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)

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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