Question

A like in the shape of a square with each diagonal 32 cm and having a tail in the shape on an isosceles triangle of base 8 cm and each side 6 cm, is made of three different shades as shown in the figure. How much paper of each shade has been used in it?

Answer 1

$$\begin{gathered} \mathrm{Every}\mathrm{square}\mathrm{is}\mathrm{a}\mathrm{rhombus}. \\ \mathrm{Area}\mathrm{of}\mathrm{rhombus}=\frac{1}{2}\left({\mathrm{I}}^{\mathrm{st}}\mathrm{diagonal}\times {\mathrm{II}}^{\mathrm{nd}}\mathrm{diagonal}\right) \\ \mathrm{Here},\mathrm{each}\mathrm{diagonal}\mathrm{measures}32\mathrm{cm}. \\ \mathrm{Therefore},\mathrm{we}\mathrm{can}\mathrm{say}: \\ \mathrm{Area}\mathrm{of}\mathrm{square}ABCD=\frac{1}{2}\times (\mathrm{Diagonal}{)}^2 \\ =\frac{1}{2}\times 32\times 32 \\ =512{\mathrm{cm}}^2 \end{gathered}$$

We know that a diagonal of a parallelogram divides into two triangles of equal areas. Also, a square is a parallelogram. Therefore, we can say that a diagonal of a square divides into two triangles of equal areas.
∴ $\mathrm{Area}\mathrm{of}△ABD=\mathrm{Area}\mathrm{of}△BDC=\frac{1}{2}\times 512=256{\mathrm{cm}}^2$
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{isosceles}△CEF=\frac{b}{4}\sqrt{4a^2-b^2} \\ \mathrm{Here}, \\ a=6\mathrm{cm}\mathrm{and}b=8\mathrm{cm} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ \frac{8}{4}\times \sqrt{4(6{)}^2-8^2} \\ =\frac{8}{4}\times \sqrt{144-64} \\ =\frac{8}{4}\times \sqrt{80} \\ =\frac{8}{4}\times 4\sqrt{5} \\ =8\sqrt{5} \\ =8\times 2.24 \\ =17.92{\mathrm{cm}}^2 \end{gathered}$$