Question

please solve it friends

Answer 1

Triangles on the same base and between the

HEY MATE HERE IS YOUR ANSWER

same parallels are equal in area.

Diagonals of a parallelogram divides it into two Triangles of equal areas

Given:

Two parallelograms ABCD and PBQR.

To show:

ar (ABCD) = ar (PBQR).

Proof:

Join AC & PQ .

Now  △ACQ & △APQ are on the same base AQ and between the

same parallel lines AQ and CP

ar(△ACQ) = ar(△APQ)

ar(△ACQ) – ar(△ABQ) = ar(△APQ) – ar(△ABQ)

[Subtracting ar(△ABQ) from both sides]

ar(△ABC) = ar(△QBP) — (i)

AC and QP are diagonals ABCD and PBQR.

Thus,

ar(ABC) = 1/2 ar(||gm ABCD) — (ii)

ar(QBP) = 1/2 ar(||gm PBQR) — (iii)

[Diagonals of a parallelogram divides it into

two Triangles of equal areas]

From (ii) and (ii),

1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR)

 ar(||gm ABCD) = ar(||gm PBQR)

Hope

this will help you...