Question 11
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
Given: A pentagon ABCDE where
BF AC
To prove: ar (ACB) = ar (ACF)
Proof :
ACB and ACF
lie on the same base AC
and are between the same parallels AC and BF.
ar( ACB) = ar( ACF)
Question 11
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(ii) ar (AEDF) = ar (ABCDE)
In part(i), we proved that
ar( ACB) = ar( ACF)
Adding ar(AEDC) both sides
ar( ACB) + ar(AEDC) = ar( ACF) + ar(AEDC)
Area (ABCDE) = Area (AEDF)
Hence proved

Made by

Davneet Singh

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 and Computer Science at Teachoo

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