if two equilateral triangles r inscribed in a unit radius,the minimum value of their common area is
1.√3
2.√3/2
3.√3/4
4.√6/4
1apply symmetry ..
no need to go into big calculation of derivatives and maxima minima etc.
area will be maximum when all the three uncommon parts are itself small equilateral triangles ..
the common area then =6/9 of total area
total area=√3/4(2cos30)2=3√3/4
hence overlapping area=2/3*3√3/4=√3/2
1see this is waht rohan meant
the area is minimum when the triangles are as shown . i.e. the resulting smaller yellow triangles are equilateral . so the common area is 6/9 of the aea of trangle DEF or ABC . hope this's fine
