In the figure (i) given below, ABC is an equilateral triangle with each side of length 10 cm. In △BCD, ∠D = 90° and CD = 6 cm. Find the area of the shaded region. Give your answer correct to one decimal place.

Given,

ABC is an equilateral triangle of side = 10 cm

We know that,

Area of equilateral triangle ABC = $\dfrac{\sqrt{3}}{4}\text{ (side)}^2$

Substituting the values we get,

$A = \dfrac{\sqrt{3}}{4} \times (10)^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 100 \\[1em] = 25\sqrt{3} \\[1em] = 43.3 \text{ cm}^2.$

In right angled triangle BCD,

⇒ BC² = BD² + CD²

⇒ 10² = BD² + 6²

⇒ BD² = 100 - 36

⇒ BD² = 64

⇒ BD = 8 cm.

We know that,

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

Area of △BCD = $\dfrac{1}{2} \times CD \times BD = \dfrac{1}{2} \times 6 \times 8$ = 24 cm²

From figure,

Area of shaded portion = Area of triangle ABC - Area of triangle BCD

Substituting the values we get,

Area of shaded portion = 43.3 - 24 = 19.3 cm².

Hence, area of shaded region = 19.3 cm².