Question

In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(APB) + ar(PCD) = ½ ar(ABCD)

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

[Hint : Through P, draw a line parallel to AB.]​

Answer 1

Given:

P is a point in the interior of a parallelogram ABCD

To show:

(i) ar(APB) + ar(PCD) = ½ ar(ABCD)  

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

Construction:

Draw a line EF parallel to AB through the point P

Solution:

(i). ar(APB) + ar(PCD) = ½ ar(ABCD)  :

We have,

EF // AB ..... [construction]

Also, AD // BC since ABCD is a parallelogram

∴ AE // BF

So, we can conclude

⇒ ABFE is a parallelogram ..... [since the opposite facing sides of a parallelogram are parallel to each other]

Similarly, we can also prove that ⇒ EFCD is a parallelogram

We know that → If a parallelogram and a triangle lie on the same base and between the same parallel lines then the area of the triangle is half the area of the parallelogram.

We have,

Δ APB and parallelogram ABFE on the same base AB and between same parallel lines AB & EF

∴ Ar(Δ APB) = ½Ar(//gm ABFE) .... (1)

and

Δ PCD and parallelogram EFCD on the same base CD and between same parallel lines CD & EF

∴ Ar(Δ PCD) = ½Ar(//gm EFCD) .... (2)

Now, on adding eq. (1) & (2), we have

Ar(Δ APB) + Ar(Δ PCD) = ½Ar(//gm ABFE) + ½Ar(//gm EFCD)

⇒ Ar(Δ APB) + Ar(Δ PCD) = ½[Ar(//gm ABFE) + Ar(//gm EFCD)]

⇒ $\boxed{\bold{Ar(\triangle APB) + Ar(\triangle PCD) = \frac{1}{2} [Ar(//gm ABCD)]}}$

Hence proved

(ii). ar(Δ APD) + ar(Δ PBC) = ar(Δ APB) + ar(Δ PCD):

L.H.S.

= Ar(Δ APD) + Ar(Δ PBC)

= Ar(//gm ABCD) - [Ar(Δ APB) + Ar(Δ PCD)]

= 2[Ar(Δ APB) + Ar(Δ PCD)] - [Ar(Δ APB) + Ar(Δ PCD)]

= Ar(Δ APB) + Ar(Δ PCD)

= R.H.S.

Thus, $\boxed{\bold{ Ar(\triangle APD) + Ar(\triangle PBC) = Ar(\triangle APB) + Ar(\triangle PCD)}}$

Hence proved

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