15 Section 8 – 3 Quadrilaterals, Perimeter, and Area Quadrilaterals: 6 Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY Rectangle = Box with four right angles. Opposite sides are equal, adjacent sides are not. Square = Rectangle with all side equal. Parallelogram = four sided figure with two sets of parallel sides. Rhombus = parallelogram with all sides equal. Trapezoid = four sided with ONE set of parallel sides. Perimeter: Square = S + S + S + S = 4S 7 Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY Rectangle = L + W + L + W = 2L + 2W What is the perimeter of a square with a side of 4.3 inches? P = 4(4.3 inches) = 17.2 inches Find the perimeter of a rectangle with a length of 3 yards and a width of 2 feet? P = 2(3 yards)+2(2 feet) P = 6 yards + 4 feet P = 18 feet + 4 feet = 22 feet Area: Rectangle = L x W Square = S x S = S2 Parallelogram = Base x Height = B x H Triangle = !" #"$%ℎ' = # Note: Height can drawn and measured outside the figure (parallelogram too). Trapezoid = 8 (!)* +, -.//"/ !$0"!1 #"$%ℎ' = Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY ( + 1 # Find the area of a parallelogram with 10.4 in and height of 3.1 in? A = BxH = (10.4 in)x(3.1 in) A = 32.24 in2 A side of a house must be painted. The length = 40 feet and the width = 12 feet. There are two 4 by 4 foot windows (not painted). What is the area to be painted? A = (40)(12) – 2(4)(4) A = 480 – 32 = 448 ft2 Figure is triangle. B = 7 m and H = 4 m A= # A= 7* A =14 m2 What is the area of this figure? 9 Chapter 8 4* Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY Figure is a trapezoid. A = 11 cm, B = 15 cm, H = 6 cm Area = ( + 1 # What is the area of this Figure? A = (11 + 1516 = (261(61 2 A = 78 cm Find the area of the shaded Region? Looks like a rectangle with a triangle cut out of the end. So, area of the rectangle minus the area of the triangle. AR = 15x10 = 150 ft2 AT = (101(61 = 30 ft2 A = 150 – 30 = 120 ft2 Paint the deck. One gallon covers 160 ft2 and cost $ 11.95. What is the cost to paint the deck? Section 8 – 4 Looks like a rectangle with a trapezoid on the end. AR = 31ft x 8 ft = 248 ft2 AT = (8 + 161(61 = 72 ft2 A = 248+72 = 320 ft2 DEFFGH 320 ft2 x M = 2 gallons IJ KL $.OP 2 gal x DEF = $23.90 Circles, Circumference, and Area Definitions: Circle – all points located the same distance from a fixed point. 10 Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY Center – that fixed point. Radius (r) – a line from the center to the circle. Diameter (d) – a line from the circle through the center on to the circle. .= 2r = d Q Circumference ( C ): - distance around a circle (perimeter). Π = (Pi – Greek Letter) – if left as the symbol it is exact. Π = 3.1415926535… (not exact – rounded) Π = 3.14 or R will normally be used. C=ST C = 2πR The line is the diameter. d =10.4 m J.U V r= = 5.2 * What is the radius? 11 Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY The line is the radius. P r = $X W P P W U d = 2( $X1 = Y0 What is the diameter? C = πD a) C = π10 = 10π ft (exact answer) b) C = 3.14(10) = 31.4 ft (approximate) Find the circumference? a) exact answer in S b) approximate answer using 3.14 C = 2πR R = 4.7 cm a) C = 2(π)(4.7 cm) = 9.4π cm b) C = 2(3.14)(4.7 cm) = 29.516 cm Find the circumference? a) exact answer in S b) approximate answer using 3.14 AREA: A = πR2 = πRxR Z Z\Z U A = π( )2 = π( 12 ) (rarely used) Chapter 8 Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY 21 in = radius A = πR2 = π(R)(R) a) A = π(21 in)(21 in) = 441π in2 b) Find the area of the circle? \ \ A= = 1386 in2 a) exact b) S = R R (approximate) 9.4 in = diameter Z O.U ]H R= = = 4.7 $X a) A = π(4.7 in)(4.7 in) A =22.09π in2 Find the area of the circle? b) a) exact A=(3.14)(4.7)(4.7)=69.3 in2 b) S = 3.14 (approximate) A = 69 in2 round to nearest whole unit A rectangle with two half circles Removed. Find perimeter and Area, use π = 3.14 Section 8 – 6 Perimeter is two staright lines and two half circles. Two half circle = one circle. C = πD A=πRR P = 2(12 cm) + πD P = 24 cm + (3.14)(6) cm P = 24 cm + 18.84 cm = 42.84 cm Area of rectangle – Area Circle A=(12)(6) – (3.14)(3)(3) A= 72 – 28.26 = 43.74 cm2 Volume 1 cm3 = 1 cc = 1ml 1 in3 (volume has units X3) “Cubic” 13 Chapter 8 Basic College Mathematics (ALEKS) Rectangular Solid V = LxWxH Cube V = S3 = SxSxS Right Circular Cylinder V = πR2H Right Circular Cone V = S_ # ^ Sphere U V = S_ ^ ^ 14 Chapter 8 Chapter Eight GEOMETRY Basic College Mathematics (ALEKS) Chapter Eight GEOMETRY What is the volume of a rectangular V = L x W x H Solid 8 in, by 11 in by 2 in? V = (8 in) x (11 in) x (2 in) V = 176 in3 Find the volume of a right cylinder V = πR2H with π = 3.14, Radius = 2 in, and V = (3.14)(2 in)(2 in) (4.5 in) Height = 4.5 in? V = 56.52 in3 Find the volume of a sphere. Use π = 3.14 and Radius = 3 cm. 4 ` = S_ ^ 3 4 ` = (3.141(3 1(31(31*^ 3 V = 113.04 cm3 Find the volume of a cone. Use π = 3.14, diameter = 8 cm and 18 cm height. = = 4 * 1 ` = S _ # 3 V = (3.141(4cm)(4cm)(18cm) ^ V = 301.44 cm3 15 R= Chapter 8 Z W aV Basic College Mathematics (ALEKS) .U VV Chapter Eight GEOMETRY = 1.2 ** a) R= b) R= c) Vouter – VInner = (3.14)(1.2)(1.2)(2) – (3.14)(0.8)(0.8)(2) .I VV = 0.8 ** V = 9.0432 – 4.0192 = 5.0240 mm3 ≈ 5 mm3 16 Chapter 8