Question

If E,F,G and H are respectively the mid-points of
the sides of a parallelogram ABCD, show that
ar (EFGH)= ar (ABCD).​

Answer 1

$\huge\red{SOLUTION:-}$

♨ CORRECT QUESTION:-

If E, F, G and H are respectively the mid-points of the sides of a ||gm ABCD, then show that ar(EFGH)= 1/2 ar(ABCD)

♨ ANSWER:-

☯ Given:-

E, F, G and H are respectively the mid-points of the sides of a ||gm ABCD.

☯ To prove:-

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

☯ Construction:-

Join OF, OG, OH and OE. Also, join AC and BD.

☯ Proof:-

In ∆BCD,

Since F and G are the mid points of BC and DC respectively.

Therefore, FG || BD -----> (i) [In a ∆le, the line segment joining the mid-points of any two sides is || to the third side]

In ∆BAD,

Since E and H are the mid-points of AB and AD respectively,

Therefore, EH || BD ----> (ii)

Now, from (i) and (ii), EH || BD. -----(iii)

Similarly, we can prove that

EH || HG -----(iv)

From (iii) and (iv),

Quad. EFGH is a IIgm.

[A quad. is a parallelogram if its opposite sides are equal]

Now,

Since, F is the mid of CB and O is the mid of CA,

Therefore,

FO || BA [In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and is half of it]

=> FO || CG ----> (v) [ Since BA || CD (opposite sides of a ||gm are parallel) and since BA || CG]

and

FO = 1/2 BA

= 1/2 CD [ Since opposite sides of a ||gm are equal]

= CG -----> (vi) [Since G is the mid of CD]

In view of (v), and (vi),

Quad. OFCG is a parallelogram.

Therefore,

OP=PC [Since, diagonals of a ||gm bisect each other]

Since ∆OPF and ∆CPF have equal bases (Since OP=PC) and have a common vertex F.

Therefore,

Their altitudes are also the same.

=> ar(∆OPF) = ar(∆CDF)

Similarly, ar(∆OQF) = ar(∆BQF)

Now, on adding, we get

ar(∆OPF) + ar(∆OQF) = ar(∆CPF) + ar(∆BQF)

=> ar(||gm OQFP) = ar (∆CPF) + ar (∆BQF) ...(vii)

Similarly,

ar( ||gm OPGS) = ar(∆GPC) + ar(∆DSG) ...(viii)

ar(IIgm OSHR) = ar(∆DSH) + ar(∆HAR) ...(ix)

ar(IIgm OREQ) = ar (∆ARE) + ar (∆EQB) ...(x)

Adding the corresponding sides of (vii), (viii), (ix) and (x), we get,

ar(||gm EFGH) = {ar (∆CPF) + ar (∆GPC)}

+ {ar(∆DSG) + ar(∆DSH)} +{ ar (∆HAR) + ar(∆ARE)} + ar(∆BQF) + ar(∆EQB)}

= ar(∆FCG) + ar(∆GDH) + ar (∆HAE) + ar(∆EBF)

= 1/2 ar (||gm ABCD)

HENCE PROVED!!!..

_________________________

$<marquee>★Hope it helps★$

Answer 2

→ ABCD is a ||gm . → E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a ||gm ABCD . → ar(EFGH) = ½ ar(ABCD).

Hope it's helpful dear.................