Example 3 - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar

Transcript

Example 3 An equilateral triangle with side a units. Let’s find Area of Triangle using both Herons formula and Height base formula Herons formula Since all sides are a, a, a So, a = a, b = a, c = a Here, s = (𝒂 + 𝒂 +𝒂)/𝟐 = 𝟑𝒂/𝟐 Now , Area of Triangle = √(𝑠 (𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) = √(3𝑎/2 (3𝑎/2−𝑎)(3𝑎/2−𝑎)(3𝑎/2−𝑎)) = √(𝟑𝒂/𝟐 ×𝒂/𝟐×𝒂/𝟐×𝒂/𝟐) = √((3𝑎^4)/2^4 ) = √((3(𝑎^2 )^2)/(2^2 )^2 ) = (√3 𝑎^2)/2^2 = (√𝟑 𝒂^𝟐)/𝟒 square units Using Height base Formula Drawing AD ⊥ BC Now, Base = BC = a Height = AD = h In an equilateral triangle, perpendicular and median are same ∴ D is mid-point of BC So, CD = 𝒂/𝟐 Now, in right ∆ ADC 𝐴𝐶^2=𝐴𝐷^2+𝐶𝐷^2 𝒂^𝟐=𝒉^𝟐+(𝒂/𝟐)^𝟐 𝑎^2=ℎ^2+𝑎^2/4 𝑎^2−𝑎^2/4=ℎ^2 〖3𝑎〗^2/4=ℎ^2 ℎ^2=〖3𝑎〗^2/4 ℎ=√(〖3𝑎〗^2/4) ℎ=√(〖3𝑎〗^2/2^2 ) 𝒉=(√𝟑 𝒂)/𝟐 Now, Area of ∆ ABC = 1/2 × Base × Height = 1/2 × 𝑎 ×(√3 𝑎)/2 = 1/2 × 𝑎 ×(√3 𝑎)/2 Area of ∆ ABC = 1/2 × Base × Height = 1/2 × 𝑎 ×(√3 𝑎)/2 = (√𝟑 𝒂^𝟐)/𝟒 square units Thus, we found same area using both formulas