regular pentagon ABCDE of side 4 cm and AD = BD = 5 cm.
(ii) regular hexagon of side 6 cm.
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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
Answer:
Step-by-step explanation: