University:
Felician University
Course:
MATH-231 | Geometry I
Academic year:

2020
Views:

467

Pages:

4
Author:

SpaceCadet101

Quadrilaterals Formulas For Triangles and Quadrilaterals Name: Key Date: Period: 1. Parallelograms The area of a parallelogram with base b and height h is A = bh. The height of a parallelogram is measured along a segment perpendicular to a line containing the base. Example #1: Find the area of the parallelogram. 1 1. 30 mm 11 mm A = bh A = (11)(16) A = 176 mm² 34mm | h 34 mm 30² + h² = 34² 900 + h² = 1156 h² = 256 h = 16 Example #2: Find the height of a rectangle in which b = 3 in. and A = (6x² + 24x - 6) in². A = bh (6x² + 24x - 6) = (3)(h) 2x² + 8x - 2 = h h = (2x² + 8x - 2) in Example #3: Find the perimeter of the rectangle, in which A = (79.8x² - 42) cm² and b = 21 cm. A = bh (79.8x² - 42) = (21)(h) 3.8x² - 2 = h P = 2(21) + 2(3.8x² - 2) P = 42 + 7.6x² - 4 P = (7.6x² + 38) cm II. Triangles and Trapezoids The area of a triangle with base b and height h is A = 1/2 bh. The area of a trapezoid with bases b₁ and b₂ and height h is A = 1/2 (b₁ + b₂)h, or A = [(b₁ + b₂)/2]h. Example #4: Find the area of a trapezoid in which b₁ = 8 in., b₂ = 5 in., and h = 6.2 in. A = (1/2)(8 + 5)(6.2) A = (13)(6.2)/2 A = 80.6/2 = 40.3 in² Example #5: Find the base of the triangle, in which A = 80.6 in² and h = (15x²) cm². A = 1/2 bh 15x² = 1/2(b)(5x) 15x²/2.5x = (2.5x)b/2.5x 6x cm = b Example #6: Find b₂ of the trapezoid, in which A = 231 mm². A = (b₁ + b₂)h/2 231 = (23 + b₂)(11)/2 462 = (23 + b₂)(11) 42 = 23 + b₂ b₂ = 19 mm III. Rhombuses and Kites Area of a Rhombus or Kite The area of a rhombus or kite with diagonals d₁ and d₂ is A = ½d₁d₂. Example #7: Find d₂ of a kite in which d₁ = 14 in. and A = 238 in². A = ½d₁d₂ 238 = ½(14)d₂ 238 = 7d₂ d₂ = 34 in Example #8: Find the area of a rhombus. 8x+7 | 14x-6 8x+7 | 14x-6 112x²-42 A = ½(8x+7)(14x-6) A = ½(112x²-48x+98x-42) A = ½(112x²+50x-42) A = (56x²+25x-21) cm² Example #9: Find the area of the kite. 29 in /|\ / | \ / | 25 in | 28 in x in y in x² + 21² = 29² x² + 441 = 841 x² = 400 x = 20 in y² + 28² = 35² y² + 784 = 1225 y² = 441 y = 21 in A = ½(48)(42) A = 1008 in² Example #10: The tile design shown is a rectangle with a base of 4 in. and a height of 2 in. Use the grid to find the perimeter and area of the leftmost shaded parallelogram. P = (2√2 + 2) in A = (1)(1) = 1 in² Example #11: In the tangram, find the perimeter and area of the large green triangle. Each grid square has a side length of 1 cm. 1² + 1² = c² 2 = c² c = √2 P = (4√2 + 4) cm A = (1/2)(4)(2) = 4 cm² or A = (1/2)(2√2)(2√2) A = (1/2)(4)(2) = 4 cm²