Composite Figures with Circles Grade 7 Composite Figures with Circles Solve each of the following problems. You must show your work including written details on your decision-­‐making, formulas used, numerical values used, calculation details and units. Find the area of the shaded section of the following figures. 8 m PUBLIC SCHOOLS OF NORTH CAROLINA State Board of Education | Department of Public Instruction AIG ~ IRP Academically and/or Intellectually Gifted Instructional Resources Project Composite Figures with Circles Grade 7 On each side of a regular hexagon is a semicircle. Find the area and perimeter of the following figure. Find the area and perimeter of the following shape. PUBLIC SCHOOLS OF NORTH CAROLINA State Board of Education | Department of Public Instruction AIG ~ IRP Academically and/or Intellectually Gifted Instructional Resources Project Composite Figures with Circles Grade 7 Composite Figures with Circles -­‐ Key Solve each of the following problems. You must show your work including written details on your decision-­‐making, formulas used, numerical values used, calculation details and units. Find the area of the shaded section of the following figures. 8m
Area of the square = bh = 8 x8 = 64 m2 Area of one circle = πr2 = 3.14 x 22 = 12.56 m2 Area of 4 circles = 50.24 m2 Area of shaded section = 64 – 50.24 = 13.76 m2 The shaded section consists of a rectangle with 4 semi-­‐circles (2 full circles cut out). 2
Area of rectangle = bh = 10 x 22 = 220 m Area of 1 circle = πr2 = 3.14 x 52 = 78.5 m2 2
Area of 2 circles = 157 m Area of shaded section = 220 – 157 = 63m2 PUBLIC SCHOOLS OF NORTH CAROLINA State Board of Education | Department of Public Instruction AIG ~ IRP Academically and/or Intellectually Gifted Instructional Resources Project Composite Figures with Circles Grade 7 On each side of a regular hexagon is a semicircle. Find the area and perimeter of the following figure. The area of the shape can be broken down into 6 semi-­‐
The area of the shape can be broken down into 6 semi-­‐circles (3 full circles) and 6 triangles. Area of 3 circles =3 πr2 = 3 x 3.14 x 42 = 150.72 mm2 Area of 6 triangles = 6(½ bh) =6(½)( 8)(8) = 192 mm2 Shape area = 150.72 + 192 = 342.72 mm2 The perimeter of the shape is 6 semi-­‐circles (3 full circles) Circumference of 3 circles = 3πd = 3 x 3.14 x 8 = 75.26 mm Find the area and perimeter of the following shape. The area of the shape is 2 semi-­‐circles of different size. r1 = radius of the larger semi-­‐circle r2 = radius of the smaller semi-­‐circle Area = ½ πr12 + ½ πr22 Area = ½ x 3.14 x 2.52 + ½ x 3.14 x 1.52 = 13.345 cm2 The perimeter of the shape includes the circumference of two different semi-­‐circles plus the lines that connect them. Perimeter = ½ πd1 + ½ πd2 + 2 = ½ x 3.14 x 5 + ½ x 3.14 x 3 + 2 = 14.56 cm PUBLIC SCHOOLS OF NORTH CAROLINA State Board of Education | Department of Public Instruction AIG ~ IRP Academically and/or Intellectually Gifted Instructional Resources Project