# 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

## Solution

Area of blue triangle = Area of **ΔPBR**

**Connect OC**

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**

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**