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Ex 9.2, 7 - Chapter 9 Class 7 Perimeter and Area
Last updated at December 16, 2024 by Teachoo
Perimeter of a combination of shapes
Chapter 9 Class 7 Perimeter and Area
Concept wise
- Basic Definitions
- Square and Rectangle
- Converting Square to Rectangle
- Area of parallelogram
- Area of triangle
- Circumference of circle
-
Perimeter of a combination of shapes
- Area of circle
- Area of combination of figures - two circles, circle and square
- Finding Area of rectangle with path outside/inside
- Finding Area of rectangle with cross roads
- Finding Area of rectangle with triangle
Transcript
Ex 9.2, 7 Find the perimeter of the adjoining figure, which is a semicircle including its diameter. Diameter = AB = 10 cm Radius = r = š·ššššš”šš/2 = 10/2 = 5 cm Now, Perimeter of semicircle = Length of semicircle part + Length of AB Thus, Perimeter of semicircle = š/š Ć 2šr + 2r = šr + 2r = 3.14 Ć 5 + 2 Ć 5 = 314/100 Ć 5 + 10 = ššš/šš + 10 = 157/10 + 10 = 15.7 + 10 = 25.7 cm ā“ Perimeter is 25.7 cm
