The width of a ring is $6 \mathrm{cm}$ and the area of inner circle is $616 {\mathrm{cm}}^2$. Find the circumference of the outer circle.

Area of a circular park is $\mathrm{P} {\mathrm{m}}^2$. A path of width $\mathrm{W} \mathrm{m}$ is laid around and outside the park. Find the area of the path.

The following steps are involved in solving the above problem. Arrange them in sequential order:

(A) Radius of the outer circle:$R=\sqrt{\frac{P}{\mathrm{\pi }}}+W$

(B) Area of the park, ${\mathrm{\pi r}}^2=\mathrm{P}$

(C) Area of the path$={\mathrm{\pi R}}^2-{\mathrm{\pi r}}^2$

(D) Radius of the park,$r=\sqrt{\frac{P}{\mathrm{\pi }}}$

(E)$\therefore$Area of the path$=\mathrm{\pi }{\left(\sqrt{\frac{\mathrm{P}}{\mathrm{\pi }}}+\mathrm{W}\right)}^2-\mathrm{P}$

Area of a rhombus is $96 {\mathrm{cm}}^2$and one of its diagonals is $12 \mathrm{cm}$. Find the side of the rhombus:

The following steps are involved  in solving the above problem. Arrange them in sequential order:

(A) Let, $\mathrm{PQRS}$ be a rhombus, $\mathrm{PR}=12 \mathrm{cm}$ and the diagonals $\mathrm{PR}$ and $\mathrm{QS}$ intersect at $\mathrm{T}$.

(B) Area of a rhombus$=\frac{1}{2}\times \mathrm{PR}\times \mathrm{QS}=96$

(C)$\Rightarrow \mathrm{QS}=16 \mathrm{cm}$

(D) $\mathrm{PTQ}$ is a right triangle and ${\mathrm{PQ}}^2=6^2+8^2$$\Rightarrow \mathrm{PQ}=10 \mathrm{cm}$

(E)$\mathrm{PT}=\frac{\mathrm{PR}}{2}=6 \mathrm{cm}$ and $\mathrm{QT}=\frac{\mathrm{QS}}{2}=8 \mathrm{cm}$