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Ex 9.1, 7 - Chapter 9 Class 7 Perimeter and Area
Last updated at Dec. 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, 7 ΔABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of Δ ABC. Also find the length of AD.We need to find area ∆ABC, As AB ⊥ AC Taking AB as height & AC as base Here, Base = AC = 12 cm Height = AB = 5 cm Area of ∆ABC = 1/2 × Base × Height = 𝟏/𝟐 × 12 × 5 = 6 × 5 = 30 cm2 Now, we have to find length of AD i.e. height corresponding to base BC So, finding Area using AD as height and BC as base Base = BC = 13 cm Height = AD = ? Area of ∆ABC = 1/2 × Base × Height 30 = 1/2 × BC × AD 30 = 𝟏/𝟐 × 13 × AD 30 = 13/2 × AD 30 × 2/13 = AD 60/13 = AD AD = 𝟔𝟎/𝟏𝟑 cm ∴ Area of ∆ABC is 30 cm2 and Length of AD is 𝟔𝟎/𝟏𝟑 cm
