Find the Area of the Triangle Inside a Semicircle

A semicircle is inscribed inside a square ABCD. Where BD is the diagonal of the square and PC is the tangent of the semicircle. Then find the area of the blue triangle shown in the figure, Here side of the square equals 8 cm

A semicircle is inscribed inside a square ABCD. Where BD is the diagonal of square and PC is the tangent of the semicircle. Then find the area of the blue triangle shown in the figure (Where side of the square equals 8 cm)

Solution

Area of blue triangle = Area of ΔPBR

Connect OC

So, triangle BOC and POC congruent right-angle triangles (ΔBOC ≅ Δ POC), Let assume ∠BOC = ∠POC = θ

From figure PC = BC (PC and BC are tangent of semicircle from C)

∠CPO = 90° (PC is tangent of the semicircle and PO is the radius of the semicircle)

So, triangle BOC and POC congruent right-angle triangles (ΔBOC ≅ Δ POC), Let assume ∠BOC = ∠POC = θ

Apply Pythagoras theorem in ΔBOC

OC^2=OB^2+BC^2
\Rightarrow OC^2=4^2+8^2
\Rightarrow OC^2=16+64
\Rightarrow OC^2=80
\Rightarrow OC=4\sqrt5 \space \mathrm{cm}
\cos \mathrm{\theta} =\dfrac{OB}{OC}
\Rightarrow \cos \mathrm{\theta} =\dfrac{4}{4\sqrt5}
\Rightarrow \cos \mathrm{\theta} =\dfrac{1}{\sqrt5}
\Rightarrow \sin \mathrm{\theta} =\dfrac{2}{\sqrt5}

Connect PB

Now we get a triangle BOP

Apply cosine rule in triangle BOP

From triangle BOP

∠OBP = ∠OPB =(180-2θ)/2 = 90 – θ

\cos \mathrm{2\theta} =2\cos^2 \mathrm{\theta} -1
\Rightarrow \cos \mathrm{2\theta} =2(\dfrac{1}{\sqrt5})^2 -1
\Rightarrow \cos \mathrm{2\theta} =\dfrac{2}{5} -1=-\dfrac{3}{5}

Apply cosine rule in triangle BOP

PB^2=OB^2+PO^2-2\times OB\times PO\times \cos \mathrm{2\theta}
\Rightarrow PB^2=4^2+4^2-2\times 4\times 4\times (-\dfrac{3}{5})
\Rightarrow PB^2=32+\dfrac{96}{5}
\Rightarrow PB^2=\dfrac{256}{5}
\Rightarrow PB=\dfrac{16}{\sqrt5}

Now we can find the area of the triangle using the sine rule

BR = 4√2 cm (BD is the diagonal of the square)

From triangle PBR

∠PBR = ∠ABD – ∠PBA

⇒ ∠PBR = 45° – (90° – θ)

so, ∠PBR = θ – 45°

⇒ sin ∠PBR = sin(θ – 45°)

⇒ sin ∠PBR = sin θ cos 45 – cos θ sin 45

Area  \space of  \space triangle = \dfrac{1}{2} \times PB \times BR \times \sin{∠PBR}
Area  \space of  \space triangle = \dfrac{1}{2} \times \dfrac{16}{\sqrt5} \times 4\sqrt2 \times (\sin θ  \cos 45  - \cos θ  \sin 45)
\Rightarrow Area  \space of  \space triangle = \dfrac{1}{2} \times \dfrac{16}{\sqrt5} \times 4\sqrt2 \times (\dfrac{2}{\sqrt5} \times \dfrac{1}{\sqrt2}  - \dfrac{1}{\sqrt5} \times \dfrac{1}{\sqrt2})
\Rightarrow Area  \space of  \space triangle = \dfrac{32}{5}\space \mathrm{cm^2 }

Area of the blue triangle = 6.4 cm²

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