Q. 34.1( 10 Votes )

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

Answer :

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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