Quantitative Aptitude & Data Interpretation

Area of a trapezium:

$$A = \frac{1}{2}(a + b) \times h$$ $$A = \frac{1}{2}(2.02 + 4.04) \times 5 = \frac{1}{2}(6.06) \times 5 = 3.03 \times 5 = 15.15\text{ m}^2.$$

Using algebraic identity $\frac{x^3 - y^3}{x^2 + xy + y^2} = x - y$:
For the first term, $x = 9.6$ and $y = 0.7$ (since $0.7^3 = 0.343$, $9.6 \times 0.7 = 6.72$, $0.7^2 = 0.49$):

$$\frac{(9.6)^3 - (0.7)^3}{(9.6)^2 + (9.6)(0.7) + (0.7)^2} = 9.6 - 0.7 = 8.9$$

For the second term, $x = 2.6$ and $y = 0.2$ (since $0.2^3 = 0.008$, $2.6 \times 0.2 = 0.52$, $0.2^2 = 0.04$):

$$\frac{(2.6)^3 - (0.2)^3}{(2.6)^2 + (2.6)(0.2) + (0.2)^2} = 2.6 - 0.2 = 2.4$$ $$\text{Total} = 8.9 + 2.4 = 11.3.$$

The centers of the four circles form a square of side $s = 2r = 2 \times 6 = 12\text{ cm}$.
Area of the square $= s^2 = 12^2 = 144\text{ cm}^2$.
The area enclosed by the 4 quadrants inside the square equals the area of one full circle of radius $r = 6\text{ cm}$:

$$\text{Area of circle} = \pi r^2 = \frac{22}{7} \times 6^2 = \frac{792}{7} = 113.14\text{ cm}^2$$ $$\text{Enclosed area} = 144 - 113.14 = 30.86\text{ cm}^2.$$