Khan Academy Pythagorean Theorem in 3D Question

[blasto]

Hello,

I am working through Khan Academy, and am having a lot of trouble figuring out how they came up with the following:

Could you explain step by step how:

(2r)^{^2} = 3^{^2} + 3^{^2}

Becomes:

r^{^2}=9/2

Which then leads to:

r=3 square root of 2 divided by 2?

I get r^{^2}=9 on the second step no matter how I try to work it out. How is it losing the second 3^{^2}? Aren't we just dividing by 2 on both sides to eliminate the "2" in (2r)^{^2}?

The context is using the pythagorean theorem to find r on the square base of a pyramid with dimensions 3x3 (the square base) as part of the calculation to find the length of the side edge of a pyramid.

Thank you for any help.

 

[Ishuda]

Hello,

I am working through Khan Academy, and am having a lot of trouble figuring out how they came up with the following:

Could you explain step by step how:

(2r)^{^2} = 3^{^2} + 3^{^2}

Becomes:

r^{^2}=9/2

Which then leads to:

r=3 square root of 2 divided by 2?

I get r^{^2}=9 on the second step no matter how I try to work it out. How is it losing the second 3^{^2}? Aren't we just dividing by 2 on both sides to eliminate the "2" in (2r)^{^2}?

The context is using the pythagorean theorem to find r on the square base of a pyramid with dimensions 3x3 (the square base) as part of the calculation to find the length of the side edge of a pyramid.

Thank you for any help.

(2r)^{^2} = = (2 r) * (2 r) = 4 r² = 3² + 3² = 2 * 3²
or
4 r² = 2 3²
Divide thru by 4 and take the square root
\(\displaystyle r = \pm \sqrt{\frac{2}{4}\, 3^2 }\, =\, \pm \frac{\sqrt{2}}{2}\, 3 \)

 

[blasto]

Thank you

(2r)^{^2} = = (2 r) * (2 r) = 4 r² = 3² + 3² = 2 * 3²
or
4 r² = 2 3²
Divide thru by 4 and take the square root
\(\displaystyle r = \pm \sqrt{\frac{2}{4}\, 3^2 }\, =\, \pm \frac{\sqrt{2}}{2}\, 3 \)

Thank you for your help!