Question

what is the length of side x on the right triangle below? image of right triangle with right angle, vertical side x, horizontal side 3, hypotenuse 4 a 5 b 7 c 25 d √7

Explanation:

Step1: Apply Pythagorean theorem

For a right triangle, the Pythagorean theorem is \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse, and \(a\), \(b\) are the legs. Here, hypotenuse \(c = 4\), one leg \(b = 3\), and the other leg is \(x\). So we have \(x^2 + 3^2 = 4^2\).

Step2: Solve for \(x^2\)

Calculate \(3^2 = 9\) and \(4^2 = 16\). Then \(x^2 = 16 - 9 = 7\)? Wait, no, wait. Wait, maybe I mixed up the legs and hypotenuse. Wait, the right angle is between \(x\) and \(3\), so the hypotenuse is \(4\)? Wait, no, that can't be, because \(3\) and \(x\) are legs, hypotenuse should be longer than either leg. Wait, maybe I made a mistake. Wait, no, the hypotenuse is the side opposite the right angle, so in the triangle, the side with length \(4\) is the hypotenuse? Wait, but \(3\) is a leg, \(x\) is a leg, and \(4\) is hypotenuse? But \(3^2 + x^2 = 4^2\), so \(x^2 = 16 - 9 = 7\), so \(x = \sqrt{7}\)? Wait, but that's option D. Wait, but maybe I misread the triangle. Wait, the triangle has legs \(x\) and \(3\), hypotenuse \(4\)? Wait, but 3-4-5 is a Pythagorean triple. Wait, maybe the hypotenuse is 5? Wait, no, the diagram shows the hypotenuse as 4? Wait, no, maybe the diagram is labeled wrong? Wait, no, let's check again. The right triangle has one leg 3, hypotenuse 4, and the other leg x. Then by Pythagoras, \(x^2 + 3^2 = 4^2\), so \(x^2 = 16 - 9 = 7\), so \(x = \sqrt{7}\), which is option D. Wait, but maybe I made a mistake. Wait, no, the Pythagorean theorem is \(a^2 + b^2 = c^2\), where \(c\) is hypotenuse. So if the legs are \(x\) and \(3\), hypotenuse \(4\), then \(x^2 = 4^2 - 3^2 = 16 - 9 = 7\), so \(x = \sqrt{7}\). So the answer is D.

Answer:

D. \(\sqrt{7}\)