Question

What is the best answer for the length of side x

Answer 1

Answer:

$5\sqrt{2}$

Step-by-step explanation:

Solution 1 (Pythagorean Theorem):

In any right triangle, the Pythagorean Theorem holds true: $a^2+b^2=c^2$, where $c$ is the hypotenuse of the triangle.

From the Isosceles-Base Theorem, the other leg of the triangle (unlabeled) must 5. Therefore, we have:

$5^2+5^2=x^2,\\25+25=x^2,\\x^2=50,\\x=\sqrt{50}=\sqrt{25}\cdot \sqrt{2}=\boxed{5\sqrt{2}}$

Solution 2 (ratios of a 45-45-90 triangle derived from the Pythagorean Theorem):

In any 45-45-90 triangle, the ratio of the sides will be $x: x:x\sqrt{2}$, where $x\sqrt{2}$ is the hypotenuse. From there, we can see that one of the legs has a length of 5, and thus the hypotenuse will be $\boxed{5\sqrt{2}}$. (derived from Pythagorean Theorem).

Solution 3 (Law of Sines):

In any triangle, the Law of Sines holds true: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$.

$\frac{5}{\sin 45^{\circ}}=\frac{x}{\sin 90^{\circ}}$, and recall $\sin 90^{\circ}=1$.

$x=\frac{5}{\sin45^{\circ}}=\boxed{5\sqrt{2}}$ (recall $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$)

Solution 4 (basic trig. for right triangles):

In any right triangle, the cosine of an angle is given by its adjacent side divided by the hypotenuse of the triangle.

$\cos45^{\circ}=\frac{5}{x},\\x=\frac{5}{\cos 45^{\circ}}=\boxed{5\sqrt{2}}$ (recall $\cos 45^{\circ}=\frac{\sqrt{2}}{2}$)

Solution 5 (Law of Cosines):

In any triangle, the Law of Cosines holds true: $c^2=a^2+b^2-2ab\cos C$.

Assigning $a=5,\: b=5, \:c=x, \:C=90$, we have:

$c^2=5^2+5^2-2\cdot5\cdot5\cdot\cos 90^{\circ},\\c^2=50,\\c=\sqrt{50}=\sqrt{25}\cdot \sqrt{2}=\boxed{5\sqrt{2}}$(recall $\cos 90^{\circ}=0$)