Question

8.) b = 4 yd c = 6 yd a = ____

Explanation:

Assuming this is a right - triangle problem and we are using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c\) is the hypotenuse and \(a,b\) are the legs of the right triangle). We want to solve for \(a\), so we can re - arrange the formula to \(a=\sqrt{c^{2}-b^{2}}\).

Step 1: Substitute the given values

We know that \(b = 4\) yd and \(c=6\) yd. Substitute these values into the formula \(a=\sqrt{c^{2}-b^{2}}\). So we have \(a=\sqrt{6^{2}-4^{2}}\).

Step 2: Calculate the squares

First, calculate \(6^{2}=36\) and \(4^{2} = 16\). Then the expression becomes \(a=\sqrt{36 - 16}\).

Step 3: Subtract inside the square root

Subtract \(16\) from \(36\): \(36-16 = 20\). So now \(a=\sqrt{20}\).

Step 4: Simplify the square root

We can simplify \(\sqrt{20}\) as \(\sqrt{4\times5}=\sqrt{4}\times\sqrt{5}=2\sqrt{5}\approx2\times2.24 = 4.48\) (if we want a decimal approximation). If we leave it in exact form, it is \(2\sqrt{5}\) yd, and if we approximate to two decimal places, it is approximately \(4.48\) yd.

Answer:

If in exact form: \(2\sqrt{5}\) yd (or approximately \(4.48\) yd)