Theorems

Areas of Parallelograms and Triangles
Serial order wise

### 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 : βADB and βABC are triangles on same base AB and between the same parallels AB and DC To Prove : πr (βADB) = πr (βABC) Proof: We know that Area of triangle = 1/2Γπ΅ππ πΓπ»πππβπ‘ Here AB is base We draw DE β₯ AB So, DE is the height of β ABD β΄ Area β ABD = 1/2Γπ΅ππ πΓπ»πππβπ‘ = 1/2Γπ΄π΅Γπ·πΈ Here AB, is base We draw CF β₯ AB So, CF is the height of β ABC β΄ Area β ABC = 1/2Γπ΅ππ πΓπ»πππβπ‘ = 1/2Γπ΄π΅ΓπΆπΉ DE & CF are perpendicular between the same parallel line β΄ DE = CF So, from (1) Area β ABD = 1/2Γπ΄π΅Γπ·πΈ = 1/2Γπ΄π΅ΓπΆπΉ = 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 : βADB and βABC are triangles on same base AB and between the same parallels AB and DC To Prove : πr (βADB) = πr (βABC) 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) πr (ABFD) = πr (ABCE) 2 πr (βABD) = 2 πr (βABC) πr (βABD) = πr (βABC) Hence, proved