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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Given :

The figure of the aeroplane made with coloured paper is given.

To find :

We have to find the total area of the colour paper used.

Solution :

The total area of the paper used $= part I + part II + part III + part IV + part V$

Part I : 

It is an isosceles triangle with sides $a = 1\ cm, b = 5\ cm, c = 5\ cm$.

Semi perimeter, $S = \frac{a+ b + c}{2}$

$S = \frac{1+ 5 + 5}{2}$ 

$S= \frac{11}{2}$

$S = 5.5\ cm$

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

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

$= \sqrt {5.5 (4.5)(0.5)(0.5)}$

$= 0.75 \sqrt{11}\ cm^2$

Part I  $= 2.488\ cm^2$.

Part II :

It is a rectangle with sides $l=6.5 \ cm, b =1\ cm$

Area of rectangle $=l\times b$

$=6.5\times 1$

Part II $=6.5\ cm^2$.

Part III :  

It is a trapezium. 

Area of trapezium $=$ area of equilateral triangle $+$ area of parallelogram

Area of equilateral triangle $=\frac{\sqrt{3}}{4}a^2$

$=\frac{\sqrt{3}}{4}\times1$

$= \frac{\sqrt{3}}{4}$ 

$=0.433\ cm^2$

Area of parallelogram$=base \times height$

Height $= \sqrt{1^2 - 0.5^2} = \sqrt{0.75} = 0.866\ cm$

Area of parallelogram$=1 \times 0.866$

$= 0.866\ cm^2$

Area of trapezium $= 0.433 + 0.866$ 

$= 1.299\ cm^2$

Part III $= 1.299\ cm^2$.

Part IV and Part V : 

These are triangles with a base of $1.5\ cm$ and a height of $6\ cm$.

Area of triangle $=\frac{1}{2} \times base \times height$

$ = \frac{1}{2} \times 1.5 \times 6$

$= 4.5\ cm^2$

Part IV $=$ Part V $= 4.5\ cm^2$.

Total area of the paper $= 2.488 + 6.5 + 1.299 + 4.5 + 4.5\ cm^2$

$ = 19.287 cm^2$

The total area of the paper is $19.287\ cm^2$. 

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Updated on: 10-Oct-2022

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