Question

ABCD is a square of side 7cm. If AE=FC=CG=HA=3cm, find the area and perimeter of EFGH.​

Answer 1

Answer:

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

∠ AEH + ∠ AHE + ∠ HAE = 180°

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

2 ∠ AEH = 90° ,

∠ AEH = 45° --- ( 1 )

And in ∆ BEF from angle sum property we get

∠ BEF + ∠ BFE + ∠ EBF = 180°

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

2 ∠ BEF = 90° ,

∠ BEF = 45° --- ( 2 )

And

∠ AEH + ∠ HEF + ∠ BEF = 180° ( Linear pair angles )

45° + ∠ HEF + 45° =180° ( From equation 1 and 2 we get )

∠ HEF = 90° , Similarly we can show : ∠ EFG = 90° , ∠ FGH = 90° and ∠ 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 2‾√ cm and FG = 3 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 2‾√ cm and GH = 4 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 2‾√ × 4 2‾√ = 24 cm2 ( Ans )

Perimeter of EFGH = 2 ( 3 2‾√ + 4 2‾√ ) = 2 ( 7 2‾√ ) = 14 2‾√ cm ( Ans )

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Step-by-step explanation:

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