Brahmaguptaโs Formula for the Area of a Cyclic 4-gon
Brahmaguptaโs Formula for the Area of a Cyclic 4-gon
Last updated at May 31, 2026 by Teachoo
Transcript
Example 7 Verify Brahmaguptaโs 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 = (๐๐ + ๐ + ๐๐ + ๐)/๐ = (2๐ + 2๐ + 2๐)/2 = a + b + c Now , Area of 4-gon = โ((๐ โ๐)(๐ โ๐)(๐ โ๐)(๐ โ๐)) = โ((๐+๐+๐โ2๐)"(๐+๐+๐โc)(๐+๐+๐โ2b)(๐+๐+๐โc)" ) = โ((๐+๐โ๐)(๐+๐)(๐+๐โ๐)(๐+๐) ) = โ((๐+๐โ๐)(๐+๐โ๐) ร (๐โ๐)^2 ) = โ((๐+[๐โ๐])(๐โ[๐โ๐] ) )ร โ((๐โ๐)^2 " " ) = โ(๐^๐โ(๐โ๐)^๐ )ร (๐โ๐) Now, formula of Trapezium is Area = ๐/๐ ร Sum of non-parallel sides ร Height = 1/2 ร (2a + 2b) ร Height = 1/2 ร 2(a + b) ร Height = (a + b) ร Height And, height using Pythagoras theorem is โ(๐^๐โ(๐โ๐)^๐ ) So, our Area matches using both formulas = 1/2 ร 2(a + b) ร Height = (a + b) ร Height And, height using Pythagoras theorem is โ(๐^๐โ(๐โ๐)^๐ ) So, our Area matches using both formulas