Question

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

Answer 1

Consider the problem:

For region $I$ :

Semi perimeter of region $I$ (triangle)

$\begin{array}{c}s=\frac{5+5+1}{2}cm\\ =\frac{11}{2}cm\\ =5.5cm\end{array}$

By using Heron's formula, then the area of region $I$ (triangle)

$\begin{array}{c}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ =\sqrt{5.5\left(5.5-5\right)\left(5.5-5\right)\left(5.5-1\right)}\\ =\sqrt{5.5\times 0.5\times 0.5\times 4.5}\\ =2.5c{m}^2\left(approx\right)..........\left(1\right)\end{array}$

For region $II$ :

Length $=6.5cm$

Breadth $=1cm$

Area of region $II$ (rectangle)

Area of rectangle $=length\times Breadth$

$=6.5\times 1c{m}^2=6.5c{m}^2..........\left(2\right)$

For region $IV$ :

Applying Pythagoras theorem on $\Delta ABC$

$A{B}^2=A{C}^2+B{C}^2$

$\begin{array}{c}\Rightarrow A{B}^2=6^2+{\left(1.5\right)}^2\\ \Rightarrow A{B}^2=36+2.25=38.25\\ \Rightarrow AB=6.2c{m}^2\left(approx\right)\end{array}$

Semi perimeter of $\Delta ABC$

$\begin{array}{c}s=\frac{6+1.5+6.2}{2}cm\\ =\frac{13.7}{2}cm\\ =6.85c{m}^2\end{array}$

Using Heron's formula, Area of $\Delta ABC$

$\begin{array}{c}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ =\sqrt{6.85\left(6.85-6\right)\left(6.85-1.5\right)\left(6.85-6.2\right)}\\ =\sqrt{6.85\times 0.85\times 5.35\times 0.65}\\ =4.5c{m}^2..............\left(3\right)\end{array}$

For region $V$ :

Area of region $V$ $=Areaof\left(\Delta ABC\right)=4.5c{m}^2.......\left(4\right)$

For region $III$,

Here it seems Region $III$ is a trapezium and we do not know the height of this trapezium,so we have to draw this trapezium separately so that, we can find its area.

So, Draw $ST\perp PQ$ and $RU\perp PQ$. doing so, we have $PT=0.5cm$ and $UQ=0.5cm$

Now, we can use Pythagoras theorem on $\Delta SPT$ to find the height of trapezium $PQRS$.

$S{P}^2=S{T}^2+P{T}^2$

$\begin{array}{c}\Rightarrow 1^2=S{T}^2+{\left(0.5\right)}^2\\ \Rightarrow S{T}^2=1-0.25=0.75\\ \Rightarrow ST=0.87cm\left(approx\right)\end{array}$

So, Area of region $III$ (Trapezium)

$\begin{array}{c}=\frac{1}{2}\left(SR+PQ\right)\times ST\\ =\frac{1}{2}\left(1+2\right)\times 0.87c{m}^2\\ =1.305c{m}^2.........\left(5\right)\end{array}$

Now, adding $\left(1\right)$,$\left(2\right)$,$\left(3\right)$,$\left(4\right)$ and $\left(5\right)$ to find the total area of aeroplane picture.

Total area$=(2.5+6.5+4.5+4.5+1.305)c{m}^2$

$=19.305c{m}^2$

Hence, the required area is approximately $19.305c{m}^2$