For surface I: It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm = (0.75 x 3.3) cm2 = 2.475 cm2 (approx.) For surface II: It is a rectangle with length 6.5 cm and breadth 1 cm. ∴ Area of surface II = Length x Breadth = (6.5 x 1) cm2 = 6.5 cm2 For surface III: It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the figure given below: For surface IV and V: Surface V is a right-angled triangle with base 6cm arid height 1.5 cm. Also, area of surface IV = area of surface V = 12 x base x height = (12 x 6 x 15) cm2 = 4.5 cm2 Thus, the total area of the paper used = (area of surface I) + (area of surface II) + (area of surface III) + (area of surface IV) + (area of surface V) = [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm2 = 19.275 cm2 = 19.3 cm2 (approx.)