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

Question 1

Question 2

Answer:

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

Explanation:

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

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

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

$\textsf{These kinds of Triangles are considered$

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

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

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

$\tt Leg : Leg : Hypotenuse$

$\tt x : x : x \sqrt2$

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

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

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

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

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

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

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

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

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

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

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

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

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