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Section 7.2 Perimeter and Area of Polygons

Section 7.2 Perimeter and Area of Polygons. The perimeter of a polygon is the sum of the lengths of all sides of the polygon. Formulas for Perimeter. Triangles Scalene: P = a + b + c Isosceles: P = b + 2s (base + 2 equal sides) Equilateral: P = 3s (3 x sides) Quadrilaterals

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Section 7.2 Perimeter and Area of Polygons

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  1. Section 7.2 Perimeter and Area of Polygons The perimeter of a polygon is the sum of the lengths of all sides of the polygon. Section 7.2 Nack

  2. Formulas for Perimeter • Triangles • Scalene: P = a + b + c • Isosceles: P = b + 2s (base + 2 equal sides) • Equilateral: P = 3s (3 x sides) • Quadrilaterals • Quadrilateral: P = a + b + c + d (sum of the sides) • Rectangle: P = 2b + 2h or P = 2(b + h) 2 x (base + height) • Square (or rhombus): P = 4s (4 x sides) • Parallelogram: P = 2b + 2s or 2(b + s) (2 bases + 2 sides) Section 7.2 Nack

  3. Heron’s Formula for the Area of a Triangle • Semiperimeter: s = ½ (a + b + c) • Heron’s Formula: A = s (s - a) (s - b) (s - c) • Example: Find the area of a triangle with sides 4, 13, 15. s = ½ (4 + 13 + 15) = 16 A = 16 (16 - 4) (16 - 13) (16 - 15) = 24 sq. units. Section 7.2 Nack

  4. Brahmagupta’s Formula for the area of a cyclic* quadrilateral • Semiperimeter = ½(a + b + c + d) • Area = A =  (s - a) (s - b) (s - c) (s – d) *cyclic quadrilateral can be inscribed in a circle so that all 4 vertices lie on the circle. Section 7.2 Nack

  5. Area of a Trapezoid • Theorem 7.2.4: The area A of a trapezoid whose bases have lengths b1 and b2 and whose altitude has length h is given by: A = ½ h (b1 + b2 ) = ½ (b1 + b2)h The average of the bases times the height Proof p. 350 Example 4 p. 350 Section 7.2 Nack

  6. Quadrilaterals with Perpendicular Diagonals • Theorem 7.2.4: The area of any quadrilateral with perpendicular diagonals of lengths d1 and d2 is given by A = ½ ( d1d2 ). • Corollary 7.2.5: The area A of a rhombus whose diagonals of lengths d1 and d2 is given by A = ½ ( d1d2 ) • Corollary 7.26: The area A of a kite whose diagonals of lengths d1 and d2 is given by A = ½ ( d1d2 ) Proof: Draw lines parallel to the diagonals to create a rectangle. The area of the rectangle A = (d1d2 ). Since the rectangle is twice the size of the kite, the area of the kite A = ½ ( d1d2 ) Section 7.2 Nack

  7. Areas of Similar Polygons • Theorem 7.2.7: The ratio of the areas of two similar triangles equals the square of the ratio of the lengths of any two corresponding sides: • Proof p. 353 • Note: This theorem can be extended to any pair of similar polygons (squares, quadrilaterals, etc.) Section 7.2 Nack

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