Question

1. use the pythagorean theorem to find x. round to the nearest tenth.

Explanation:

Step1: Recall Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\).

Step2: Solve for part a

In the right - triangle with legs 6 and 8, and hypotenuse \(x\). By the Pythagorean theorem, \(x^{2}=6^{2}+8^{2}\). Calculate \(6^{2}=36\) and \(8^{2}=64\), then \(x^{2}=36 + 64=100\), so \(x=\sqrt{100}=10\).

Step3: Solve for part b

In the right - triangle with legs 3 and \(x\) and hypotenuse 5. By the Pythagorean theorem, \(x^{2}=5^{2}-3^{2}\). Calculate \(5^{2}=25\) and \(3^{2}=9\), then \(x^{2}=25 - 9 = 16\), so \(x=\sqrt{16}=4\).

Step4: Solve for part c

In the right - triangle with legs 5 and \(x\) and hypotenuse 10. By the Pythagorean theorem, \(x^{2}=10^{2}-5^{2}\). Calculate \(10^{2}=100\) and \(5^{2}=25\), then \(x^{2}=100 - 25=75\), so \(x=\sqrt{75}\approx8.7\).

Step5: Solve for part d

In the right - triangle with legs 11 and \(x\) and hypotenuse 13. By the Pythagorean theorem, \(x^{2}=13^{2}-11^{2}\). Calculate \(13^{2}=169\) and \(11^{2}=121\), then \(x^{2}=169 - 121 = 48\), so \(x=\sqrt{48}\approx6.9\).

Answer:

a. \(x = 10\)
b. \(x = 4\)
c. \(x\approx8.7\)
d. \(x\approx6.9\)