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# Unit 6.1 Pythagorean Theorem:

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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)

Jan 23, 2018

#1
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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

Jan 23, 2018

#1
+2

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

hectictar Jan 23, 2018