Question

Class-9
Exercise 9.2

4. In Fig. 9.16, P is a point in the interior of a
parallelogram ABCD. Show that
1
i)
2
(ii) ar (APD) +ar (PBC)= ar (APB)+ar (PCD)
[Hint: Through P, draw a line parallel to AB.]
Fig. 9.16​

Answer 1

Answer:

ye pura figer bana kar show kar na pade ga

Answer 2

Answer:

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

In the parallelogram,

AD || EF (by construction) --- (i)

∴,

AB || CD ⇒ AE || DF --- (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

ΔAPD and parallelogram AEFD are lying on the same base AD and in-between the same

parallel lines AD and EF.

∴ar(ΔAPD) = ½ ar(AEFD) --- (iii)

also,

ΔPBC and parallelogram BCFE are lying on the same base BC and in-between the same

parallel lines BC and EF.

∴ar(ΔPBC) = ½ ar(BCFE) --- (iv)

Adding equations (iii) and (iv),

ar(ΔAPD)+ ar(ΔPBC) = ½ {ar(AEFD)+ar(BCFE)}

⇒ar(APD)+ar(PBC) = ar(APB)+ar(PCD)