To the nearest tenth of an inch, find the length of a diagonal of a square whose side length is 8 inches.
Marissa/Mathew -- solve this problem the same way we showed you in your last post.
Okay thanks, Marissa is my sister.
To find the length of the diagonal of a square, you can use the Pythagorean theorem. The Pythagorean theorem relates the lengths of the sides of a right triangle.
In a square, the diagonal forms a right triangle with two sides equal to the lengths of the square's sides. Let's call the side length of the square "s" and the length of the diagonal "d."
According to the Pythagorean theorem, the sum of the squares of the two shorter sides of a right triangle is equal to the square of the hypotenuse. In this case, the two shorter sides are s (side length of the square) and s (again, side length of the square), and the hypotenuse is d (length of the diagonal).
So, we have the equation:
s^2 + s^2 = d^2
Since the side length of the square is given as 8 inches, we substitute s = 8 into the equation:
8^2 + 8^2 = d^2
Simplifying:
64 + 64 = d^2
128 = d^2
To find the length of the diagonal, we take the square root of both sides:
√128 = √d^2
Approximately:
11.31 = d
Therefore, to the nearest tenth of an inch, the length of the diagonal of a square with side length 8 inches is 11.3 inches.