In the figure given below, ABCD is a square of side 7 cm. If

AE = FC = CG = HA = 3 cm,

(i) prove that EFGH is a rectangle.

(ii) find the area and perimeter of EFGH.

(i) Given, ABCD is a square of side 7 cm and AE = FC = CG = HA = 3 cm.

From figure,

DH = DA - HA = 7 - 3 = 4 cm,

GD = DC - CG = 7 - 3 = 4 cm,

EB = AB - AE = 7 - 3 = 4 cm,

FB = BC - FC = 7 - 3 = 4 cm.

Since in square, sides are perpendicular to each other,

By pythagoras theorem,

In right angle triangle HAE,

⇒ HE² = HA² + AE²

⇒ HE² = 3² + 3²

⇒ HE² = 9 + 9

⇒ HE² = 18

⇒ HE = $\sqrt{18} = 3\sqrt{2}$ cm.

In right angle triangle FCG,

⇒ GF² = GC² + FC²

⇒ GF² = 3² + 3²

⇒ GF² = 9 + 9

⇒ GF² = 18

⇒ GF = $\sqrt{18} = 3\sqrt{2}$ cm.

In right angle triangle GDH,

⇒ GH² = GD² + DH²

⇒ GH² = 4² + 4²

⇒ GH² = 16 + 16

⇒ GH² = 32

⇒ GH = $\sqrt{32} = 4\sqrt{2}$ cm.

In right angle triangle EBF,

⇒ EF² = EB² + FB²

⇒ EF² = 4² + 4²

⇒ EF² = 16 + 16

⇒ EF² = 32

⇒ EF = $\sqrt{32} = 4\sqrt{2}$ cm.

In isosceles triangle AEH,

∠E = ∠H = x (let) (As angles opposite to equal side in isosceles triangle are equal).

So, ∠A + ∠E + ∠H = 180°

90° + x + x = 180°

2x = 90°

x = 45°

Similar is the case for triangles EBF, GDH, FCG.

From figure,

∠AEH + ∠HEF + ∠FEB = 180° (Linear pairs)

45° + ∠HEF + 45° = 180°

∠HEF = 90°

Since, angles of EFGH = 90° and EH = GF and HG = EF.

Hence, proved that EFGH is a rectangle.

(ii) Area of EFGH = EF × GF = $4\sqrt{2} \times 3\sqrt{2}$ = 24 cm².

Perimeter of EFGH = 2(EF + GF) = $2(4\sqrt{2} + 3\sqrt{2}) = 14\sqrt{2}$ cm.

Hence, area = 24 cm² and perimeter = $14\sqrt{2}$ cm.