Question

HELP PLEASE !!!!!!! Pythagorean theorem

Answer 1

We are missing one of the legs. Thus, we rearrange the original formula of the pythagorean theorem...

`\boxed{A^2+B^2=C^2}` →
`\boxed{A^2=C^2-B^2}`

Let's solve it...

`Plug\;in\;the\;given\;numbers\;to\;the\;formula...\\\\A^2=16^2-8^2\\\\Find\;what\;the\;squared\;numbers\;equal\\\\\left[\begin{array}{ccc}16^2(16*16)=256\\8^2(8*8)=64\end{array}\right]\\\\\\Now,\;since\;we\;are\;trying\;to\;find;\a\;missing\;leg\;we\;subtract\;256\;from\;64...\\\\\boxed{256-64=192}\\\\\\Then,\;we\;find\;the\;square\;root\;of\;192\;(√(x) )\\\\\boxed{√(192)=13.856 }\\\\\\Since\;we\;have\;a\;decimal\;we\;need\;to\;round\;it\;to\;the\;nearest\;tenths...\\\\\boxed{13.856=13.9}`

Answer 2

`\large\boxed{x=8\sqrt3}`

Explanation:

The Pythagorean theorem:

`leg^2+leg^2=hypotenuse^2`

We have:

`leg=8, leg=x,\ hypotenuse=16`

Substitute:

`8^2+x^2=16^2`

`64+x^2=256` subtract 64 from both sides

`x^2=192\to x=√(192)\\\\x=√((64)(3))\\\\x=√(64)\cdot\sqrt3\\\\x=8\sqrt3`