Question

homework: unit 3 - geometry question 41, 6.6.31 hw score: 86.21%, 50 of 58 points points: 0 of 1 find the unknown length in the right triangle. image of right triangle with 15 cm (vertical leg), 18 cm (hypotenuse), and ? (horizontal leg) the unknown length is approximately (type a whole number or decimal rounded to the nearest thousandth as needed.)

Explanation:

Step1: Identify the formula

We use the Pythagorean theorem for a right triangle, which is \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse and \(a\), \(b\) are the legs. Here, the hypotenuse \(c = 18\) cm and one leg \(a = 15\) cm. We need to find the other leg \(b\). Rearranging the formula for \(b\), we get \(b=\sqrt{c^{2}-a^{2}}\).

Step2: Substitute the values

Substitute \(c = 18\) and \(a = 15\) into the formula: \(b=\sqrt{18^{2}-15^{2}}\). First, calculate \(18^{2}=324\) and \(15^{2}=225\). Then, \(18^{2}-15^{2}=324 - 225 = 99\). So, \(b=\sqrt{99}\).

Step3: Calculate the square root

\(\sqrt{99}\approx9.949874371\). Rounding to the nearest thousandth, we get approximately \(9.950\). But if we round to a whole number or a reasonable decimal, \(\sqrt{99}\approx9.95\) (or 10 if rounded to the nearest whole number, but let's check the precise calculation). Wait, \(9.94987\) is approximately \(9.950\) when rounded to the nearest thousandth. But maybe the problem expects a simpler approach. Wait, \(99 = 9\times11\), so \(\sqrt{99}=3\sqrt{11}\approx3\times3.3166 = 9.9498\), so approximately \(9.95\) (or 10 if to the nearest whole number). But let's follow the steps.

Answer:

The unknown length is approximately \(\boldsymbol{9.95}\) cm (or \(\boldsymbol{10}\) cm if rounded to the nearest whole number; the precise value to the nearest thousandth is \(9.950\)).