Call $\displaystyle \begin{align*} l \end{align*}$ the length and $\displaystyle \begin{align*} w \end{align*}$ the width, then the area of the rectangle is $\displaystyle \begin{align*} l\,w\end{align*}$
The area of triangle CPS is $\displaystyle \begin{align*} \frac{1}{2}\,\frac{w}{2}\,l = \frac{l\,w}{4} \end{align*}$
The area of triangle CQA is $\displaystyle \begin{align*} \frac{1}{2}\,\frac{w}{2}\,\frac{l}{3} = \frac{l\,w}{12} \end{align*}$
The area of triangle ARS is $\displaystyle \begin{align*} \frac{1}{2}\,\frac{2\,l}{3}\,w = \frac{l\,w}{3} \end{align*}$
The area of triangle ACS is $\displaystyle \begin{align*} 10\,\textrm{cm}^2 \end{align*}$.
All these triangles together make the full rectangle, so
$\displaystyle \begin{align*} \frac{l\,w}{4} + \frac{l\,w}{12} + \frac{l\,w}{3} + 10 &= l\,w \\ \frac{3\,l\,w}{12} + \frac{l\,w}{12} + \frac{4\,l\,w}{12} + 10 &= l\,w \\ \frac{8\,l\,w}{12} + 10 &= l\,w \\ \frac{2\,l\,w}{3} + 10 &= l\,w \\ 10 &= \frac{l\,w}{3} \\ 30 &= l\,w \end{align*}$
So yes I agree that the area of the rectangle is $\displaystyle \begin{align*} 30\,\textrm{cm}^2 \end{align*}$.
My method ignores the areas of the three surrounding triangles.
Let m(QA) = m(AB) = m(BR) = x
Let m(QC) = m(CP) = y
Locate point D, the midpoint of line segment AS, which is the diagonal of a rectangle having points
A, R, S, and some fourth unnamed point as vertices. Point D is the center of that rectangle.
CD is parallel to both QR and PS. Then CD is perpendicular to RS. So, line segment CD is a base
of triangles ACD and CDS.
Area of triangle ACS = area of triangle ACD + area of triangle CDS
= $\tfrac{1}{2}(2x)y + \tfrac{1}{2}(2x)y = $
$xy + xy = 2xy = 10$
Area of rectangle PQRS = (3x)(2y) = 6xy = 3(2xy) = 3(10) = 30
Area of rectangle PQRS = $ \ 30\ cm^2$