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Answer :

For part I :

It is a triangle with sides 5 cm, 5 cm and 1 cm.

Thus, We know that,

\(s = \frac{a + b + c}{2}\)

\(\therefore \) \(s = \frac{5 + 5 + 1}{2} = \frac{11}{2}cm\)

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 = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{\frac{11}{2} × \frac{1}{2} × \frac{1}{2} × \frac{9}{2}}\)

= \(\frac{3}{4}\sqrt{11} {cm}^2\)

= \(\frac{3}{4} × 3.31 {cm}^2\) = 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

\(\therefore \) 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.

\(\triangle{EBC}\) is an equilateral with side 1 cm.

\(\therefore \) Area of \(\triangle{EBC}\) = \(\frac{1}{2}\) × EB × CF = \(\frac{\sqrt{3}}{4} × {1}^2\)

(Since, Area of triangle = \(frac{1}{2}\) × Base × Height and Area of equilateral triangle = \(frac{\sqrt{3}}{4} × (side)^2\))

\(\therefore \)
\(\frac{1}{2}\) × 1 × CF = \(\frac{\sqrt{3}}{4}\)

\(\Rightarrow \) 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.

\(\therefore \) 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.)

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