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# 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.)