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To determine the length of one of the legs in a 45-45-90 triangle with a hypotenuse of 10 units, we'll use the properties of this specific type of triangle. In a 45-45-90 triangle, the legs are of equal length and the hypotenuse is related to the legs by the following relationship:
[tex]\[ \text{Hypotenuse} = \text{Leg} \times \sqrt{2} \][/tex]
Let's denote the length of one leg as \( x \). Therefore, we have:
[tex]\[ 10 = x \times \sqrt{2} \][/tex]
To solve for \( x \), we need to isolate \( x \):
[tex]\[ x = \frac{10}{\sqrt{2}} \][/tex]
To rationalize the denominator, multiply both the numerator and the denominator by \( \sqrt{2} \):
[tex]\[ x = \frac{10 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{10 \sqrt{2}}{2} \][/tex]
Simplify the fraction:
[tex]\[ x = \frac{10 \sqrt{2}}{2} = 5 \sqrt{2} \][/tex]
Thus, the length of each leg in this 45-45-90 triangle is \( 5 \sqrt{2} \) units. Given the options in the question, the correct answer is:
D. [tex]\( 5 \sqrt{2} \)[/tex] units
[tex]\[ \text{Hypotenuse} = \text{Leg} \times \sqrt{2} \][/tex]
Let's denote the length of one leg as \( x \). Therefore, we have:
[tex]\[ 10 = x \times \sqrt{2} \][/tex]
To solve for \( x \), we need to isolate \( x \):
[tex]\[ x = \frac{10}{\sqrt{2}} \][/tex]
To rationalize the denominator, multiply both the numerator and the denominator by \( \sqrt{2} \):
[tex]\[ x = \frac{10 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{10 \sqrt{2}}{2} \][/tex]
Simplify the fraction:
[tex]\[ x = \frac{10 \sqrt{2}}{2} = 5 \sqrt{2} \][/tex]
Thus, the length of each leg in this 45-45-90 triangle is \( 5 \sqrt{2} \) units. Given the options in the question, the correct answer is:
D. [tex]\( 5 \sqrt{2} \)[/tex] units
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