# 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 is 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?

**Solution:**

Let's draw the figure as shown below.

From the figure, it can be observed that point A divides the parallelogram-shaped field into three parts. These parts are triangular in shape: ΔPSA, ΔPAQ, and ΔQRA

From the figure, we can observe that:

Area of ΔPSA + Area of ΔPAQ + Area of ΔQRA = Area of parallelogram PQRS .....(1)

We know that if a parallelogram and a triangle are on the same base and between the same set of parallel lines, then the area of the triangle is half the area of the parallelogram.

∴ Area (ΔPAQ) = 1/2 Area (PQRS) ......(2)

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

Area (ΔPSA) + Area (ΔQRA) = 1/2 Area (PQRS) .....(3)

Clearly, it can be observed that the farmer must sow wheat in the triangular part ΔPAQ and pulses in the other two triangular parts - ΔPSA and ΔQRA, or wheat in triangular parts - ΔPSA and ΔQRA and pulses in the triangular part ΔPAQ.

In this manner, the farmer can sow wheat and pulses in equal portions of the field separately.

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 9

**Video Solution:**

## 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 field is 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?

NCERT Maths Solutions Class 9 Chapter 9 Exercise 9.2 Question 6

**Summary:**

A farmer was having a field in the form of a parallelogram PQRS, and she took any point A on RS and joined it to points P and Q, then point A divides the field into three triangular parts, and the farmer must sow wheat in triangular part ΔPAQ and pulses in other two triangular parts ΔPSA and ΔQRA or wheat in triangular parts ΔPSA and ΔQRA and pulses in the triangular part ΔPAQ.

**☛ Related Questions:**

- If E, F, G, and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar (ABCD).
- P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
- 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.]
- 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)

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