Q2 of 8 Page 244

Sahana drew a quadrilateral ABCD of which AB = 4 cm, BC = 5 cm, CD = 4.8 cm, DA = 4.2 cm and diagonal AC = 6 cm. I draw a triangle with equal area of that quadrilateral.

A. Draw the given quadrilateral ABCD.


B. Draw the diagonal AC of quadrilateral ABCD.



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


D. DF||AC, ABF is the required triangle.



Proof:


∆ADC = ∆ACF (on same base AC and between same parallels AC and DF)


∴ ∆ ADC = ∆ ACF


∆ABC + ∆ADC = ∆ACF + ∆ABC (adding area of ∆ABC on both sides)


∴ quadrilateral ABCD = ∆ABF


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