In the given figure, △ABC is an equilateral triangle having each side equal to 10 cm and △PBC is right angled at P in which PB = 8 cm. Find the area of the shaded region.

Given,

△ABC is an equilateral triangle.

Each side = 10 cm

By formula,

$\Rightarrow \text{Area of equilateral △ABC} = \dfrac{\sqrt{3}}{4} \times \text{ (side)}^2 \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} \times 10^2 \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} \times 100 \\[1em] \Rightarrow 25\sqrt{3} \\[1em] \Rightarrow 43.3 \text{ cm}^2.$

Given,

PB = 8 cm and BC = 10 cm

By using the Pythagoras theorem in △PBC,

⇒ BC² = PB² + PC²

⇒ 10² = 8² + PC²

⇒ 100 = 64 + PC²

⇒ PC² = 36

⇒ PC = $\sqrt{36}$

⇒ PC = 6 cm.

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

Area of △PBC = $\dfrac{1}{2}$ × PB × PC

= $\dfrac{1}{2}$ × 8 × 6

= 4 × 6 = 24 cm².

Shaded region = Area of △ABC − Area of △PBC

= 43.3 - 24

= 19.3 cm².

Hence, area of shaded region = 19.3 cm².