Ex 9.1, 8 - Chapter 9 Class 7 Perimeter and Area
Last updated at April 16, 2024 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, 8 ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of Δ ABC. What will be the height from C to AB i.e., CE? Finding area of ∆ABC Using BC as base & AD as height Here, Base = BC = 9 cm Height = AD = 6 cm Area of ∆ABC = 1/2 × Base × Height = 1/2 × BC × AD = 𝟏/𝟐 × 9 × 6 = 9 × 3 = 27 cm2 Now, we have to find EC which is height corresponding to base AB So, finding Area of Δ ABC using EC as height and AB as base So, Base = AB = 7.5 cm Height = EC = ? Area of ∆ABC = 1/2 × Base × Height 27 = 𝟏/𝟐 × 7.5 × EC (27 × 2)/7.5 = EC (𝟐𝟕 × 𝟐 × 𝟏𝟎)/𝟕𝟓 = EC (9 × 2 × 10)/25 = EC (9 × 2 × 2)/5 = EC 𝟑𝟔/𝟓 = EC 7.2 = EC EC = 7.2 cm Therefore, Area of ∆ABC is 27 cm2 & Height CE is 7.2 cm
