To the nearest tenth of an inch, find the lengths of a diagonal of a square whose side lengths are 8 inches.
a. 8
b. 9.5
c. 11.3
d. 16
Can someone explain this to me I don't understand the question?
The answer is C
Your welcome also sorry I didn't have an explanation for this but I hope this helped :)
thank you
Certainly! The question is asking for the length of the diagonal of a square. A diagonal is a line segment that connects two opposite corners of a square.
To solve this problem, you can use the Pythagorean theorem. According to the theorem, in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
In this case, the diagonal of the square is the hypotenuse of a right triangle, and the two sides are the lengths of the square's sides.
The formula for the Pythagorean theorem is:
c² = a² + b²
Where:
- c is the length of the hypotenuse (the diagonal in this case)
- a and b are the lengths of the sides of the square
Using the given information, we know that the side length of the square is 8 inches. So, we can substitute this value into the formula:
c² = 8² + 8²
c² = 64 + 64
c² = 128
To find the length of the diagonal (c), we need to take the square root of both sides of the equation:
c = √(128)
c ≈ 11.3 (rounded to the nearest tenth)
Therefore, the length of the diagonal of the square, to the nearest tenth of an inch, is approximately 11.3 inches. So the correct answer is option c.
oops! misread the question....
we know the sides are 8
so d^2 = 8^2 + 8^2 = 128
d = √128 = ....
Just giving the answer does NOT help!
If you study this topic you must be learning about Pythagoras
In this case you have a square, where the sides are equal, so
x^2 + x^2 = 8^2
2x^2 = 64
x^2 = 32
x = ....