Question

ABCDE is a pentagon.A line through B II to AC meets DC produced at F.Show that:-
(i)ar(ACB)=ar(ACF)
(ii)ar(AEDF)=ar(ABCDE)

Answer 1

answer ma question plzzzzzzzzzzz

Answer 2

Sorry -- I don't know what "ar" as in "ar(ACB)" means. Also, I am uncertain about "produced at F." Line AC intersected line CD at point C when I drew the diagram.

Answer 3

ar(ACB) means area of triangle ACB

Answer 4

ya

Answer 5

the distance between two parallel lines remains constant at every point...... so the hieght of triangles ACB and ACF is same.... and the base is common.... so the area 1/2 * AC * hieght is same.....

Answer 6

will u show me the figure

Answer 7

BF and AC are parallel... got it
???

Answer 8

ya

Answer 9

understood 1st part???

Answer 10

ya

Answer 11

now for the second part.....

Answer 12

k

Answer 13

the area of pentagon ABCDE = ar(ACDE) + ar(ABC)
but in 1st part we proved that ar(ABC)=ar(ACF)
so ar(ABCDE) = ar(ACDE) + ar(ACF) = ar(AFDE)

Answer 14

this is an NCERT question of class 9th i suppose

Answer 15

ya

Answer 16

then for the first part.... Area of triangles is same if they lie on the same base and between parallel lines....... THEorum

Answer 17

ya

Answer 18

wch class r u