Theorem 9.2 - Areas of Parallelograms and Triangles

Last updated at April 16, 2024 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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