Denote the area of quadilateral MBQP as A, the area of the triangle MBP as [tex]A_1[/tex] and the area of the triangle BPQ as [tex]A_2.[/tex]
Note that:
1.
[tex]A_{\triangle CMB}=\dfrac{1}{3}A_{\triangle ABC}=\dfrac{1}{3}\cdot 30=10\ cm^2.[/tex]
2.
[tex]A_{\triangle CPQ}=A_2.[/tex]
3.
[tex]A_{\triangle ABQ}=\dfrac{1}{2}A_{\triangle ABC}=\dfrac{1}{2}\cdot 30=15\ cm^2.[/tex]
4.
[tex]A_{\triangle APM}=2A_1.[/tex]
Now
[tex]A+A_{\triangle CPQ}=A_{\triangle CMB}\Rightarrow A+A_2=10\ cm^2.[/tex]
[tex]A+A_{\triangle AMP}=A_{\triangle ABQ}\Rightarrow A+2A_1=15\ cm^2.[/tex]
You get the system of three equations:
[tex]\left\{\begin{array}{l}A=A_1+A_2\\A+A_2=10\\A+2A_1=15\end{array}\right..[/tex]
Substitute the first equation into the last two:
[tex]\left\{\begin{array}{l}A_1+A_2+A_2=10\\A_1+A_2+2A_1=15\end{array}\right.\Rightarrow \left\{\begin{array}{l}A_1+2A_2=10\\A_2+3A_1=15\end{array}\right..[/tex]
From the first equation [tex]A_1=10-2A_2[/tex] and then
[tex]A_2+3(10-2A_2)=15,\\ \\A_2+30-6A_2=15,\\ \\-5A_2=-15,\\ \\A_2=3\ cm^2.[/tex]
Thus,
[tex]A_1=10-2\cdot 3=10-6=4\ cm^2[/tex]
and
[tex]A=4+3=7\ cm^2.[/tex]
Answer: 7 sq. cm
