G is the centroid of triangle ABC and A1 and .B1 are the midpoints of sides AB and AC, respectively. If ∆1 is the area of quadrilateral GA1,AB1 and ∆ is the area of triangle ABC then  ∆1∆2 is equal to

Let P.V. of A,B and C be 0→,b→ and c→ respectivelyTherefore, G→=b→+c→3A→1=b→2,B→1=c→2ΔAB1G=12AG→×AB1→=12b→+c→3×c→2=112|b→×c→|ΔAA1G=12AG→×AA1→=12b→+c→3×b→2=112|b→×c→|⇒ ΔGA1AB1=16|b→×c→|=13⋅12|b→×c→|=13ΔABC⇒ ΔΔ1=3