In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangles; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.

In △AEB,

Let AE = BE = x cm, then from right angled triangle AEB,

⇒ AB² = AE² + EB²

⇒ 10² = x² + x²

⇒ 2x² = 100

⇒ x² = 50

⇒ x = $\sqrt{50} = 5\sqrt{2}$ cm.

Area of right angled △AEB = $\dfrac{1}{2}$ × base × height

$= \dfrac{1}{2} \times x \times x = \dfrac{1}{2}x^2 \\[1em] = \dfrac{1}{2} \times 50 \\[1em] = 25 \text{ cm}^2$

In △DGC,

Let DG = GC = y cm, then from right angled triangle DGC,

⇒ DC² = DG² + GC²

⇒ 10² = y² + y²

⇒ 2y² = 100

⇒ y² = 50

⇒ y = $\sqrt{50} = 5\sqrt{2}$ cm

Area of right angled △DCG = $\dfrac{1}{2}$ × base × height

$= \dfrac{1}{2} \times y \times y = \dfrac{1}{2}y^2 \\[1em] = \dfrac{1}{2} \times 50 \\[1em] = 25 \text{ cm}^2$

Since, HAD and BFC are equilateral triangle with side = 8 cm.

Area of HAD = Area of BFC = $\dfrac{\sqrt{3}}{4}$ × (side)²

= $\dfrac{\sqrt{3}}{4} \times (8)^2$

= $16\sqrt{3}$ cm²

Area of rectangle ABCD = l × b = AB × CD

= 10 × 8 = 80 cm²

From figure,

Area of shaded region = Area of (△DGC + △BFC + △AEB + △HAD + rectangle ABCD)

= $25 + 16\sqrt{3} + 25 + 16\sqrt{3} + 80 = 130 + 32\sqrt{3}$ cm².

Perimeter of figure = (AE + EB + BF + FC + CG + GD + DH + HA)

= $(5\sqrt{2} + 5\sqrt{2} + 8 + 8 + 5\sqrt{2} + 5\sqrt{2} + 8 + 8)$

= $20\sqrt{2} + 32$ cm.

Hence, area of shaded region = $130 + 32\sqrt{3}$ and perimeter = $20\sqrt{2} + 32$ cm.