Question

find each missing length to the nearest tenth.
7)
8)
9)
10)
11)
12)

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), the Pythagorean theorem is \(a^{2}+b^{2}=c^{2}\). If we need to find a leg, say \(a\), then \(a=\sqrt{c^{2}-b^{2}}\); if we need to find the hypotenuse \(c\), then \(c = \sqrt{a^{2}+b^{2}}\)

Step2: Solve problem 7

Let the two legs be \(a = 4\) and \(b\), and the hypotenuse \(c=8\). Using \(a^{2}+b^{2}=c^{2}\), we substitute \(a = 4\) and \(c = 8\) into the formula:

$$b=\sqrt{c^{2}-a^{2}}=\sqrt{8^{2}-4^{2}}=\sqrt{64 - 16}=\sqrt{48}\approx6.9$$

Step3: Solve problem 8

Let the two legs be \(a\) and \(b = 3\), and the hypotenuse \(c = 6\). Using \(a^{2}+b^{2}=c^{2}\), we substitute \(b = 3\) and \(c = 6\) into the formula:

$$a=\sqrt{c^{2}-b^{2}}=\sqrt{6^{2}-3^{2}}=\sqrt{36-9}=\sqrt{27}\approx5.2$$

Step4: Solve problem 9

Let the two legs be \(a = 7\) and \(b\), and the hypotenuse \(c = 10\). Using \(a^{2}+b^{2}=c^{2}\), we substitute \(a = 7\) and \(c = 10\) into the formula:

$$b=\sqrt{c^{2}-a^{2}}=\sqrt{10^{2}-7^{2}}=\sqrt{100 - 49}=\sqrt{51}\approx7.1$$

Step5: Solve problem 10

Let the two legs be \(a = 3\) and \(b = 7\). Using \(c=\sqrt{a^{2}+b^{2}}\), we substitute \(a = 3\) and \(b = 7\) into the formula:

$$c=\sqrt{3^{2}+7^{2}}=\sqrt{9 + 49}=\sqrt{58}\approx7.6$$

Step6: Solve problem 11

Let the two legs be \(a\) and \(b = 2\), and the hypotenuse \(c = 7\). Using \(a^{2}+b^{2}=c^{2}\), we substitute \(b = 2\) and \(c = 7\) into the formula:

$$a=\sqrt{c^{2}-b^{2}}=\sqrt{7^{2}-2^{2}}=\sqrt{49-4}=\sqrt{45}\approx6.7$$

Step7: Solve problem 12

Let the two legs be \(a = 2\) and \(b\), and the hypotenuse \(c = 6\). Using \(a^{2}+b^{2}=c^{2}\), we substitute \(a = 2\) and \(c = 6\) into the formula:

$$b=\sqrt{c^{2}-a^{2}}=\sqrt{6^{2}-2^{2}}=\sqrt{36 - 4}=\sqrt{32}\approx5.7$$

Answer:

1. \(6.9\); 8) \(5.2\); 9) \(7.1\); 10) \(7.6\); 11) \(6.7\); 12) \(5.7\)