polygons. 5 cm 4 cm 4 cm 2 cm 12 cm 25 cm
Answers
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Answer:
Step-by-step explanation:
Area of regular pentagon A B C DE of side 5 cm and AD = B D =4 cm .
→ The sides AD and BD will convert polygon AB C D E in three triangles namely ΔADE, ΔADB, and ΔDCB.
→ΔADE≅ΔDCB ⇒ [SAS, AE=DE=BC=DC= 5 cm, AD=DB=4 cm]
→Draw EM ⊥ AD, and D N ⊥ AB
In an isosceles triangle perpendicular from opposite vertex divides the side on which perpendicular is falling into two equal parts.
Using pythagoras theorem In Δ EMD,
ED = 5 cm, EM =?, DM = 2 cm
EM² = ED² - DM²
= 5² - 2²
= 25 - 4
=21
EM =√ 21
Area of a triangle = × Base × Height
Area (Δ ADE) =
→Similarly , In ΔADB , length of perpendicular DN is given by the method used above ==
Area ( ΔDAB)= cm²
Area of pentagon A B CD E = Ar(ΔADE) + Ar(DAB) + Ar(ΔDBC)
=2× 2√21 + ⇒Ar(ΔADE) =Ar(ΔDBC)
= 4 × 4.5 + 2.5 × 6.24
= 18 + 15.60
= 33.60 cm²
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