Question

If ABCD is a parallelogram, then prove that

ar(Δ ABD) = ar(Δ BCD) = ar(Δ ABC)=ar(Δ ACD) = 1/2 ar(||gm ABCD)

Answer 1

ABCD is a parallelogram.

When we join the diagonal of parallelogram, it divides it into two quadrilaterals.

$Step 1: Let \: \: \: AC \: \: is \: \: the \: \: diagonal, \: \: then, \: \: Area \: \: (ΔABC) = Area \: \: (ΔACD) = 1/2 \: \: (Area \: \: of \: \: {ll}^{gm}  ABCD)$

Step 2: Let BD be another diagonal

Area (ΔABD) = Area (ΔBCD) = 1/2( Area of llgm ABCD)

Now,

From Step 1 and step 2, we have

Area (ΔABC) = Area (ΔACD) = Area (ΔABD) = Area (ΔBCD) = 1/2(Area of llgm ABCD)

Hence Proved.

Answer 2

Answer:

If ABCD is a parallelogram, then prove that

ar(Δ ABD) = ar(Δ BCD) = ar(Δ ABC)=ar(Δ ACD) = 1/2 ar(||gm ABCD)

Step-by-step explanation:

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