Question

if E,F,G,H are respectively the mid points of sides of a parallelogram ABCD , show that ar (EFGH)= 1/2ar (ABCD).

Answer 1

Step-by-step explanation:

▶ Given :-

→ 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 .

▶ To Prove :-

→ ar(EFGH) = ½ ar(ABCD).

▶ Construction :-

→ Join FH , such that FH || AB || CD .

Since, FH || AB and AH || BF .

→ •°• ABFH is a ||gm .

∆EFH and ||gm ABFH are on same base FH and between the same parallel lines .

•°• ar( ∆ EFH ) = ½ ar( ||gm ABFH )............. (1) .

▶ Now,

Since, FH || CD and DH || FC .

→ •°• DCFH is a ||gm .

∆FGH and ||gm DCFH are on same base FH and between the same parallel lines .

•°• ar( ∆ FGH ) = ½ ar( ||gm DCFH )............. (2) .

▶ On adding equation (1) and (2), we get

==> ar( ∆ EFH ) + ar( ∆ FGH) = ½ ar( ||gm ABFH ) + ½ ar( ||gm DCFH ) .

==> ar( ∆ EFH ) + ar( ∆ FGH) = ½ [ ar( ||gm ABFH ) + ar( ||gm DCFH ) ] .

•°• ar(EFGH) = ½ ar(ABCD).

✔✔ Hence, it is proved ✅✅.

THANKS

Answer 2

Answer:

  ar (EFGH) = $\frac{1}{2}$ ar (ABCD)  proved

Step-by-step explanation:

Explanation:

Given , ABCD is a parallelogram in which ,

E,F,G and H are the mid point of side AB , BC , CD and AD .

oin EF , FG , GH and EH .

And also join AC and HF .

Step 1:

In ΔABC ,

E  and F is the mid-point of AB and BC      [Given]

⇒EF || AC and

EF = $\frac{1}{2} AC$  ........(i)

Similarly in ΔADC

G and H are the mid-point of CD and AD    [Given ]

⇒ HG || AC  and HG = $\frac{1}{2}AC$  .......(ii)

From (i) and (ii) we get ,

EF || HG and EF = HG   ,

Therefore , EFGH is a parallelogram .

Step 2:

Now , in quadrilateral ABFH  we have

AD = BC  ⇒$\frac{AD}{2} = \frac{BC}{2}$

⇒ HA = FB and  HA ||FB .

So , ABFH  is a parallelogram

[one  pair of  opposite side are equal and parallel]

Now , ΔHEF  and parallelogram HABF are on the same base HF .

ar(ΔHEF) = $\frac{1}{2}$ ar(HFCD)   ..........(iii)

Similarly ,

ar(ΔHGF) = $\frac{1}{2}$ar(HFCD) .........(iv)

On adding (iii) and (iv ) we get ,

ar(ΔHEF) + ar(ΔHGF) = $\frac{1}{2}$ [ar(HABF) + ar(HFCD)] .

⇒ ar (EFGH) = $\frac{1}{2}$ ar (ABCD)

Final answer:

Hence , here we proved that  ar (EFGH) = $\frac{1}{2}$ ar (ABCD) .

#SPJ2