Which special version of the Pythagorean Theorem can you use to find the length of any square's diagonal, d, using only the length of its side, s?
is this a^2+b^2=c^2?
yes, but you are dealing with a square, where a=b. Plug that in and see what you get.
great
(leg)^2 + (leg)^2 = (hypotenuse)^2
Let leg = s
Let d = hypotenuse
s^2 + s^2 = d^2
2s^2 = d^2
sqrt{2s^2} = sqrt{d^2}
s•sqrt{2} = d
You can use d = s•sqrt{2} to find the length of a diagonal of any square.
I hope this helps.
No, the special version of the Pythagorean Theorem that can be used to find the length of any square's diagonal, d, using only the length of its side, s, is actually derived from the standard Pythagorean Theorem.
The standard Pythagorean Theorem states that for any right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be expressed as:
a^2 + b^2 = c^2
However, in the case of a square, all sides are equal in length, so we don't have the distinction of a "base" and "height" as we do in a general right-angled triangle. Instead, we can consider the diagonal of the square as the hypotenuse.
To find the length of the square's diagonal (d) using only the length of its side (s), we can create a right-angled triangle within the square. The hypotenuse of this triangle will be the diagonal of the square, and the other two sides will be two adjacent sides of the square.
Let's label one side of the square as s, which represents the length of the side. Then, we can label the other sides of the right-angled triangle formed within the square as s and s. Now we can apply the Pythagorean Theorem to this triangle:
s^2 + s^2 = d^2
2s^2 = d^2
To solve for d, we take the square root of both sides:
d = √(2s^2)
So, the special version of the Pythagorean Theorem to find the length of a square's diagonal using only the length of its side is d = √(2s^2).