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