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Developing Formulas for Triangles and Quadrilaterals

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Presentation on theme: "Developing Formulas for Triangles and Quadrilaterals"β€” Presentation transcript:

1 Developing Formulas for Triangles and Quadrilaterals
Geometry CP1 (Holt 10-1) K. Santos

2 Area of a Parallelogram
Area = product of its base and height A= bh Base must be perpendicular to the height b h 5cm 3cm 9cm A = 9(3) A = 27 π‘π‘š 2

3 Example Find the perimeter of a parallelogram, in which the base is 4ft and the area is 12 𝑓𝑑 2 . Need to find height A = bh 12 = 4h h = 3 ft P = 2b + 2h P = 2(6) + 2(3) P = 18 ft

4 Area of a Triangle Area = one half of the product of its base and height A= 1 2 bh or A = π‘β„Ž 2 Base perpendicular to height h h h b b b If b = 4” and h = 6” Then A = 1 2 (4)(6) A = 1 2 (24) A = 12 π‘–π‘›π‘β„Žπ‘’π‘  2

5 Exampleβ€”finding a side
The area of a triangle is 24 π‘π‘š 2 and its height is 3 cm. Find the length of its corresponding base. A = π‘β„Ž 2 for a triangle 24= 3𝑏 2 48 = 3b b = 16 cm

6 Area of a Trapezoid Area = (average of the bases)(height) A = 𝑏 1 + 𝑏 2 2 h 𝑏 1 h 𝑏 2 Remember: height is perpendicular to both bases

7 Example 1--Trapezoid Find the area of the trapezoid. 20 in 25 in 18 in 36 in A = ( 𝑏 1 + 𝑏 2 ) 2 h A = ( ) 2 18 A = (56) 2 18 A = 28(18) A = 504 𝑖𝑛 2

8 Example 2--Trapezoid A = ( 𝑏 1 + 𝑏 2 ) 2 h find the missing height
Find the area of the trapezoid ft 13 ft 16 ft A = ( 𝑏 1 + 𝑏 2 ) 2 h find the missing height Find the hypotenuse Use Pythagorean theorem 13 2 = β„Ž when solved you get …… h = 12 ft A = (11+16) 2 12 A =( ) 12 A = 27(6) A = 162 𝑓𝑑 2

9 Area of a Rhombus The area of a rhombus is half the product of the lengths of its diagonals. A = 𝑑 1 𝑑 2 2 𝑑 2 𝑑 1 Example: Find the area if the diagonals are: 6 in and 8 in A = 𝑑 1 𝑑 2 2 A = 6(8) 2 A = 48 2 A = 24 𝑖𝑛 2

10 Area of a Kite The area of a kite is half the product of the lengths of its diagonals. 𝑑 1 A = 𝑑 1 𝑑 2 2 𝑑 2 Example 1: Kite with diagonals 9 cm & 8 cm A = 𝑑 1 𝑑 2 2 A = 9(8) 2 A = 72 2 A = 36 π‘π‘š 2

11 Example 2--Kite Find the area of the kite. 5” 4” A = 𝑑 1 𝑑 2 2 6” Find diagonals Watch for right triangles 5 2 = π‘₯ 2 so x = 3” One diagonal is 2(4) = 8 inches Other diagonal = = 9 inches A = 8(9) 2 A = 72 2 A = 36 𝑖𝑛 2

12 Formulas Square: A = bh Rectangle: A = bh Parallelogram: A = bh Trapezoid: A = 𝑏 1 + 𝑏 2 2 h Triangle: A = Β½ bh Rhombus: A = 𝑑 1 𝑑 2 2 Kite: A = 𝑑 1 𝑑 2 2

13 Area Addition Postulate
The area of a region is equal to the sum of the areas of its nonoverlapping parts. Best way to find this area is to find the area of rectangle + area of triangle

14 Exampleβ€”Partitioning Shapes
Find the area of the shape below: Find the sum of the areas of the rectangle and the triangle A = bh A = π‘β„Ž 2 A = 4(14) A = (12)(5) 2 A = 56 A = 30 total area: = 86 𝑒𝑛𝑖𝑑𝑠 2


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