2. The triangles BOC, OQC, on the same base BQ, have the same "height" (namely the perpendicular distance from C to BQ). Since the area of BOC is twice the area of OQC, it follows that $BO = 2OQ$. By a similar argument from the triangles COB, OBP, it follows that $CO = \frac54OP$.

Now let x be the area of triangle PAO, and let y be the area of triangle OAQ. Since $BO = 2OQ$, it follows (by the same reasoning as in the previous paragraph) that the area of triangle BAO is twice the area of triangle OAQ. In other words, $8+x = 2y$. A similar argument in triangles PAO and OAC shows that $5+y = \frac54x$.

That gives you two simultaneous equations for x and y. The answer to the problem is the sum x+y.

3. Thank you again Opalg, realy it is a logical solution.
The answer to the problem is the sum x+y as you say
x = 12 , y = 10
and x + y = 22

4. Originally Posted by razemsoft21
Alternatively, you can use ratios...

Using [BQ]

$5:10=yx+8)" alt="5:10=yx+8)" />

Using [PC]

$(5+y):x=10:8$

This is because......

$h_1=perpendicular\ height\ of\ triangles\ OAB\ and\ OAQ$

$\displaystyle\huge\frac{1}{2}|OQ|h_1=y,\ \frac{1}{2}|OB|h_1=x+8\ \Rightarrow\frac{h_1}{2}=\frac{y}{|OQ|}=\frac{x+8} {|OB|}\Rightarrow\frac{|OB|}{|OQ|}=\frac{x+8}{y}$

$h_2=perpendicular\ height\ of\ triangles\ OQC\ and\ OBC$

$\displaystyle\huge\frac{1}{2}|OQ|h_2=5,\ \frac{1}{2}|OB|h_2=10\Rightarrow\frac{h_2}{2}=\fra c{5}{|OQ|}=\frac{10}{|OB|}\Rightarrow\frac{|OB|}{| OQ|}=\frac{10}{5}$

$h_3=perpendicular\ height\ of\ triangles\ OAP\ and\ OAC$

$\displaystyle\huge\frac{1}{2}|OP|h_3=x,\ \frac{1}{2}|OC|h_3=y+5\Rightarrow\frac{h_3}{2}=\fr ac{x}{|OP|}=\frac{y+5}{|OC|}\Rightarrow\frac{|OC}{ OP|}=\frac{y+5}{x}$

$h_4=perpendicular\ height\ of\ triangles\ OBP\ and\ OBC$

$\displaystyle\huge\frac{1}{2}|OP|h_4=8,\ \frac{1}{2}|OC|h_4=10\Rightarrow\frac{h_4}{2}=\fra c{8}{|OP|}=\frac{10}{|OC|}\Rightarrow\frac{|OC|}{| OP|}=\frac{10}{8}$