Answer:
[tex]PQ^2+QR^2=PR^2[/tex]
Step-by-step explanation:
The picture of the question in the attached figure
we know that
In the right triangle PQR of the figure
Applying the Pythagorean Theorem
[tex]PR^2=PQ^2+QR^2[/tex]
Remember that the area of square is equal to
[tex]A=b^2[/tex]
where
b is the length side of the square
so
Area of square C
[tex]A_c=PR^2[/tex]
[tex]A_c=25\ units^2[/tex] ---> is given
so
[tex]PR^2=25\ units^2[/tex]
Area of square A
[tex]A_a=PQ^2[/tex]
[tex]A_a=9\ units^2[/tex] ---> is given
so
[tex]PQ^2=9\ units^2[/tex]
Area of square B
[tex]A_b=QR^2[/tex]
[tex]A_b=16\ units^2[/tex] ---> is given
so
[tex]QR^2=16\ units^2[/tex]
substitute the given values in the Pythagorean Theorem
[tex]PR^2=PQ^2+QR^2[/tex]
[tex]25=9+16[/tex]
[tex]25=25[/tex] ---> is true
The Pythagorean Theorem is satisfied (PQR is a right triangle)
we have
[tex]PR^2=PQ^2+QR^2[/tex]
rearrange
[tex]PQ^2+QR^2=PR^2[/tex]
