Q3.    Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

                

Answers (1)
D Divya Prakash Singh

The total area of the paper used will be the sum of the area of the sections I, II, III, IV, and V. i.e., 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 = \frac{a+b+c}{2} = \frac{5+5+1}{2} = 5.5\ cm

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} = 2.5\ cm^2\ \ \ \ \ \ (Approx.)

For section II:

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

Therefore, the area of the rectangle is = l\times b = 6.5\times 1 = 6.5\ cm^2.

For section III:

From the figure: quadrilateral 4

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

\left [ \because Opposite\ sides\ of\ a\ parallelogram \right ]

Therefore, BF = BC -FC =2-1 = 1\ cm.

\implies ABF is an equilateral triangle.      \left [ \because AB = BF =AF = 1\ cm \right ]

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

 

 

 

 

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