For the triangle marked I : a = 5 cm, b = 5 cm, c = 1 cm

∴ s = (a +b + c)/ 2 = (5 +5+ 1) /2 cm = 11 / 2 cm =5.5 cm

**For the rectangle marked II : **

Length = 6.5 cm, Breadth = 1 cm

Area of the rectangle = 6.5 × 1 cm^{2} = 6.5 cm^{2}

**For the trapezium marked III : **

Draw AF || DC and AE ⊥ BC.

AD = FC = 1 cm, DC = AF = 1 cm

∴ BF = BC – FC = (2 – 1) cm = 1 cm

Hence, ∆ABF is equilateral.

Also, E is the mid-point of BF.

∴ BE = 1/ 2 cm = 0.5 cm

Also, AB^{2} = AE^{2} + BE^{2} [Pythagoras theorem]

⇒ AE^{2} = 1^{2 }– (0.5)^{2} = 0.75

⇒ AE = 0.9 cm (approx.)

Area of the trapezium = 1/ 2 (sum of the parallel sides) × distance between them.

= 1/ 2 × (BC + AD) × AE = 1/ 2 × (2 + 1) × 0.9 cm^{2} = 1.4 cm^{2}.

**For the triangle marked IV : **

It is a right-triangle

∴ Area of the triangle = 1/ 2 × base × height

= 1/ 2 × 6 × 1.5 cm cm^{2} = 4.5 cm^{2}.

**For the triangle marked V : **

This triangle is congruent to the triangle marked IV.

Hence, area of the triangle = 4.5 cm^{2}

Total area of the paper used = (2.5 + 6.5 + 1.4 + 4.5 + 4.5) cm^{2}