Question

How do I solve this 45-45-90 triangle?

Answer 1

`\large\boxed{\tt x \ and \ y = 4}`

Explanation:

`\textsf{We are asked to solve for x and y of a 45-45-90 degree triangle.}`

`\large\underline{\textsf{What is a 45}^(\circ)\textsf{- 45}^(\circ) \textsf{- 90}^(\circ) \textsf{Triangle?}}`

`\textsf{A 45}^(\circ) \textsf{- 45}^(\circ) \textsf{- 90}^(\circ) \ \textsf{triangle is a special right triangle with legs that are congruent.}`

`\textsf{These kinds of Triangles are considered`

`\textsf{Right Triangles, and secondly they have special side lengths that can be figured}`

`\textsf{out with only one side length given.}`

`\underline{\textsf{What are the Ratios between the sides?}}`

`\tt Leg : Leg : Hypotenuse`

`\tt x : x : x \sqrt2`

`\textsf{If we are given a leg, the other leg will equal the same since these kinds of triangles}`

`\textsf{have 2 equal sides, which are considered \underline{Isosceles Triangles}. Given that 4 is one}`

`\textsf{leg, this means that the other leg is 4, and the Hypotenuse is 4} \tt \sqrt 2. \ \textsf{The}`

`\textsf{Hypotenuse is always multiplied by} \ \tt \sqrt 2 \ \textsf{to equal the sum of the legs. Finding}`

`\textsf{the legs are different however, as we need to divide the Hypotenuse by} \ \tt \sqrt2.`

`\large\underline{\textsf{Solving;}}`

`\textsf{We are given the Hypotenuse, and a 45}^(\circ) \ \textsf{angle to show that it's a 45}^(\circ)\textsf{- 45}^(\circ) \textsf{- 90}^(\circ)`

`\textsf{triangle. Let's divide the Hypotenuse by} \ \tt \sqrt 2 \ \textsf{to find the value of x and y. Remember}`

`\textsf{that the legs are equal to each other.}`

`\tt (Hypotenuse)/(\sqrt2) = Leg.`

`\tt \frac{4 √(\\ot2)}{\sqrt{\\ot{2}}} = x \ and \ y.`

`\large\boxed{\tt x \ and \ y = 4}`