Answer :

For part I :
It is a triangle with sides 5 cm, 5 cm and 1 cm.

Thus, We know that,

$s=\frac{a+b+c}{2}$
$?$ $s=\frac{5+5+1}{2}=\frac{11}{2}cm$

Now, Area of part I triangle
= $\sqrt{\frac{11}{2}(\frac{11}{2}?5)(\frac{11}{2}?5)(\frac{11}{2}?1)}$
(Since, Heron's formula [area = $\sqrt{s(s?a)(s?b)(s?c)}$])

= $\sqrt{\frac{11}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{9}{2}}$
= $\frac{3}{4}\sqrt{11}{cm}^2$
= $\frac{3}{4}\times 3.31{cm}^2$ = 3 × 0.829${cm}^2$ = 2.487${cm}^2$(approx)

For part II :

It is a rectangle with sides 6.5 cm and 1 cm

$?$ Area of part II = 6.5 × 1
(Since, Area of rectangle = Lenght × Breadth)

= 6.5${cm}^2$

For part III :

It is a trapezium ABCD.



$\mathrm{?}EBC$ is an equilateral with side 1 cm.
$?$ Area of $\mathrm{?}EBC$ = $\frac{1}{2}$ × EB × CF = $\frac{\sqrt{3}}{4}\times 1^2$
(Since, Area of triangle = $frac12$ × Base × Height and Area of equilateral triangle = $frac\sqrt{3}4\times (side{)}^2$)

$?$ $\frac{1}{2}$ × 1 × CF = $\frac{\sqrt{3}}{4}$
$?$ CF = $\frac{\sqrt{3}}{2}cm$

Now, Area of trapezium
= $\frac{1}{2}$ × Sum of parallel sides × Height
= $\frac{1}{2}$ × (AB + CD) × CF
= $\frac{1}{2}$ × (2 + 1) × $\frac{\sqrt{3}}{2}$
= $\frac{3}{4}$ × $\sqrt{3}$
= 3 × 0.433
= 1.299 ${cm}^2$

Therefore, Area of part III is 1.299 ${cm}^2$

For part IV & V :

Both the parts IV and V are the same.

$?$ it is a right triangle with sides 6 cm and 1.5 cm.

Area of part IV & V = $\frac{1}{2}$ × 1.5 × 6 = $\frac{9}{2}$ = 4.5 ${cm}^2$

So, we get,

Total area of paper used = Area of (I + II + III + IV + V)
= (2.487 + 6.5 + 1.299 + 4.5 + 4.5)${cm}^2$
= 19.286${cm}^2$
= 19.3${cm}^2$(approx.)