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

Solution:

If a triangle and parallelogram are on the same base and between the same parallel lines, then the area of the triangle will be half of the area of the parallelogram.

i) Let us draw a line segment EF, passing through the point P and parallel to line segment AB in parallelogram ABCD.

AB || EF (By construction)  .....(1)

We know that ABCD is a parallelogram.

∴ AD || BC (Opposite sides of a parallelogram are parallel)

⇒ AE || BF  ..... (2)

From Equations (1) and (2), we obtain

AB || EF and AE || BF

Therefore, quadrilateral ABFE is a parallelogram.

Similarly, it can be deduced that quadrilateral EFCD is a parallelogram.

It can be observed that ΔAPB and parallelogram ABFE is lying on the same base AB and between the same set of parallel lines AB and EF.

∴ Area (ΔAPB) = 1/2 Area (ABFE)  .....(3)

Similarly, for ΔPCD and parallelogram EFCD, Area (ΔPCD) = 1/2 Area (EFCD)  .....(4)

Adding Equations (3) and (4), we obtain Area (ΔAPB) + Area (ΔPCD) = 1/2 [Area (ABFE) + Area (EFCD)]

Area (ΔAPB) + Area (ΔPCD) = 1/2 Area (ABCD)  .....(5)

ii)

Let us draw a line segment MN, passing through point P and parallel to line segment AD.

In parallelogram ABCD, MN || AD (By construction)  .....(6)

We know that ABCD is a parallelogram.

∴ AB || DC (Opposite sides of a parallelogram are parallel)

⇒ AM || DN.....(7)

From Equations (6) and (7), we obtain

MN || AD and AM || DN

Therefore, quadrilateral AMND is a parallelogram.

It can be observed that ΔAPD and parallelogram AMND is lying on the same base AD and between the same parallel lines AD and MN.

∴ Area (ΔAPD) = 1/2 Area (AMND)  .....(8)

Similarly, for ΔPCB and parallelogram MNCB,

Area (ΔPCB) = 1/2 Area (MNCB)  .....(9)

Adding Equations (8) and (9), we obtain,

Area (ΔAPD) + Area (ΔPCB) = 1/2 [Area (AMND) + Area (MNCB)]

Area (ΔAPD) + Area (ΔPCB) = 1/2 Area (ABCD)  .....(10)

On comparing equations (5) and (10), we obtain Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area(ΔPCD)

☛ Check: NCERT Solutions for Class 9 Maths Chapter 9

---

Video Solution:

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

NCERT Maths Solutions Class 9 Chapter 9 Exercise 9.2 Question 4

Summary:

If P is a point in the interior of a parallelogram ABCD, we have proved that Area of (ΔAPB) + Area of (ΔPCD) = 1/2 Area of (ABCD), and Area of (ΔAPD) + Area of (ΔPBC) = Area of (ΔAPB) + Area of (ΔPCD).

---

☛ Related Questions:

• In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) ar (PQRS) = ar (ABRS) (ii) ar (AXS) = 1/2 ar (PQRS)
• A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields are divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
• In Fig.9.23, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).
• In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4 ar(ABC).