Question

1. solve for the unknown variable for a right triangle: c = 16in, b = 12in, a = __

Explanation:

Step1: Recall Pythagorean theorem

For a right triangle, \(a^{2}+b^{2}=c^{2}\), so we can solve for \(a\) as \(a = \sqrt{c^{2}-b^{2}}\).

Step2: Substitute values

Given \(c = 16\) in and \(b = 12\) in, substitute into the formula: \(a=\sqrt{16^{2}-12^{2}}\).

Step3: Calculate squares

\(16^{2}=256\) and \(12^{2} = 144\), so \(a=\sqrt{256 - 144}\).

Step4: Subtract inside the square root

\(256-144 = 112\), so \(a=\sqrt{112}\).

Step5: Simplify the square root

\(\sqrt{112}=\sqrt{16\times7}=4\sqrt{7}\approx4\times2.6458 = 10.583\) (or keep it as \(4\sqrt{7}\) in exact form). But if we calculate the numerical value, \(\sqrt{112}\approx10.58\) in (or more precisely, let's check the calculation again: \(16^{2}-12^{2}=(16 + 12)(16 - 12)=28\times4 = 112\), so \(a=\sqrt{112}=4\sqrt{7}\approx10.58\) in. Wait, but maybe I made a mistake, wait \(16^{2}-12^{2}=256 - 144 = 112\), but wait, maybe the triangle is is there a typo? Wait no, wait \(12 - 16 - \) wait, maybe I miscalculated. Wait \(12^{2}+10^{2}=144 + 100 = 244
eq256\), \(12^{2}+10.58^{2}\approx144 + 112 = 256\), yes. So \(a=\sqrt{16^{2}-12^{2}}=\sqrt{256 - 144}=\sqrt{112}=4\sqrt{7}\approx10.58\) inches. But let's do it step by step.

Wait, maybe the problem is that I misread the triangle. Wait, the right triangle has legs \(a\) and \(b\), hypotenuse \(c\). So Pythagorean theorem: \(a^{2}+b^{2}=c^{2}\), so \(a=\sqrt{c^{2}-b^{2}}\). So \(c = 16\), \(b = 12\), so \(c^{2}=256\), \(b^{2}=144\), \(256 - 144 = 112\), so \(a=\sqrt{112}=4\sqrt{7}\approx10.58\) inches. Or maybe the question expects a simplified radical form or a decimal. Let's compute \(4\sqrt{7}\): \(\sqrt{7}\approx2.6458\), so \(4\times2.6458\approx10.58\) inches.

Answer:

\(4\sqrt{7}\) inches (or approximately \(10.58\) inches)