Question

In Fig. 9.16, P is a point in the interior of aparallelogram ABCD. Show that
(1) ar (APB) + ar (PCD) = - ar (ABCD)
(ii) ar (APD) +ar (PBC)= ar (APB)+ar​

Answer 1

Answer:

ABCD is a ∥gm such that AB∥CD & AD∥BC. Draw line GM passing through P, parallel to AD & BC. i.e. GH∥AD∥BC.

Here, GH∥AD (By construction) & AG∥DH (∵ AB∥CD & G & H are points on AB & CD respectively).

∴ AGHD is a ∥gm.

Now, △APD and ∥gm AGHD are on the same base AD and between the bank ∥als AD & Gm.

Therefore,

ar △APD =

2

1

ar (AGHD) ⟶(1)

Similarly ar (△PCB) =

2

1

ar (GBCH) ⟶(2)

Adding equations (1) & (2), we get

ar △APD + ar △PCB =

2

1

(ar AGHD + ar GBCH)

⇒ ar △APD + ar △PCB =

2

1

ar ABCD

Also,

2

1

ar ABCD = ar △APB + ar △PCD

∴ ar △APD + ar △PBC = ar △APB + ar △PCD

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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