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Answer :

Given: E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD

Construction: Let us join HF.

To prove: ar(EFGH) = \(\frac{1}{2} \) ar(ABCD)

Proof: In parallelogram ABCD,

AD = BC and AD || BC

( Opposite sides of a parallelogram are equal and parallel)

Multiplying both sides by \(\frac{1}{2} \) we get,

\(\Rightarrow \) \(\frac{1}{2} \) x AD = \(\frac{1}{2} \) x BC

\(\Rightarrow \) AH = BF and AH || BF

\(\therefore \) ABFH is a parallelogram.

Since, \(\triangle{HEF}\) and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,

\(\therefore \) Area (\(\triangle{HEF}\)) = \(\frac{1}{2} \) Area (ABFH) ...(i)

Similarly, it can be proved that,

Area (\(\triangle{HGF}\)) = \(\Rightarrow \) Area (HDCF) ...(ii)

On adding equations (i) and (ii), we get,

\(\Rightarrow \) Area (\(\triangle{HEF}\)) + Area (\(\triangle{HGF}\)) = \(\frac{1}{2} \) Area (ABFH) + \(\frac{1}{2} \) Area (HDCF)

\(\Rightarrow \) Area [(\(\triangle{HEF}\)) + (\(\triangle{HGF}\))] = [Area (ABFH) + Area (HDCF)]

\(\therefore\) Area (EFGH) = \(\frac{1}{2} \) Area (ABCD)

Hence, proved.

- In Figure, ABCD is a parallelogram, \(AE\perp{DC}\) and \(CF\perp{AD}\). If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
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- In Figure, P is a point in the interior of a parallelogram ABCD. Show that(i) ar (\(\triangle{APB}\) + \(\triangle{PCD}\)) = \(\frac{1}{2} \) ar (ABCD) (ii) ar (\(\triangle{APD}\) + \(\triangle{PBC}\)) = ar(APB) + ar(PCD)[Hint : Through P, draw a line parallel to AB.]
- In Figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show thati)ar(PQRS) = ar(ABRS)ii) ar(AXS) = \(\frac{1}{2} \) 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 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?

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