How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 60 cm ?

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{ABCD} \mathrm{is} \mathrm{a} \mathrm{square} \mathrm{with} \mathrm{diagonal} 60 \mathrm{cm} \\ \mathrm{So} \mathrm{Area} \mathrm{of} \mathrm{each} \mathrm{part} \mathrm{in} \mathrm{ABCD}=\frac{1}{4}\times \mathrm{area} \mathrm{of} \mathrm{ABCD} \left(\mathrm{because} \mathrm{in} \mathrm{square} \mathrm{diagonlas} \mathrm{divide} \mathrm{in} \mathrm{same} \mathrm{parts}\right) \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{each} \mathrm{part} \mathrm{in} \mathrm{ABCD}=\frac{1}{4}\times \left[\frac{1}{2}\times {\left(\mathrm{diagonal}\right)}^2\right]=\frac{1}{4}\times \left[\frac{1}{2}\times 60\times 60\right]=\frac{1800}{4}=450 {\mathrm{cm}}^2 \\ \mathrm{So} \mathrm{Paper} \mathrm{for} \mathrm{Blue}=450 {\mathrm{cm}}^2 , \mathrm{for} \mathrm{Black}=450 {\mathrm{cm}}^2 , \mathrm{for} \mathrm{blue}=450 {\mathrm{cm}}^2 \mathrm{Answer} \\ \mathrm{And} \mathrm{for} \mathrm{orange}1=450 {\mathrm{cm}}^2 \\ \mathrm{In} \mathrm{ECF}, \mathrm{It}\text{'}\mathrm{s} \mathrm{a} \mathrm{isosceles} \mathrm{triangle} \mathrm{with} \mathrm{CE}=\mathrm{CF} \\ \mathrm{So} \\ \mathrm{Put} \mathrm{perpendicular} \mathrm{from} \mathrm{C} \mathrm{to} \mathrm{EF} \mathrm{which} \mathrm{divides} \mathrm{EF} \mathrm{in} \mathrm{same} \mathrm{parts} \\ \mathrm{So} \mathrm{height}=\sqrt{{20}^2-{15}^2}=\sqrt{400-225}=\sqrt{175}=13.22 \mathrm{cm} \\ \mathrm{So} \mathrm{area} \mathrm{of} \mathrm{ECF}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}=\frac{1}{2}\times 30\times 13.22=198.3 {\mathrm{cm}}^2 \\ \mathrm{So} \mathrm{total} \mathrm{paper} \mathrm{for} \mathrm{ornage}=\mathrm{orange}1+198.3=450+198.3=648.3 {\mathrm{cm}}^2 \mathrm{Answer} \end{gathered}$$

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