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.

Answer 1

Area of region $I:$

$\Rightarrow$ $2S=5+5+1=11cm$ (Isosceles triangle)
$S=5.5cm$
Area of $I=\sqrt{s(s-a)(s-b)(s-c)}$

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

$=.75\sqrt{11}=2.48{cm}^2$

Area $II:$ or Area of rectangle $=l\times b$

$=6.5\times 1$
$=6.5{cm}^2$

Area $III:$ Its a trapezium
So, we need to calculate the perpendicular height
$={\left(1\right)}^2-{\left(0.5\right)}^2=\sqrt{0.75}$
$=0.86cm$

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

$=0.866\times 1+\frac{\sqrt{3}}{4}{\left(1\right)}^2$

$=0.86+0.43=1.29{cm}^2$

Area of region $IV=$ Area of region $V=\frac{1}{2}\times 6\times 1.5$
$=4.5{cm}^2$

Total area of paper used $=2.48+6.5+1.29+4.5+4.5$
$=19.27{cm}^2$