Question

find the length of the third side. if necessary, write in simplest radical form. image of a right triangle with one leg 4, hypotenuse 2√5? wait no, wait the triangle has a right angle, one leg 4, one leg? wait no, the labels: one side is 4, another is 2√5, and right angle. wait, maybe 4 is a leg, 2√5 is the hypotenuse? wait no, lets re-express: the triangle is right-angled, with one leg 4, another side 2√5, find the third side. wait the ocr text: \find the length of the third side. if necessary, write in simplest radical form.\ then the image is a right triangle with one side labeled 4, another labeled 2√5, and the right angle. so ocr text is: \find the length of the third side. if necessary, write in simplest radical form.\

Explanation:

Step1: Identify the triangle type

This is a right - triangle, so we can use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse and \(a\), \(b\) are the legs. Here, the hypotenuse \(c = 2\sqrt{5}\)? Wait, no, wait. Wait, in a right - triangle, the hypotenuse is the longest side. Let's check the lengths. The two legs: let's assume one leg \(a = 4\), and the hypotenuse \(c=2\sqrt{5}\)? Wait, no, \(2\sqrt{5}\approx2\times2.236 = 4.472\), and \(4\) is less than \(4.472\). Wait, maybe I got the hypotenuse wrong. Wait, the right - angle is between the side of length \(4\) and the unknown side \(x\), and the other side is \(2\sqrt{5}\). So according to the Pythagorean theorem, if the hypotenuse is \(2\sqrt{5}\), then \(4^{2}+x^{2}=(2\sqrt{5})^{2}\)? Wait, no, that would give \(16 + x^{2}=20\), then \(x^{2}=4\), \(x = 2\). But wait, if the hypotenuse is the side with length \(2\sqrt{5}\), but \(2\sqrt{5}\approx4.47\) and \(4\) is a leg. Wait, maybe I mixed up. Let's re - examine. The right - angle is at the vertex where the side of length \(4\) and the unknown side \(x\) meet. So the two legs are \(4\) and \(x\), and the hypotenuse is \(2\sqrt{5}\)? Wait, no, because \(2\sqrt{5}\approx4.47\), and \(4\) is a leg, so the hypotenuse should be longer than the leg. Wait, maybe the hypotenuse is the side with length \(2\sqrt{5}\)? No, that can't be. Wait, maybe I made a mistake. Wait, let's calculate:

If we consider the Pythagorean theorem: Let the two legs be \(a = 4\) and \(b=x\), and the hypotenuse \(c = 2\sqrt{5}\). Then \(a^{2}+b^{2}=c^{2}\) would be \(16+x^{2}=20\), so \(x^{2}=4\), \(x = 2\). But wait, if the hypotenuse is \(2\sqrt{5}\), and one leg is \(4\), but \(4>2\sqrt{5}\) (since \(4 = \sqrt{16}\) and \(2\sqrt{5}=\sqrt{20}\)? Wait, no, \(\sqrt{16}=4\) and \(\sqrt{20}=2\sqrt{5}\approx4.47\), so \(\sqrt{20}>\sqrt{16}\), so \(2\sqrt{5}>4\). So the hypotenuse is \(2\sqrt{5}\), and one leg is \(4\), and the other leg is \(x\). Then by Pythagorean theorem: \(4^{2}+x^{2}=(2\sqrt{5})^{2}\)

Step2: Apply the Pythagorean theorem

We know that for a right - triangle, \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse, and \(a\) and \(b\) are the legs. Let \(a = 4\), \(b=x\) (the unknown side), and \(c = 2\sqrt{5}\).

Substitute the values into the formula:

\(4^{2}+x^{2}=(2\sqrt{5})^{2}\)

Calculate \(4^{2}=16\) and \((2\sqrt{5})^{2}=2^{2}\times(\sqrt{5})^{2}=4\times5 = 20\)

So the equation becomes:

\(16+x^{2}=20\)

Subtract \(16\) from both sides:

\(x^{2}=20 - 16=4\)

Take the square root of both sides:

\(x=\sqrt{4}=2\) (we take the positive value since length can't be negative)

Answer:

\(2\)