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Question 1:

In this 45-45-90 right triangle, the hypotenuse has a length of [tex]x \sqrt{2} [/tex] and the leg has a length of [tex]x[/tex]

We want to solve for x. Let's start with the equation of [tex]x \sqrt{2} =10[/tex] and solve for x.

[tex]x \sqrt{2} =10[/tex]

Divide both sides by the square root of 2.

[tex]x=\dfrac{10}{ \sqrt{2} }[/tex]

Simplify the fraction by multiply both the numerator and denominator by the sqaure root of 2.

[tex]x=\dfrac{10 \sqrt{2}}{2} [/tex]
[tex]=5 \sqrt{2} [/tex]

Question 2:

Let's imagine a right triangle where the legs are 250 and we want to solve for the hypotenuse. Since the legs are congruent, this is a 45-45-90 right triangle.

Multiply 250 by the square root of 2 and use a calculator.

[tex]250 \sqrt{2} \approx 353.6[/tex]

Hope this helps! :)

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