Question

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

Class 9 - Math - Areas of Parallelograms and Triangles Page 159"

Answer 1

If a parallelogram and a triangle are on the same base and between the same parallels then area of the triangle is half the area of the parallelogram.

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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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Hope this will help you...

 

Answer 2

Answer:

(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 1