Radha made a picture of an aeroplane with coloured papers as shown in the given figure.

Solution:



For triangle I

This triangle is an isosceles triangle.

Perimeter $=2 s=(5+5+1) \mathrm{cm}=11 \mathrm{~cm}$

$s=\frac{11 \mathrm{~cm}}{2}=5.5 \mathrm{~cm}$

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

$=[\sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}] \mathrm{cm}^{2}$

$=[\sqrt{(5.5)(0.5)(0.5)(4.5)}] \mathrm{cm}^{2}$

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

$=(0.75 \times 3.317) \mathrm{cm}^{2}$

$=2.488 \mathrm{~cm}^{2}$ (approximately)

For quadrilateral II

This quadrilateral is a rectangle.

Area $=1 \times b=(6.5 \times 1) \mathrm{cm}^{2}=6.5 \mathrm{~cm}^{2}$

For quadrilateral III

This quadrilateral is a trapezium.

Perpendicular height of parallelogram $=\left(\sqrt{1^{2}-(0.5)^{2}}\right) \mathrm{cm}$

$=\sqrt{0.75} \mathrm{~cm}=0.866 \mathrm{~cm}$

Area = Area of parallelogram + Area of equilateral triangle

$=(0.866) 1+\frac{\sqrt{3}}{4}(1)^{2}=0.866+0.433=1.299 \mathrm{~cm}^{2}$



Area of triangle (IV) = Area of triangle in (V)

$=\left(\frac{1}{2} \times 1.5 \times 6\right) \mathrm{cm}^{2}=4.5 \mathrm{~cm}^{2}$

Total area of the paper used $=2.488+6.5+1.299+4.5 \times 2$

$=19.287 \mathrm{~cm}^{2}$