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# 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).

**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 a parallelogram.

We can observe in the figure shown below that triangle GHF and parallelogram DHFC are on the same base and between parallel lines.

Similarly, triangle EHF and Parallelogram AHFB are on the same base and between parallel lines. By using these results, we can find areas and add them to get the required result.

Let us join HF.

In parallelogram ABCD,

AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)

AB = CD (Opposite sides of a parallelogram are equal)

⇒ 1/2 AD = 1/2 BC and AH || BF

⇒ AH = BF and AH || BF (Since H and F are the mid-points of AD and BC)

Therefore, ABFH is a parallelogram.

Since ∆HEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,

∴ Area (ΔHEF) = 1/2 Area (ABFH) .....(1)

Similarly, Area (ΔHGF) = 1/2 Area (HDCF) .....(2)

On adding Equations (1) and (2), we obtain

Area (ΔHEF) + Area (ΔHGF) = 1/2 Area (ABFH) + 1/2 Area (HDCF) ...... (3)

From the figure, we can see that,

Area (ΔHEF) + Area (ΔHGF) = Area (EFGH) ....... (4)

1/2 Area (ABFH) + 1/2 Area (HDCF) = 1/2 Area (ABCD) ..... (5)

From (3) & (4) and (5) we have,

Area (EFGH) = 1/2 Area (ABCD)

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

**Video Solution:**

## 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).

NCERT Maths Solutions Class 9 Chapter 9 Exercise 9.2 Question 2

**Summary:**

If ABCD is a parallelogram such that E, F, G, and H are respectively the mid-points of the sides of a parallelogram, we have proved that area of (EFGH) = 1/2 area of (ABCD).

**☛ Related Questions:**

- 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)
- 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?

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