Last updated at Aug. 10, 2018 by Teachoo

Transcript

Theorem 9.2 (Method 1) Two triangles on the same base (or equal bases) and between the same parallels are equal in area. Given : βADB and βABC are triangles on same base AB and between the same parallels AB and DC To Prove : πr (βADB) = πr (βABC) Proof: We know that Area of triangle = 1/2Γπ΅ππ πΓπ»πππβπ‘ Here AB is base We draw DE β₯ AB So, DE is the height of β ABD β΄ Area β ABD = 1/2Γπ΅ππ πΓπ»πππβπ‘ = 1/2Γπ΄π΅Γπ·πΈ Here AB, is base We draw CF β₯ AB So, CF is the height of β ABC β΄ Area β ABC = 1/2Γπ΅ππ πΓπ»πππβπ‘ = 1/2Γπ΄π΅ΓπΆπΉ DE & CF are perpendicular between the same parallel line β΄ DE = CF So, from (1) Area β ABD = 1/2Γπ΄π΅Γπ·πΈ = 1/2Γπ΄π΅ΓπΆπΉ = Area β ABC β΄ Area β ABD = Area of β ABC Hence proved. Theorem 9.2 Two triangles on the same base (or equal bases) and between the same parallels are equal in area. Given : βADB and βABC are triangles on same base AB and between the same parallels AB and DC To Prove : πr (βADB) = πr (βABC) Construction : Construct a line through A parallel to be meeting DC at E i.e. AE β₯ BC & Construct a line through B parallel to AD meeting DC at F i.e. BF β₯ AD Proof : From (1) πr (ABFD) = πr (ABCE) 2 πr (βABD) = 2 πr (βABC) πr (βABD) = πr (βABC) Hence, proved

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.