Answer:

2. Regular hexagon of side 6 cm.

CONSIDER a regular hexagon P Q R S T U in which PQ=QR=RS=ST=TU=UP= 6 cm.

Join Q and U , then T and R.

Sum of all angles of Regular hexagon = 180° × (6-2)

= 180° × 4

= 720°

All interior angles of regular hexagon = 720° ÷ 6

= 120°

As, PU = QP=6 cm

→∠PUQ = ∠PQU [ if SIDES are equal then angle opposite to them are equal]

→ ∠P + ∠PUQ + ∠PQU = 180° → [Angle sum property of triangle]

→ 120° + 2∠PUQ = 180°

→ 2∠PUQ = 180°- 120°

→ ∠PUQ = 60° ÷ 2 = 30°

Draw , PH ⊥ UQ and SJ⊥TR.→[ Perpendicular from opposite vertex in an ISOSCELES triangle divides the side on which perpendicular is falling in two equal parts.]

Cos 30° =

→

→ UH = 3 √3 cm , So U Q = 2 × UH =2 ×3 √3 cm= 6√3 cm

Sin 30° =

As, sin 30° =

PH = 3 cm

→Area (ΔPUQ) = cm²

Area(ΔPUQ) = Area(ΔTRS)= 18 √3 cm² ∵ [ΔPUQ and Δ TRS are congruent by SAS, PU=TS, PQ=SR, and UQ= TR]

Now consider rectangle URTQ

→Area (Rectangle UQRT) = UQ × QR → [LENGTH × Breadth=Area of Rectangle]

= 6 √3 × 6

= 36 √3 cm²

→Area Hexagon (P Q R STU)

= Area(ΔPQU) + Area rectangle (UQRT) + Area(ΔTRS)

= 18 √3 + 36 √3 +18 √3

= 72 √3 cm²

= 124 .704 cm²