Option 2 : 108

**Formula used:**

**Area of equilateral triangle** **with side 'a'**

\(A\ = \frac{\sqrt 3}{4}a^2\)

If a, b and c are three side of triangle, then area can be calculated by using **Heron's formula** of triangle.

\(A \ = \ \sqrt {s(s - a)(s - b)(s - c)}\)

Where, 's' is semi-perimeter and given by

\(s \ =\ \frac{a\ \ +\ \ b\ \ +\ \ c}{2} \)

**Calculation:**

According to question, AB = 6 cm, BC = 5 cm, CD = 5 cm, AD = 6 cm and BD = 6 cm

Area of equiletral triangle ABD,

\(A_1 = \frac{√ 3}{4}a^2 \ =\ \frac{√ 3}{4}6^2\)

⇒ A_{1} = 9√3 .....(1)

Area of Δ BCD

\(A_2 \ = \ \sqrt {s(s - 5)(s - 6)(s - 5)}\)

⇒ A_{2} = 12 .......(2)

Hence, area of quadiletral ABCD = area of Δ BCD + area of Δ ABD

⇒ A = A_{1} + A_{2}

⇒ A = 12 + 9√3

On compairig this from a + b√3, we will get

a = 12 and b = 9

Hence, the value of 'ab' is 108.