Question

ex 2 find x. a right triangle with one leg labeled x, the other leg labeled 8 feet, and the hypotenuse labeled 13 feet. answer:

Explanation:

Step 1: Apply Pythagorean theorem

For a right triangle, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse, and \(a\), \(b\) are the legs. Here, \(c = 13\), \(b = 8\), \(a = x\). So \(x^2+8^2 = 13^2\).

Step 2: Solve for \(x^2\)

Calculate \(8^2 = 64\), \(13^2 = 169\). Then \(x^2=169 - 64=105\)? Wait, no, wait: Wait, \(13^2 - 8^2=(13 - 8)(13 + 8)=5\times21 = 105\)? Wait, no, wait, 13 squared is 169, 8 squared is 64, 169 - 64 is 105? Wait, no, wait, 1313 is 169, 88 is 64, 169 - 64 = 105? Wait, no, that can't be. Wait, no, wait, maybe I made a mistake. Wait, 12^2 + 5^2 = 13^2? No, 5^2+12^2=13^2. Wait, 8 is not 12. Wait, let's recalculate: \(x^2=13^2 - 8^2=169 - 64 = 105\)? Wait, no, 169 - 64 is 105? Wait, 64 + 105 is 169? 64 + 100 is 164, plus 5 is 169. Yes. Wait, but 105 is not a perfect square. Wait, maybe the triangle is different? Wait, no, the hypotenuse is 13, one leg is 8, so the other leg is \(\sqrt{13^2 - 8^2}=\sqrt{169 - 64}=\sqrt{105}\approx10.25\)? Wait, no, wait, maybe I misread the triangle. Wait, the right angle is at the left, so the legs are x and 8, hypotenuse 13. So Pythagorean theorem: \(x^2 + 8^2 = 13^2\). So \(x^2=13^2 - 8^2=169 - 64 = 105\), so \(x=\sqrt{105}\approx10.25\) feet? Wait, but maybe the problem has a typo? Wait, no, maybe I made a mistake. Wait, 5-12-13 triangle: 5^2 + 12^2 = 13^2. So if one leg is 8, then the other leg is \(\sqrt{13^2 - 8^2}=\sqrt{169 - 64}=\sqrt{105}\approx10.24695\) feet.

Wait, but let's check again. \(x^2 + 8^2 = 13^2\)

\(x^2 = 13^2 - 8^2\)

\(x^2 = (13 - 8)(13 + 8)\) (using difference of squares: \(a^2 - b^2=(a - b)(a + b)\))

\(x^2 = 5\times21 = 105\)

So \(x=\sqrt{105}\approx10.25\) feet.

Answer:

\(\sqrt{105}\) feet (or approximately \(10.25\) feet)