Please answer both questions

Part a : 20 units², Part d : 12 units²

Step-by-step explanation:

We are given that the side lengths of each square is 6 units. In this case we can calculate the area of the triangles, subtracting from the area of the outer square, such that we calculate the remaining, area of the shaded region / inscribed square.

Area of Outer Square ( Same in all Cases ) = 6 6 = 36 units²,

Area of Common Triangles ( 4 Triangles ) = 1 / 2 Base Height = 1 / 2 2 ( 6 - 2 ) = 1 / 2 2 4 = 4,

4 4 = 16 units²

Area of inscribed Square = 36 units² - 16 units² = 20 units²

For this second case the inscribed shape is not a square, but it can be calculated through a similar approach,

Area of Outer Square ( Same in all Cases ) = 6 6 = 36 units²,

Area of Common Triangles ( 2 Triangles ) = 1 / 2 Base Height = 1 / 2 4 6 = 2 6 = 12,

12 2 = 24 units²

Area of inscribed shape = 36 units² - 24 units² = 12 units²

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b