Brahmaguptaβs Formula for the Area of a Cyclic 4-gon
Brahmaguptaβs Formula for the Area of a Cyclic 4-gon
Last updated at May 31, 2026 by Teachoo
Transcript
Example 7 Verify Brahmaguptaβs formula for the case of an isosceles trapezium. Since sides are 2a, c, 2b, c So, a = 2a, b = c, c = 2b, d = c Here, s = (ππ + π + ππ + π)/π = (2π + 2π + 2π)/2 = a + b + c Now , Area of 4-gon = β((π βπ)(π βπ)(π βπ)(π βπ)) = β((π+π+πβ2π)"(π+π+πβc)(π+π+πβ2b)(π+π+πβc)" ) = β((π+πβπ)(π+π)(π+πβπ)(π+π) ) = β((π+πβπ)(π+πβπ) Γ (πβπ)^2 ) = β((π+[πβπ])(πβ[πβπ] ) )Γ β((πβπ)^2 " " ) = β(π^πβ(πβπ)^π )Γ (πβπ) Now, formula of Trapezium is Area = π/π Γ Sum of non-parallel sides Γ Height = 1/2 Γ (2a + 2b) Γ Height = 1/2 Γ 2(a + b) Γ Height = (a + b) Γ Height And, height using Pythagoras theorem is β(π^πβ(πβπ)^π ) So, our Area matches using both formulas = 1/2 Γ 2(a + b) Γ Height = (a + b) Γ Height And, height using Pythagoras theorem is β(π^πβ(πβπ)^π ) So, our Area matches using both formulas