Dear Student,

Given :  ABCD is a square of side 7 cm. If AE = FC = CG = HA = 3 cm, So

BE  =  BF = DG  = DH  = 4  ( We get from : 7 - 3 = 4 )  , So

$∆$ AEH , $∆$ BEF , $∆$ CGF and $∆$ DGH are isosceles right angle triangle ( As two sides equal and all have one angle at 90$°$ from vertex of square ABCD )

i ) In $∆$ AEH from angle sum property we get

$\angle$ AEH + $\angle$ AHE  + $\angle$ HAE  =  180$°$

$\angle$ AEH  + $\angle$AEH  + 90$°$ = 180$°$                   ( Here $\angle$ AEH  = $\angle$AHE from base angle theorem as we know AE = HA and $\angle$ HAE = 90$°$ vertex angle of square ABCD )

2 $\angle$ AEH = 90$°$ ,

$\angle$ AEH = 45$°$                         --- ( 1 )

And in $∆$ BEF from angle sum property we get

$\angle$ BEF + $\angle$ BFE  + $\angle$ EBF  =  180$°$
$\angle$ BEF +  $\angle$BEF + 90$°$ = 180$°$                   ( Here $\angle$ BEF = $\angle$ BFE from base angle theorem as we know BE = BF and $\angle$ EBF = 90$°$ vertex angle of square ABCD )

2 $\angle$ BEF = 90$°$ ,

$\angle$ BEF = 45$°$                         --- ( 2 )

And

$\angle$ AEH  + $\angle$ HEF + $\angle$ BEF  = 180$°$  ( Linear pair angles )

45$°$ + $\angle$ HEF + 45$°$   =180$°$  ( From equation 1 and 2 we get )

$\angle$ HEF  = 90$°$ , Similarly we can show : $\angle$ EFG = 90$°$ , $\angle$ FGH = 90$°$ and $\angle$ GHE = 90$°$

Therefore,

EFGH is a rectangle .                                           ( Hence proved )

ii ) From Pythagoras theorem we get in $∆$ AEH  :

HE2 = AE2 + HA2 , Substitute values we get

HE2 = 32 + 32 ,

HE2 = 9 + 9 ,

HE2 = 18 ,

HE = 3 $\sqrt{2}$ cm   and FG = 3 $\sqrt{2}$  ( Opposite sides of rectangle EFGH as we proved in previous part )

And now we apply Pythagoras theorem and get in $∆$ BEF  :

EF2 = BE2 + BF2 , Substitute values we get

EF2 = 42 + 42 ,

EF2 = 16 + 16 ,

EF2 = 32 ,

EF = 4 $\sqrt{2}$ cm   and GH = 4 $\sqrt{2}$  ( Opposite sides of rectangle EFGH as we proved in previous part )

Then,

We know area of rectangle  =  Length $×$ Breadth  and Perimeter of rectangle =  2 ( Length  + Breadth ) , So

Area of EFGH = 3 $\sqrt{2}$  $×$ 4 $\sqrt{2}$ = 24 cm2                                                                           ( Ans )

Perimeter of EFGH = 2 (  3 $\sqrt{2}$  + 4 $\sqrt{2}$ ) =  2 ( 7 $\sqrt{2}$  )  =  14 $\sqrt{2}$ cm                             ( Ans )