Situation:
I have a $8\times 8$ board (sic), but two squares from it's one diagonal are removed
(Black colored squares are removed)
I'm given with plenty of(Rectangular) bricks having dimensions $2 $ Units and $1$ units ( I don't need to use all of them ).
I've to find a way to cover all of the remaining(white) squares without overlapping bricks , and any brick used should be fully utilized ( Bricks should not cover any excess area except White squares) and I cannot break bricks.
Otherwise I've to prove that it's not possible.
What I did :
(In image Squares having 'X' written on them are removed)
If I transform given board into a chessboard Then I'll have $32$ white and $30$ Black squares, every brick is made up of two unit squares ( one is black and other is white ) Then clearly it's not possible,
Is there any another Interesting way to do this
The OP is asking for a proof that doesn't involve coloring the grid. Here's one that comes to mind.
Suppose you can cover the $8\times8$ grid with two opposite corners removed. It's easy to see that the number of vertically oriented dominoes that "hang" from the first row into the second must be odd (because there's an odd number of squares in the first row, and any horizontally oriented dominoes cover an even number of them). By the same reasoning, there must be an odd number of vertically oriented dominoes that hang from the second row into the third, and, by simple induction, for each row thereafter up to row $7$. In all, the number of vertically oriented dominoes is the sum of $7$ odd numbers, hence is odd.
By geometric symmetry, the same is true of the horizontally oriented dominoes: There must be an odd number of them. Thus the total number of dominoes in the covering is the sum of two odd numbers, hence even. But covering $62$ squares requires $31$ dominoes, which is odd.