Question

1. $a = 5$ cm, $c = 13$ cm
2. $b$
3. $a = 8$ ft, $b = 6.2$ ft
4. $a$
5. $a = 1$ m, $b = \sqrt{3}$ m
6. $b$

Explanation:

Since the problem isn't fully clear (it seems to be about right - triangle side calculations, maybe finding the missing side using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), but the specific question for each sub - problem isn't stated), let's assume we need to find the missing side for each of these right - triangle problems.

Problem 16: \(a = 5\space cm\), \(c = 13\space cm\) (finding \(b\))

Step 1: Recall the Pythagorean theorem

For a right - triangle, \(a^{2}+b^{2}=c^{2}\), so we can solve for \(b\) as \(b=\sqrt{c^{2}-a^{2}}\)

Step 2: Substitute the values of \(a\) and \(c\)

Substitute \(a = 5\) and \(c = 13\) into the formula: \(b=\sqrt{13^{2}-5^{2}}=\sqrt{169 - 25}=\sqrt{144}\)

Step 3: Simplify the square root

\(\sqrt{144}=12\)

Step 1: Recall the Pythagorean theorem

\(a^{2}+b^{2}=c^{2}\), so \(c=\sqrt{a^{2}+b^{2}}\)

Step 2: Substitute the values of \(a\) and \(b\)

Substitute \(a = 8\) and \(b = 6.2\) into the formula: \(c=\sqrt{8^{2}+6.2^{2}}=\sqrt{64 + 38.44}=\sqrt{102.44}\)

Step 3: Calculate the square root

\(\sqrt{102.44}\approx10.1\) (rounded to the nearest tenth)

Step 1: Recall the Pythagorean theorem

\(a^{2}+b^{2}=c^{2}\), so \(c=\sqrt{a^{2}+b^{2}}\)

Step 2: Substitute the values of \(a\) and \(b\)

Substitute \(a = 1\) and \(b=\sqrt{3}\) into the formula: \(c=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{1 + 3}=\sqrt{4}\)

Step 3: Simplify the square root

\(\sqrt{4}=2\)

Answer:

\(b = 12\space cm\)

Problem 19: \(a = 8\space ft\), \(b = 6.2\space ft\) (finding \(c\))