Figure Matrix

Given,

There are 26 triangles in the given figure, they are $\triangle \mathrm{ABK}, \triangle \mathrm{BKC}, \triangle \mathrm{ABC}, \triangle \mathrm{DCL}, \triangle \mathrm{DEL}, \triangle \mathrm{CDE}, \triangle \mathrm{AJK}, \triangle \mathrm{KJC}, \triangle \mathrm{AJC}, \triangle \mathrm{CJL}, \triangle \mathrm{LJE}, \triangle \mathrm{CJE}, \triangle \mathrm{ACE}, \triangle \mathrm{HGI}, \triangle \mathrm{GIF}, \triangle \mathrm{HGF}, \triangle \mathrm{AGJ}, \triangle \mathrm{EGJ}, \triangle \mathrm{AGE}, \triangle \mathrm{AGC}, \triangle \mathrm{ECG}, \triangle \mathrm{JCD}, \triangle \mathrm{JED}, \triangle \mathrm{ABJ}, \triangle \mathrm{BJC}$ and $\triangle \mathrm{BJD}$.

Clearly, in the first row, the number of dots in the second figure is twice the number of dots in the first figure. Similarly, in the second row, the number of dots in the second figure must be twice the number of dots in the first figure.

So, number of dots in missing segment = 2 $\times$ 3 = 6

The contents of the third figure in each row (and column) are determined by the contents of the first two figures. Lines are carried forward from the first two figures to the third one, except where two lines appear in the same position, in which they are cancelled out.

In figure (A), the dot is placed in the region which is common to the circle and triangle. Now, we have to search similar common region in the four options. Only in figure (c), we find such a region which is common to the circle and triangle.

Clearly, there are three types of shadings of circles, one is unshaded another has its right half shaded and yet another has its upper half shaded. Also, two of the figures in each row have one triangle shaded.