Example 4 - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar

Transcript

Example 4 An isosceles triangle with equal sides a units and base 2b units. Let’s find Area of Triangle using both Herons formula and Height base formula Herons formula Since sides are a, a, 2b So, a = a, b = a, c = 2b Here, s = (𝒂 + 𝒂 +𝟐𝒃)/𝟐 = (2𝑎 + 2𝑏)/2 = a + b Now , Area of Triangle = √(𝑠 (𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) = √((𝑎+𝑏)(𝑎+𝑏−𝑎)(𝑎+𝑏−𝑎)(𝑎+𝑏−2𝑏)) = √((𝒂+𝒃) × 𝒃 × 𝒃 × (𝒂−𝒃)) = √((𝑎+𝑏)(𝑎−𝑏) × 𝑏^2 ) = √((𝑎^2−𝑏^2) × 𝑏^2 ) = √(𝑎^2−𝑏^2 ) × 𝑏 = 𝒃√(𝒂^𝟐−𝒃^𝟐 ) square units Using Height base Formula Drawing AD ⊥ BC Now, Base = BC = 2b Height = AD = h In an equilateral triangle, perpendicular and median are same ∴ D is mid-point of BC So, CD = 𝟐𝒃/𝟐=𝒃 Now, in right ∆ ADC 𝐴𝐶^2=𝐴𝐷^2+𝐶𝐷^2 𝒂^𝟐=𝒉^𝟐+𝒃^𝟐 𝑎^2−𝑏^2=ℎ^2 ℎ^2=𝑎^2−𝑏^2 𝒉=√(𝒂^𝟐−𝒃^𝟐 ) Now, Area of ∆ ABC = 1/2 × Base × Height = 1/2 × 2𝑏 ×√(𝑎^2−𝑏^2 ) = 𝐛√(𝒂^𝟐−𝒃^𝟐 ) square units Thus, we found same area using both formulas