Answer:
4
Step-by-step explanation:
Let x be the side of the square, BP=6/x ( area = 3 sq.units) ; CQ = 2/x (area = 1 sq. unit); AE = 2/x ( area = 1 sq. unit); BP = 3CQ or 3 AE.
Draw perpendicular AED from vertex A to side BC,AE being the altitude for Δ ARO (since RO parallel to BC).
ΔARO similar to ΔABC ( since RO parallel BC angle ARO = angle ACB and angle AOR = angle ABP)
AR/AC = AO/AB=RO/BC=AE/AD
BC/RO = AD/AE
(BC-RO) / RO = (AD-AE)/AE
(BP + CQ)/RO = x / AE
(4AE) / x = x/AE , x = 2 AE;
Area of ARO = 1/2 * AE* x = 1; AE = 1
AE=1; x =2; Area = 4 sq units.
