Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DEFG is an isosceles trapezium. All measurements are in centimeters.

In right angle triangle ECF,

Using pythagoras theorem,

⇒ EF² = EC² + CF²

⇒ 5² = EC² + 3²

⇒ EC² = 5² - 3²

⇒ EC² = 25 - 9 = 16

⇒ EC = $\sqrt{16}$ = 4 cm.

Since, DEFG is an isosceles trapezium.

∴ GD = EF= 5 cm.

Since, BDEC is a rectangle,

∴ BD = EC = 4 cm and BC = DE = 6 cm.

In right angle triangle DBG,

Using pythagoras theorem,

⇒ GD² = BD² + GB²

⇒ 5² = 4² + GB²

⇒ GB² = 5² - 4²

⇒ GB² = 25 - 16 = 9

⇒ GB = $\sqrt{9}$ = 3 cm.

In trapezium,

GF = GB + BC + CF = 3 + 6 + 3 = 12 cm.

Area of trapezium DEFG = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2} \times (12 + 6) \times 4$

= 18 × 2

= 36 cm².

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

= $\dfrac{\sqrt{3}}{4} \times (6)^2$

= $\dfrac{\sqrt{3}}{4} \times 36$

= 1.732 × 9

= 15.59 cm²

Area of figure = Area of trapezium DEFG + Area of equilateral triangle ABC

= 36 + 15.59 = 51.59 cm².

Hence, area of figure = 51.59 cm².