Question

find the area of : (1) regular pentagon ABCDE of side 5 cm and AD = BD =4 cm (2)regular hexagon of side 6 cm

Answer 1

Solution:

→The Meaning of word regular in polygon is used that a polygon having all sides equal and all angles (exterior and interior) equal.

1. 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 = $\frac{1}{2}$ × Base × Height

Area (Δ ADE) = $\frac{1}{2}\times \sqrt{21}\times4= 2\sqrt{21}  (cm)^{2}$

→Similarly , In ΔADB , length of perpendicular DN is given by the method used above =$\sqrt{4^{2} -  [\frac{5}{2}]^{2}}$= $\frac{\sqrt{39}} {2}$

Area ( ΔDAB)= $\frac{5}{2} \times\sqrt{39}$ cm²

Area of pentagon A B CD E = Ar(ΔADE) + Ar(DAB) + Ar(ΔDBC)

                                              =2× 2√21 + $\frac{5}{2} \times\sqrt{39}$ ⇒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° =$\frac{UH}{PU} = \frac{UH}{6}$

→$\frac{\sqrt{3}} {2} = \frac{UH}{6}$

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

Sin 30° = $\frac{PH}{UP}=\frac{PH}{6}$

As, sin 30° =$\frac{1}{2}$

$\frac{PH}{6}=\frac{1}{2}$

PH = 3 cm

→Area (ΔPUQ) = $\frac{1}{2} \times (PH)\times(UQ)=\frac{1}{2} \times(6\sqrt{3})\times6 = 18\sqrt{3}$ 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