Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

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.)