Question

i. determine whether each set of measures can be the sides of a right triangle. 1. 3, 4, 5 2. 5, 5, 10 3. 8, 12, 13 4. 26, 24, 10 ii. use the pythagorean theorem to find the length of the missing third side. round to the nearest tenth if necessary. 5. a = 5, b = 7, c = 8.6 6. a = 6, b = 8, c = 10 7. a = 9, b = 12, c = 15 8. a = , b = 4/12, c = 5/12

Explanation:

Step1: Recall Pythagorean Theorem

For a right - triangle, \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse (the longest side).

Step2: Solve for problem 5

Given \(a = 5\) and \(b = 7\), we find \(c\) using \(c=\sqrt{a^{2}+b^{2}}\). So \(c=\sqrt{5^{2}+7^{2}}=\sqrt{25 + 49}=\sqrt{74}\approx8.6\).

Step3: Solve for problem 6

Given \(b = 8\) and \(c = 10\), we find \(a\) using \(a=\sqrt{c^{2}-b^{2}}\). So \(a=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6\).

Step4: Solve for problem 7

Given \(a = 9\) and \(c = 15\), we find \(b\) using \(b=\sqrt{c^{2}-a^{2}}\). So \(b=\sqrt{15^{2}-9^{2}}=\sqrt{225 - 81}=\sqrt{144}=12\).

Step5: Solve for problem 8

Given \(b=\frac{4}{12}\) and \(c=\frac{5}{12}\), we find \(a\) using \(a=\sqrt{c^{2}-b^{2}}\). So \(a=\sqrt{(\frac{5}{12})^{2}-(\frac{4}{12})^{2}}=\sqrt{\frac{25}{144}-\frac{16}{144}}=\sqrt{\frac{25 - 16}{144}}=\sqrt{\frac{9}{144}}=\frac{3}{12}=\frac{1}{4}\).

Answer:

1. \(8.6\)
2. \(6\)
3. \(12\)
4. \(\frac{1}{4}\)