Q1 of 8 Page 244

Pritam drew a quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm and ∠ABC = 60°, I draw a triangle with equal area of that quadrilateral.

A. Draw the given quadrilateral ABCD.



B. Draw the diagonal DB of quadrilateral ABCD.


C. Draw a parallel line through point A to diagonal DB of quadrilateral ABCD which intersects at F produced BC.


D. Join FD, AF||BD


ΔDFC is the required triangle.



Proof:


∆ABD = ∆BFD (on same base DB and between same parallels DB and AF)


∴ ∆ABD = ∆BFD


∆DBC + ∆ABD = ∆BFD + ∆DBC (adding area of ∆DBC on both sides)


∴ quadrilateral ABCD = ∆DFC


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