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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.

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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:

quadrilateral 4

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 $AD = FC = 1\ cm$

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

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:

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

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

Posted by

Divya Prakash Singh

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