Question

part 2 of 4
substitute 10 for a and 11 for c into the equation given by the pythagorean theorem and simplify the squares.
$a^2 + b^2 = c^2$
$10^2 + b^2 = 11^2$
$100 + b^2 = 121$

part 3 of 4
solve for b. use a calculator if necessary. (round your answer to one decimal place.)
$100 + b^2 = 121$
$b^2 = 121 - 100$
$b^2 = \square$
$\sqrt{b^2} = \sqrt{\square}$
$b = \square$

Explanation:

Step1: Calculate \(121 - 100\)

We know that \(b^{2}=121 - 100\), so \(121-100 = 21\), thus \(b^{2}=21\).

Step2: Take square root of both sides

To solve for \(b\), we take the square root of both sides: \(\sqrt{b^{2}}=\sqrt{21}\). Since \(b\) represents a length (in the context of the Pythagorean theorem, assuming it's a side of a triangle), we consider the positive square root, so \(b = \sqrt{21}\approx4.6\) (rounded to one decimal place).

Answer:

First blank (for \(b^{2}\)): \(21\)
Second blank (inside the square root): \(21\)
Third blank (for \(b\)): \(\approx4.6\)