A geometry problem by Ajay Sambhriya

By pythagorean theorem: $AC=\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt{2}$

$AC=AO+OC$ but $AO=3(OC)$ . Therefore, $AC=3(OC)+OC=4(OC)$ . We find $OC=\dfrac{5\sqrt{2}}{4}$ . It follows that $AO=3\left(\dfrac{5\sqrt{2}}{4}\right)=\dfrac{15\sqrt{2}}{4}$ .

$A_{AOB}=\dfrac{1}{2}(5)\left(\dfrac{15\sqrt{2}}{4}\right)(\sin 45)=9.375$

$A_{DOC}=\dfrac{1}{2}(5)\left(\dfrac{5\sqrt{2}}{4}\right)(\sin 45)=3.125$

Finally,

$A_{AOB}+A_{DOC}=9.375+3.125=12.5$

Draw two perpendicular lines each parallel to the sides of the square through $O$ . The two lines divide the square into 4 pairs of equal area triangles $a$ , $b$ , $c$ and $d$ . Implying that the area of the square $[ABCD]=2(a+b+c+d)=5^2 = 25$ .

We note that $[AOB]+[DOC] = a+b+c+d = \dfrac {25}2 = \boxed{12.5}$ .

Draw a line through point O parallel to AD. Let the intersection of this line and AB be denoted F, and define FO = h. Then OG = 5 - h, where G is the extension of FO until it intersects CD. Then the sum of areas is (1/2)(5)(h) + (1/2)(5 -h)(5) = 12.5. Ed Gray