A.
The length of the diagonal is given by the Pythagorean theorem therefore
[tex]d=\sqrt[]{4^2+4^2}=\sqrt[]{16+16}=\sqrt[]{32}[/tex]
The length of the diagonal is ) units
B.
The area of the square is given by the next formula
[tex]A=s^2[/tex]
where s is the side
s=4
[tex]A=(4)^2=16units^2[/tex]
The area of the square is 16 units^2
C.
For the area of the triangle we will use the next formula
[tex]A=\frac{1}{2}b\times h[/tex]
where b is the base and h is the height
b=4 units
h=4units
[tex]A=\frac{1}{2}(4)(4)=\frac{1}{2}(16)=8units^2[/tex]
The area of the triangle formed by a diagonal and two of the sides is 8 units^2
D.
For the area of this triangle, we will use the same formula that we use in C. but in this case
b=sqrt(32)/2
h=sqrt(32)/2
We substitute the values
[tex]A=\frac{1}{2}(\frac{\sqrt[]{32}}{2})(\frac{\sqrt[]{32}}{2}))=4units^2[/tex]
The area of one of these triangles is 4 units^2
ANSWER
A. sqrt(32) units
B.16 units^2
C. 8 units^2
D. 4 units^2
