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Sagot :
Alright, let's solve this step by step.
Given:
- We have an isosceles right triangle.
- The hypotenuse ([tex]\( c \)[/tex]) measures 6 inches.
- We need to find the length of one of the legs ([tex]\( a \)[/tex]).
In an isosceles right triangle, the relationship between the legs and the hypotenuse is derived from the Pythagorean theorem. For such a triangle:
[tex]\[ c = a\sqrt{2} \][/tex]
Given [tex]\( c = 6 \)[/tex] inches, we substitute the value into the equation:
[tex]\[ 6 = a\sqrt{2} \][/tex]
To isolate [tex]\( a \)[/tex], we divide both sides of the equation by [tex]\( \sqrt{2} \)[/tex]:
[tex]\[ a = \frac{6}{\sqrt{2}} \][/tex]
To rationalize the denominator, we multiply the numerator and the denominator by [tex]\( \sqrt{2} \)[/tex]:
[tex]\[ a = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \][/tex]
Thus, the length of each leg ([tex]\( a \)[/tex]) of the isosceles right triangle is:
[tex]\[ 3\sqrt{2} \][/tex]
Given:
- We have an isosceles right triangle.
- The hypotenuse ([tex]\( c \)[/tex]) measures 6 inches.
- We need to find the length of one of the legs ([tex]\( a \)[/tex]).
In an isosceles right triangle, the relationship between the legs and the hypotenuse is derived from the Pythagorean theorem. For such a triangle:
[tex]\[ c = a\sqrt{2} \][/tex]
Given [tex]\( c = 6 \)[/tex] inches, we substitute the value into the equation:
[tex]\[ 6 = a\sqrt{2} \][/tex]
To isolate [tex]\( a \)[/tex], we divide both sides of the equation by [tex]\( \sqrt{2} \)[/tex]:
[tex]\[ a = \frac{6}{\sqrt{2}} \][/tex]
To rationalize the denominator, we multiply the numerator and the denominator by [tex]\( \sqrt{2} \)[/tex]:
[tex]\[ a = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \][/tex]
Thus, the length of each leg ([tex]\( a \)[/tex]) of the isosceles right triangle is:
[tex]\[ 3\sqrt{2} \][/tex]
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