The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.

For triangle ABC (right-angled triangle),

Using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ 8² + height² = 16²

⇒ 64 + height² = 256

⇒ height² = 256 - 64

⇒ height² = 192

⇒ height = $\sqrt{192}$

⇒ height = 8 $\sqrt{3}$

Area of triangle ABC = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2} \times 8 \times 8 \sqrt{3}$

= $4 \times 8 \sqrt{3}$

= $32 \sqrt{3}$ cm²

For triangle BCD,

Area of an equilateral triangle = $\dfrac{\sqrt{3}}{4}$ x side²

= $\dfrac{\sqrt{3}}{4}$ x 8²

= $\dfrac{\sqrt{3}}{4}$ x 64

= 16 $\sqrt{3}$ cm²

Area of Δ ABD (shaded portion) = Area of Δ ABC - Area of Δ BDC

= 32 $\sqrt{3}$ - 16 $\sqrt{3}$ cm²

= 16 $\sqrt{3}$ cm²

Hence, the area of the shaded portion is 16$\sqrt{3}$ cm² = 27.712 cm².