Area of I in figure
It is triangle part and its sides are 5 cm, 5 cm, 1 cm. Here, semiperimeter of the triangle
$=\frac{5\mathrm{cm}+5\mathrm{cm}+1\mathrm{cm}}{2}=\frac{11}{2}\mathrm{cm}$
Area of part I $=\sqrt{\frac{11}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{9}{2}}{\mathrm{cm}}^2=\frac{3}{4}\sqrt{11}{\mathrm{cm}}^2$
$=\frac{3}{4}$× 3.31 cm² = 2.482 (approx)
Area of II in figure = area of rectangle = L × B = 6.5 × 1 = 65 cm²
Area of III in figure
Area of $\mathrm{\Delta BEC}=\frac{\sqrt{3}}{4}{\left(1\right)}^2{\mathrm{cm}}^2=\frac{\sqrt{3}}{4}{\mathrm{cm}}^2$
Let h be the height of the $∆$BEC
$\frac{1}{2}\times \mathrm{BE}\times \mathrm{h}=\frac{\sqrt{3}}{4}\Rightarrow \frac{1}{2}\times 1\times \mathrm{h}=\frac{\sqrt{3}}{4}\Rightarrow \mathrm{h}=\frac{\sqrt{3}}{2}\mathrm{cm}$
Area of Region III $=\frac{1}{2}(1+2)\times \frac{\sqrt{3}}{2}{\mathrm{cm}}^2=\frac{3}{4}\sqrt{3}{\mathrm{cm}}^2$
$=\frac{3}{4}$ × 1.732 cm² = 1.3 cm² (approx)
Area of IV in figure $=\frac{1}{2}\times 6\times \frac{3}{2}{\mathrm{cm}}^2=\frac{9}{2}{\mathrm{cm}}^2$
Area of V in figure $=\frac{1}{2}\times 6\times \frac{3}{2}{\mathrm{cm}}^2=\frac{9}{2}{\mathrm{cm}}^2$
Total area of the paper used = 2.482 cm² + 6.5 cm² + 1.3 cm² $+\frac{9}{2}{\mathrm{cm}}^2+\frac{9}{2}{\mathrm{cm}}^2$
= (10.282 + 9) cm² (approx.) = 19.282 cm² (approx.) = 19.3 cm² (approx.)