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

We know that, $\times$

The Heron's formula for calculating the area of the triangle,

Area of triangle = $\sqrt{s (s - a) (s - b) (s -c)}$

where, s is the semi-perimeter = half of the perimeter

and  a, b and c are the sides of the triangle 

a = 5 cm, b = 5 cm, c = 1 cm

Semi Perimeter: (s) = (Perimeter/2)

s = (a $+$ b $+$ c)/2

= (5 $+$ 5 $+$ 1)/2

= 11/2

= 5.5 cm

Area of triangle marked I = $\sqrt{s (s - a) (s - b) (s -c)}$

= $\sqrt{5.5 (5.5 - 5) (5.5 - 5) (5.5 - 1)}$

= $\sqrt{5.5 \times 0.5 \times 0.5 \times 4.5}$

= $\sqrt{6.1875}$ 

= 2.5 cm² 

Area of triangle (I) = 2.5 cm²

(ii) For the rectangle marked as II:

The measures of the sides are 6.5 cm, 1 cm,  6.5 cm, and 1 cm.

Area of rectangle = length $\times$ breadth

= 6.5 cm $\times$ 1 cm

= 6.5 cm²

Area of a rectangle (II) = 6.5 cm²

(iii) For the trapezium marked as III

so name  ABCD.

Draw AE ⊥ BC from A on BC and AF parallel to DC

AD = FC = 1 cm (opposite sides of a parallelogram)

AB = DC = 1 cm

BC = 2 cm

AF = DC = 1 cm 

BF = BC - FC = 2 cm - 1 cm = 1 cm

Here ∆ABF is an equilateral triangle, Hence E will be the mid-point of BF.

So, BE = EF = BF/2

EF = 1/2 = 0.5 cm

In ∆AEF, by using pythagoras theoren 

AF² = AE² $+$ EF²  

1² = AE² $+$ 0.5²

AE² = $\sqrt{1^2}$ - 0.5²

AE² = $\sqrt{0.75}$

AE = 0.9 cm 

Area of trapezium = 1/2 $\times$ sum of parallel sides $\times$ distance between them

= 1/2 $\times$ (BC $+$ AD)$\times$ AE

= 1/2 $\times$ (2 + 1) $\times$ 0.9

= 1/2$\times$ 3 $\times$ 0.9

= 1.4 cm² 

Area of trapezium (III) = 1.4 cm²

(iv) For the triangle marked as IV and V

Triangles IV and V are congruent right-angled triangles with base 6 cm and height 1.5 cm.

Area of the triangle = 1/2 $\times$ base $\times$height

= 1/2 $\times$ 6 $\times$ 1.5

= 4.5 cm²

Area of two triangles (IV and V) = 4.5  $+$ 4.5  = 9 cm²

Thus, we get the total area as,

Total area of the paper used = Area I$+$ Area II $+$ Area III $+$Area IV $+$Area V

= 2.5  $+$ 6.5 $+$1.4  $+$ 9  = 19.4 cm²

Thus, the total area of the paper used is 19.4 cm²