Theorems

Chapter 9 Class 9 - Areas of Parallelograms and Triangles [Deleted]
Serial order wise

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12

### 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

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.