Question

4.
In Fig. 9.16, P is a point in the interior of a
parallelogram ABCD. Show that
ma (APB) + 2 (PCD)= *(ABCD)
G (APD) +ar (PBC)=2(APB) + 2 (PCD)
(Hint: Through P. draw a line parallel to AB.)
Fig. 9.16​

Answer 1

Given: ABCD is a parallelogram

So, AB||CD & AD|| BC

To show:

 (i) ar (APB) + ar (PCD) = ar(ABCD) 

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) 

Proof:

(i)

Through the point P ,draw GH parallel to AB.

In a parallelogram,

AB || GH (by construction) — (i)

Thus,

AD || BC ⇒ AG || BH — (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now,

In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.

∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii)

also,

In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.

∴ ar(ΔPCD) = 1/2 ar(CDGH) — (iv

Adding equations (iii) and (iv),

ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}

 ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii)

A line EF is drawn parallel to AD passing through P.

In a parallelogram,

AD || EF (by construction) — (i)

Thus,

AB || CD ⇒ AE || DF — (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.

∴ ar(ΔAPD) = 1/2 ar(AEFD) — (iii)

also,

In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.

∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}

ar(ΔAPD) + ar(ΔPBC) =1/2 ar(ABCD)

 ar(APD) + ar(PBC) = ar(APB) + ar(PCD)  ( From part i)

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Answer 2

Answer:

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