Question

Radha made a picture of an airplane with colored paper as shown in Fig. Find the total area of the paper used.

Answer 1

For the triangle( section I ):

The area of a triangle can be calculated by Heron’s Formula, where

$\mathrm{a}=5\mathrm{cm},\mathrm{b}=5\mathrm{cm},\mathrm{c}=1\mathrm{cm}$

Semi Perimeter $s=\frac{a+b+c}{2}=\frac{5+5+1}{2}=\frac{11}{2}cm=5.5cm$

$ar(△)=\sqrt{s(s-a)(s-b)(s-c)}$

$$\begin{gathered} =\sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}=\sqrt{5.5\times (0.5)\times (0.5)\times (4.5)} \\ =\sqrt{\frac{11}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{9}{2}}=\frac{3}{4}\sqrt{11}\approx 2.5c{m}^2..........................\left(1\right) \end{gathered}$$

Area of Rectangle ( Section II )

Area of Rectangle $=\mathrm{length}\times \mathrm{Breadth}$

$=1\times 6.5=6.5{\mathrm{cm}}^2\dots \dots \dots \left(\mathrm{ii}\right)$

Area of Trapezium ( Section III )

Let's draw two altitudes in trapezium such that it will be divided into three parts: two right-angle triangles and a rectangle.

The breadth of the rectangle will be equal to the height of a triangle that's equal to the height of the trapezium

$\Rightarrow$Height of trapezium $=\sqrt{1^2-0.5^2}=0.86cm$

Now the area of the trapezium $=\frac{1}{2}\times \mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\times \mathrm{height}=\frac{1}{2}\times 3\times 0.86\approx 1.3{\mathrm{cm}}^2$

Area of triangles ( Section IV and V )

These triangles are two congruent right-angled triangles having a base as $6cm$ and height of $1.5cm$

$\Rightarrow$Area triangles IV and V $=2\times (\frac{1}{2}\times 6\times 1.5)c{m}^2=9c{m}^2$

So, the total area of the paper used $=(2.5+6.5+1.3+9)c{m}^2=19.3c{m}^2$

Hence the total area of the paper used is $19.3c{m}^2$