Unit 6.1 Pythagorean Theorem:
The question I'm having difficulty is a square. I'm trying to find the legs with the hypotenuse being 5sqrt(2)
+9465 Remember, the Pythagorean theorem says...
(leg)2 + (other leg)2 = (hypotenuse)2
Since the legs of this triangle are sides of a square, they must be the same length.
Let's call the length of each leg s , and plug in 5√2 for the hypotenuse.
s2 + s2 = ( 5√2 )2
Now we can solve this equation for s .
2s2 = ( 5√2 )2
2s2 = 5√2 * 5√2
Multiplication can be done in any order.
2s2 = 5 * 5 * √2 * √2
And √2 * √2 = 2
2s2 = 25 * 2
Divide both sides of the equation by 2 .
s2 = 25
Take the positive, since s is side length, square root of both sides.
s = 5
+9465 Remember, the Pythagorean theorem says...
(leg)2 + (other leg)2 = (hypotenuse)2
Since the legs of this triangle are sides of a square, they must be the same length.
Let's call the length of each leg s , and plug in 5√2 for the hypotenuse.
s2 + s2 = ( 5√2 )2
Now we can solve this equation for s .
2s2 = ( 5√2 )2
2s2 = 5√2 * 5√2
Multiplication can be done in any order.
2s2 = 5 * 5 * √2 * √2
And √2 * √2 = 2
2s2 = 25 * 2
Divide both sides of the equation by 2 .
s2 = 25
Take the positive, since s is side length, square root of both sides.
s = 5
