Q3 of 8 Page 244

Sahana drew a rectangle ABCD of which AB = 4 cm and BC = 6 cm. I draw a triangle with equal area of that rectangle.

A. Draw the given rectangle ABCD.



B. Draw the diagonal AC of rectangle ABCD.


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


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



Proof:


∆ACE = ∆ADC (on same base AC and between same parallels AC and DE)


∴ ∆ACE = ∆ADC


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


∴ rectangle ABCD = ∆ABE


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