University:
North Central College
Course:
MATH 140 | Precalculus
Academic year:

2019
Views:

433

Pages:

12
Author:

SpaceCadet101

Section 1: Solving Equations for x and y 15. 5x + y = 130 10x = 80 Solving for x: x = 8 Substituting x = 8 into the first equation: 5(8) + y = 130, y = 76 16. x + 43 + 3y = 180 2y - 5 + 3y = 180 Solving for y: y = 34 Substituting y = 34 into the first equation: x + 43 + 3(34) = 180, x = 104 Section 2: Parallel Lines 17. Determine which lines are parallel (if any) based on the given angle relationships. a) ∠1 and ∠5: parallel b) ∠4 and ∠9: parallel c) ∠6 and ∠11: parallel d) ∠12 and ∠14: parallel e) ∠10 and ∠15: parallel 18. Complete the following proof: Given: ∠1 ≅ ∠2 and ∠3 ≅ ∠4 Prove: AB || CD Statements Reasons ∠1 ≅ ∠2 and ∠3 ≅ ∠4 Given ∠2 ≅ ∠3 Vertical Angles Theorem ∠1 ≅ ∠4 Substitution AB || CD Alternate Interior Angles Converse Section 3: Finding Slope 19. Find the slope of the line that passes through the given points. (2, 4) and (8, 4): m = 0 (5, 0) and (5, -4): m is undefined 20. Using slopes, determine if the lines are parallel, perpendicular, or neither. Line 1: (1, 1) and (3, 3) Line 2: (2, 2) and (0, 4) m₁ = 1, m₂ = -1, perpendicular Line 1: (-2, 3) and (-5, 2) Line 2: (4, 1) and (5, 3) m₁ = 1/3, m₂ = 2, neither Line 1: (-3, -2) and (1, 2) Line 2: (1, 3) and (4, 6) m₁ = 1, m₂ = 1, parallel Section 4: Graphing Lines 21. Graph the line with the given slope that passes through the given point. Line A: (1, 2), slope: 3 Line B: (-2, 1), slope: -1/2 Line C: (0, 2), slope: 1/2 Section 5: Writing Equations of Lines 27. Determine an equation for each line. Passes through (0, 4) with a slope of 2: y = 2x + 4 Passes through (2, 6) and (4, 12): y = 3x + 6 Passes through (2, 6) with a slope of -1/2: y = -1/2x + 7 Problem 1: Given: Triangle ABC with sides 5, 5, and 2√3 Task: Find the area of triangle ABC Solution: Using Heron's formula: s = (5 + 5 + 2√3) / 2 = 5 + √3 Area = √(s(s-a)(s-b)(s-c)) = √((5+√3)(5+√3-5)(5+√3-5)(5+√3-2√3)) = (50√3 + 75) / 6 Problem 2: Given: Area of a figure = Area of an equilateral triangle with side 2√11 Task: Find the value of x Solution: Area of equilateral triangle = (√3/4) * (side)^2 = (√3/4) * (2√11)^2 = 11√3 Equating the areas: (2√3)(7.5 + x) = 11√3 Solving for x: x = 3.5 Problem 3: Given: A figure divided into three triangles (I, II, III) Task: Find the area of triangle III Solution: Area(I) = 4 Area(II) = 7 Area(III) = 2 * Area(I) = 8 Problem 4: Given: A rectangle with area 48 sq. in Task: Find the value of x Solution: Area of rectangle = length * width = 48 6 * (x + 4) = 48 x = 4 Problem 5: Given: A figure divided into three triangles (I, II, III) Task: Find the area of triangle III Solution: Area(I) = 4 Area(II) = 7 Area(III) = 2 * Area(I) = 8 Problem 1: Given: An equilateral triangle and a regular hexagon with equal perimeters. Task: Find the ratio of their areas. Solution: Let the side length of the equilateral triangle be s. Then, the side length of the hexagon is s/3. Area of equilateral triangle = (sqrt(3)/4) * s^2 Area of hexagon = 6 * (sqrt(3)/4) * (s/3)^2 Ratio of areas: (sqrt(3)/4) * s^2 / (6 * (sqrt(3)/4) * (s/3)^2) = 3:2 Problem 2: Given: Two similar triangles with areas 100 and 25. Task: a) Find the ratio of the corresponding sides. b) Find an altitude of the smaller triangle. Solution: a) Ratio of areas = (ratio of sides)^2 100/25 = (4/x)^2 x = 2 b) Area of smaller triangle = (1/2) * base * height 25 = (1/2) * 2 * h h = 25 Problem 3: Given: A hexagon inscribed in a circle and another hexagon circumscribed around the same circle. Task: a) Find the ratio of the areas of the two hexagons. b) If the sum of the areas of the two hexagons is 56√3, find the radius of the circle. Solution: a) Ratio of areas = 4:9 b) 4x + 9x = 56√3, x = 8√3, radius = 4/√3 Problem 4: Given: A square inscribed in a semicircle and a square inscribed in the whole circle. Task: How does the area of the square in the semicircle compare with the square in the whole circle? Solution: The ratio of the areas is 2:5. Problem 7: Given: A rectangle with base b and height h. Task: Find the change in area if the base is increased by 30% and the height is decreased by 10%. Solution: New area = (1.3b)(0.9h) = 1.17bh The area increases by 17% (1.17 - 1 = 0.17) Problem 8: Given: A rectangle with base b and height h. Task: Find the change in area if the base is decreased by 25% and the height is increased by 25%. Solution: New area = (0.75b)(1.25h) = 0.9375bh The area decreases by 6.25% (1 - 0.9375 = 0.0625) Problem 9: Given: A rectangle with base b and height h. Task: Find the percentage change in the base to keep the same area if the height is increased by 20%. Solution: New base = (1/1.2)b = 0.833b The base needs to be decreased by 16.67% (1 - 0.833 = 0.1667) Problem 10: Given: A rectangle with base b and height h. Task: Find the percentage increase in the height to triple the area if the base is decreased by 25%. Solution: 3bh = (0.75b)(xh) x = 4 The height needs to be increased by 300% (4 - 1 = 3) Problem 11: Given: Two triangles with equal heights and bases of 6 inches and 45 inches. Task: Find the ratio of the area of triangle A to the area of triangle B. Solution: Ratio of areas = (6/45)^2 = 4/225 = 8/45 Problem 12: Given: A parallelogram with sides 8 and 11 and an included angle of 45°, and an equilateral triangle with a perimeter of 24. Task: Find the ratio of the area of the parallelogram to the area of the triangle. Solution: Area of parallelogram = 8 * 11 * sin(45°) = 44√2 Area of equilateral triangle = (sqrt(3)/4) * (8)^2 = 16√3 Ratio of areas = (44√2) / (16√3) = 11√2 / 4√3

Math Geometry Chapter 11 Worksheet Solutions

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