Q10 of 25 Page 9

ABCD is a trapezium in which AB||DC. DC is produced to E such that CE=AB, Prove that ar(Δ ABD) = ar(Δ BCE)

Construction:

Draw DG on BA produced and BF⊥ DC.



Proof:


We have,


Area ΔABD = 1/2 × AB × DG …. (1)


Area ΔBCE = 1/2 × CE × BF ….. (2)


Since Area ΔABD and Area ΔBCE are between same parallels so their heights are equal


i.e. DG = BF


Also,


AB = CE (given)


1/2 × AB × DG = 1/2 × CE × BF


∴ from (1) and (2),


Area ΔABD = Area ΔBCE


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8

ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC, such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.

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In a Δ ABC, if L and M are points on AB and AC respectively such that LM||BC. Prove that:

(i) ar(Δ LCM) = ar(Δ LBM)


(ii) ar(Δ LBC) = ar(Δ MBC)


(iii) ar(Δ ABM) = ar(Δ ACL)

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If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.

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In ΔABC, D is the mid-point of AB. P is any point of BC. CQ||PD meets AB in Q. Show that ar (Δ BPQ) = 1/2 ar(Δ ABC)