Find the Area of the Star Inside a Regular Pentagon

Find the Area of a five-pointed star which is inscribed inside a regular pentagon with sides of 8 cm

Solution: Area of the Star

Area of the star = Area of the pentagon – Area of the five white triangles

We can divide a regular pentagon into five equal triangles as shown below, where O is the centre of the pentagon

From figure

Area of the Pentagon = 5 × Area of the triangle AOB

∠OBA = ∠OBC = 108/2 = 54°

⇒ ∠OAB = 54°

∠AOB = 180 – (∠OBA + ∠OAB)

⇒ ∠AOB = 180 – (54 + 54) = 72°

Apply sine rule in triangle AOB

AO / sin ∠OBA = AB / sin ∠AOB

\Rightarrow \ AO=\dfrac{AB \times \sin ∠OBA}{\sin ∠AOB}
\Rightarrow \ AO=\dfrac{8 \times \sin 54}{\sin 72}
\Rightarrow \ AO=\dfrac{8 \times \dfrac{\sqrt{5}+1}{4}}{\dfrac{\sqrt{10+2\sqrt5}}{4}}
\Rightarrow \ AO=\dfrac{8 \times (\sqrt{5}+1)}{\sqrt{10+2\sqrt5}}
\Rightarrow \ AO=\dfrac{8\sqrt{5}+8}{\sqrt{10+2\sqrt5}} \ \mathrm{cm}
\mathrm{Area\ of \ the\ triangle \ AOB} = \dfrac{1}{2}\times AB \times AO \times \sin 54
\Rightarrow \ \mathrm{Area\ of \ the\ triangle \ AOB} = \dfrac{1}{2}\times 8 \times \dfrac{8\sqrt{5}+8}{\sqrt{10+2\sqrt5}} \times \dfrac{\sqrt{5}+1}{4}
\Rightarrow \ \mathrm{Area\ of \ the\ triangle \ AOB} = \dfrac{48+16\sqrt{5}}{\sqrt{10+2\sqrt5}} \ \mathrm{cm^2}
\Rightarrow \ \mathrm{Area\ of \ the\ pentagon} = 5 \times \dfrac{48+16\sqrt{5}}{\sqrt{10+2\sqrt5}} \ \mathrm{cm^2}
\Rightarrow \ \mathrm{Area\ of \ the\ pentagon} = \dfrac{240+80\sqrt{5}}{\sqrt{10+2\sqrt5}} \ \mathrm{cm^2}
\Rightarrow \ \mathrm{Area\ of \ the\ pentagon} = \dfrac{240+80\sqrt{5}}{\sqrt{10+2\sqrt5}}  \times \dfrac{\sqrt{10-2\sqrt5}}{\sqrt{10-2\sqrt5}}
\Rightarrow \ \mathrm{Area\ of \ the\ pentagon} = \dfrac{240\sqrt{10-2\sqrt5}+80\sqrt{5} \times \sqrt{10-2\sqrt5}}{4\sqrt{5}}
\Rightarrow \ \mathrm{Area\ of \ the\ pentagon} = {12\sqrt{5} \times \sqrt{10-2\sqrt5}+20\sqrt{10-2\sqrt5}}
\Rightarrow \ \mathrm{Area\ of \ the\ pentagon} = {12 \sqrt{50-10\sqrt5}+20\sqrt{10-2\sqrt5}}

From Figure

∠ATB = ∠ATB = 108°

Then

∠TAB = ∠TBA = (180 – 108)/36 = 72/2 = 36°

Apply sine rule in triangle ABT, then

AT / sin ∠TBA = AB / sin ∠ATB

AT=\dfrac{AB \times  \sin ∠TBA}{ \sin ∠ATB}
\Rightarrow AT=\dfrac{8 \times  \sin 36}{ \sin 108}
\Rightarrow AT=\dfrac{8 \times  \dfrac{\sqrt{10-2\sqrt5}}{4}}{ \dfrac{\sqrt{10+2\sqrt5}}{4}}
\Rightarrow AT=\dfrac{8 \times  \sqrt{10-2\sqrt5}}{\sqrt{10+2\sqrt5}}
\Rightarrow AT=8 \times  \sqrt{\dfrac{(10-2\sqrt5)(10-2\sqrt5)}{(10+2\sqrt5)(10-2\sqrt5)}}
\Rightarrow AT=8 \times  {\dfrac{10-2\sqrt5}{4\sqrt{5}}}
\Rightarrow AT={\dfrac{20-4\sqrt5}{\sqrt{5}}}
\Rightarrow AT=4\sqrt5-4
\mathrm{Area\ of \ the\ triangle \ ABT} = \dfrac{1}{2}\times AB \times AT \times \sin 36
\Rightarrow \mathrm{Area\ of \ the\ triangle \ ABT} = \dfrac{1}{2}\times 8 \times (4\sqrt5-4) \times \dfrac{\sqrt{10-2\sqrt5}}{4}
\Rightarrow \mathrm{Area\ of \ the\ triangle \ ABT} =(4\sqrt5-4) \times \sqrt{10-2\sqrt5}
\Rightarrow \mathrm{Area\ of \ the\ 5 \ White \ triangles} =20\sqrt{50-10\sqrt5}-20\sqrt{10-2\sqrt5} \  \mathrm{cm^2}

Area of the star = Area of the pentagon – Area of the five white triangles

\mathrm{Area\ of \ the\ star} ={12 \sqrt{50-10\sqrt5}+20\sqrt{10-2\sqrt5}}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   \ \ \ \ \ \ \ \ \ \ \ \ \ \   \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(20\sqrt{50-10\sqrt5}-20\sqrt{10-2\sqrt5})
\Rightarrow \mathrm{Area\ of \ the\ star} =40\sqrt{10-2\sqrt5}-8 \sqrt{50-10\sqrt5} \ \  \mathrm{cm^2}

That is

Area of the star ≈ 51.987 cm²

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