Answer to Question #103975 in Geometry for Alexandra

Question

It's later going to happen that you will need to know the side lengths of special right triangles where the hypotenuse is equal to 1. Sal just mentioned that the ratio of the sides of a 45-45-90 triangle is 1:1:LaTeX: \sqrt{2}2 and the ratio of the sides of a 30-60-90 triangle is 1: LaTeX: \sqrt{3}3 : 2.

For the assignment for this section, find the ratios in which the longest side is 1, that is each ratio should look like ___: ___: 1.

Answer 1

Solution

Take the first 45-45-90 triangle which is isosceles. The ratio of sides is $1 : 1: \sqrt{2}$ . It means that the hypotenuse, which is the biggest side in right triangle as it is lying against an angle of 90 degrees, has $\sqrt{2}$ parts and in the same time its length is 1. From this point, we can say that 1 part is equal to $\frac{1}{\sqrt{2}}$ . Therefore, legs, that are equal to each other, are equal to $1* \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$ . For this 45-45-90 triangle we have sides $\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1$

Take the second 30-60-90 triangle. The ratio of sides is $1 : \sqrt[3]{3} : 2$ . It means that the hypotenuse, which is the biggest side in right triangle as it is lying against an angle of 90 degrees, has 2 parts and in the same time its length is 1. From this point, we can say that 1 part is equal to $\frac{1}{2}$ . Therefore, one leg, that is equal to 1 part, is equal to $1 * \frac{1}{2} = \frac{1}{2}$ and other leg, that is equal to $\sqrt[3]{3}$ part, is equal to $\sqrt[3]{3} * \frac{1}{2} = \frac{\sqrt[3]{3}}{2}$ . For this 30-60-90 triangle we have sides $\frac{1}{2} : \frac{\sqrt[3]{3}}{2} : 1$ .

Answer

For 45-45-90 triangle: $\frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1$

For 30-60-90 triangle: $\frac{1}{2} : \frac{\sqrt[3]{3}}{2} : 1$