Answer:
c 11 and 56 A 8 and 58
Step-by-step explanation:
B. 7 and 60
C. 11 and 56
see the attached figure to better understand the problem
we know that
The area of the three squares must satisfy the Pythagorean Theorem
so
where
c^2 is the area of the largest square
a^2 and b^2 are the areas of the smaller squares
The sum of the areas of the smaller squares must be equal to 67
therefore
7 and 60 ----> could be the areas of the smaller squares (60+7=67)
11 and 56 ----> could be the areas of the smaller squares (11+56=67)
Firstly, subtract 2 from all sides of this inequality:
Next, divide by 2 on all sides of the inequality and <u>your answer will be </u>