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Published byErika Houston Modified over 8 years ago

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**Developing Formulas for Triangles and Quadrilaterals**

Geometry CP1 (Holt 10-1) K. Santos

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**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 ππ 2

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

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Area of a Triangle Area = one half of the product of its base and height A= 1 2 bh or A = πβ 2 Base perpendicular to height h h h b b b If b = 4β and h = 6β Then A = 1 2 (4)(6) A = 1 2 (24) A = 12 πππβππ 2

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**Exampleβfinding a side**

The area of a triangle is 24 ππ 2 and its height is 3 cm. Find the length of its corresponding base. A = πβ 2 for a triangle 24= 3π 2 48 = 3b b = 16 cm

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

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

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**Example 2--Trapezoid A = ( π 1 + π 2 ) 2 h find the missing height**

Find the area of the trapezoid ft 13 ft 16 ft A = ( π 1 + π 2 ) 2 h find the missing height Find the hypotenuse Use Pythagorean theorem 13 2 = β when solved you get β¦β¦ h = 12 ft A = (11+16) 2 12 A =( ) 12 A = 27(6) A = 162 ππ‘ 2

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

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

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

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

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**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

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**ExampleβPartitioning Shapes**

Find the area of the shape below: Find the sum of the areas of the rectangle and the triangle A = bh A = πβ 2 A = 4(14) A = (12)(5) 2 A = 56 A = 30 total area: = 86 π’πππ‘π 2

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