Question

What is the length of a diagonal of a square with a side length of 8? A.4 B.4√2 C.4√3 D.8√2

Answer 1

Answer:

Step-by-step explanation:

A square by definition has sides all the same length.  I know you know this.  The diagonal of a square forms the hypotenuse of a right triangle that is, to be specific, a 45-45-90 right triangle.  The height of the right triangle is 8 (one of the side lengths of the square), and the base of the right triangle is 8 (another of the side lengths of the square), and the diagonal, as already stated, is the hypotenuse.  We can use Pythagorean's Theorem to solve for the length of the hypotenuse:

$8^2+8^2=c^2$  and

$64+64=c^2$  and

$c^2=128$ so

$c=\sqrt{128}$

Now we just need to find the largest perfect square that multiplied by another number (preferably a prime number) that gives us 128.  I do believe that 64 * 2 = 128, and 64 is the largest perfect square that goes into 128 evenly.  2 is prime.  Simplifying our value for c:

$c=\sqrt{64*2}$  which simplifies finally to

$c=8\sqrt{2}$

That is the length of the diagonal of that square.

Answer 2

Answer:

Step-by-step explanation:

A square by definition has sides all the same length.  I know you know this.  The diagonal of a square forms the hypotenuse of a right triangle that is, to be specific, a 45-45-90 right triangle.  The height of the right triangle is 8 (one of the side lengths of the square), and the base of the right triangle is 8 (another of the side lengths of the square), and the diagonal, as already stated, is the hypotenuse.  We can use Pythagorean's Theorem to solve for the length of the hypotenuse:

 and

 and

so

Now we just need to find the largest perfect square that multiplied by another number (preferably a prime number) that gives us 128.  I do believe that 64 * 2 = 128, and 64 is the largest perfect square that goes into 128 evenly.  2 is prime.  Simplifying our value for c:

 which simplifies finally to

That is the length of the diagonal of that square.