Question

Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find
the total area of the paper used. See the figure above by clicking it. ​

Answer 1

Answer:

Step-by-step explanation:

The total area of the paper used will be the sum of the area of the sections I, II, III, IV, and V.

Total\ area = I +II+III+IV+V

For section I:

Here, the sides are a = 1cm and b =c = 5cm.

So, the Semi-perimeter will be:  

s ={a+b+c}/{2}

  ={5+5+1}/{2}

  = 5.5cm

Therefore, the area of section I will be given by Heron's Formula,

A =  $\sqrt{s(s-a)(s-b)(s-c)}$

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

     =$\sqrt{5.5(4.5)(0.5)(0.5)}$

   ={$\sqrt{6.1875}$

    = 2.5cm²(Approx.)

For section II:

Here the sides of the rectangle are l =6.5 cm and b =1cm.

Therefore, the area of the rectangle is = l*b

                                                                 = 6.5*1

                                                                  = 6.5cm².

For section III:

From the figure:

Drawing the parallel line AF to DC and a perpendicular line AE to BC.

We have the quadrilateral ADCF,

AF || DC      (by construction. )

AD || FC       (because ABCD is a trapezium)

So, ADCF is a parallelogram.

Therefore, AF = DC = 1 cm  and  AD = FC = 1 cm

Therefore, BF = BC -FC =2-1 = 1 cm.

this implies ABF is an equilateral triangle.  

because AB = BF =AF = 1cm

Then, the area of the equilateral triangle ABF is given by:

\implies \frac{\sqrt3}{4}a^2 = \frac{\sqrt3}{4}1^2 = \frac{\sqrt3}{4}

= \frac{1}{2}\times BF \times AE

= \frac{1}{2}\times 1cm \times AE = \frac{\sqrt3}{4}

\implies AE = \frac{\sqrt3}{2} = \frac{1.732}{2} = 0.866 \approx 0.9

Hence, the area of trapezium ABCD will be:

= \frac{1}{2}\times(AD+BC)\times AE

= \frac{1}{2}\times(1+2)\times 0.9

=1.35 =1.4\ cm^2\ \ \ \ (Approx.)

For Section IV:

Here, the base is 1.5 cm and the height is 6 cm.

Therefore, the area of the triangle is :

= \frac{1}{2}\times base\times height

= \frac{1}{2}\times 1.5\times 6 = 4.5\ cm^2

For section V:

The base length = 1.5cm and the height is 6cm.

Therefore, the area of the triangle will be:

= \frac{1}{2}\times 1.5\times 6 = 4.5\ cm^2

Hence, the total area of the paper used will be:

Total\ area = I +II+III+IV+V

= 2.5+6.5+1.4+4.5+4.5 = 19.4\ cm^2