Question

How much paper of each shade is needed to make a kite given in the figure in which $ \mathrm{ABCD}$ is a square with a diagonal $ 44 \mathrm{cm}$.

14 cm 

Answer 1

Step 1: Find the area of triangles $I,II,III,IV$:

Given that: Diagonal of the square $\mathrm{ABCD}$ is $44\mathrm{cm}$

$$\begin{gathered} \mathrm{AC}=\mathrm{BD}=44\mathrm{cm} \\ \mathrm{AO}=\frac{\mathrm{AC}}{2}=\frac{44}{2}=22\mathrm{cm} \\ \mathrm{BO}=\frac{\mathrm{BD}}{2}=\frac{44}{2}=22\mathrm{cm} \end{gathered}$$

Since $AOB$ is a right-angled triangle.

$$\begin{gathered} \mathrm{AB}=\sqrt{{\mathrm{AO}}^2+{\mathrm{BO}}^2}[\mathrm{Pythagoras}\mathrm{Theorem}] \\ \mathrm{AB}=\sqrt{{22}^2+{22}^2} \\ \mathrm{AB}=\sqrt{2\times {22}^2} \\ \mathrm{AB}=22\sqrt{2}\mathrm{cm} \end{gathered}$$

Area of square $\mathrm{ABCD}={\mathrm{AB}}^2$

$$\begin{gathered} ={\left(22\sqrt{2}\right)}^2 \\ =968{\mathrm{cm}}^2 \end{gathered}$$

Area of triangle $=\frac{1}{4}\times \mathrm{Area}\mathrm{of}\mathrm{square}$

$$\begin{gathered} =\frac{1}{4}\times 968{\mathrm{cm}}^2 \\ =242{\mathrm{cm}}^2 \end{gathered}$$

Area of triangles $I,II,III,IV=242{\mathrm{cm}}^2$

Step 2: Find the area of the lower triangle:

The sides of the lower triangle are $\mathrm{a}=20\mathrm{cm},\mathrm{b}=20\mathrm{cm},\mathrm{c}=14\mathrm{cm}$

Heron's Formula: $\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$ $s$ is the perimeter and $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ are the sides of the triangle.

$\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{20+20+14}{2}=27\mathrm{cm}$

$$\begin{gathered} \mathrm{Area}=\sqrt{27(27-20)(27-20)(27-14)} \\ =\sqrt{27\times 7\times 7\times 13} \\ =\sqrt{17199} \\ =131.14{\mathrm{cm}}^2 \end{gathered}$$

Step 3: Area of shaded parts:

Are of yellow $=$Area of the triangle $I$ $+$Area of triangle $III$

$$\begin{gathered} =242+242 \\ =484{\mathrm{cm}}^2 \end{gathered}$$

Area of green $=$Area of the triangle $II$ $+$Area of the lower triangle

$$\begin{gathered} =242+131.14 \\ =373.14{\mathrm{cm}}^2 \end{gathered}$$

Area of red $=$Area of triangle $IV$

$=242{\mathrm{cm}}^2$

Hence, the area of the yellow is $484{\mathrm{cm}}^2$ the area of the green is $373.14{\mathrm{cm}}^2$, and the area of the red is $242{\mathrm{cm}}^2.$