Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 9.1, 5 - Chapter 9 Class 7 Perimeter and Area
Last updated at May 18, 2023 by Teachoo
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 9.1, 5 PQRS is a parallelogram (Fig 11.23). QM is the height from Q to SR & QN is the height from Q to PS. If SR = 12 cm & QM = 7.6 cm. Find: (a) the area of the parallelogram PQRS (b) QN, if PS = 8 cm Finding area of parallelogram Using QM as height and SR as base Here, Base = SR = 12 cm Height = QM = 7.6 cm Area of parallelogram = Base × Height = SR × QM = 12 × 7.6 = 12 × 76/10 = 912/10 = 91.2 cm2 Now, we have to find QN, if PS = 8 cm Our figure looks like So, Base = PS = 8 cm Height = QN = ? Area of parallelogram = Base × height 91.2 = PS × QN 91.2 = 8 × QN 91.2/8 = QN 912/(8 × 10) = QN 114/10 = QN 11.4 = QN QN = 11.4 cm ∴ Area of parallelogram is 91.2 cm2 and length of QN is 11.4 cm
