An isosceles right triangle is a special type of triangle where two of the sides (the legs) are equal in length, and the angles opposite these sides are each [tex]\(45^\circ\)[/tex]. The third side of the triangle, the hypotenuse, is opposite the right angle ([tex]\(90^\circ\)[/tex]).
To understand the relationship between the legs and the hypotenuse in an isosceles right triangle, let's consider the Pythagorean theorem. For any right triangle, the Pythagorean theorem states:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
In the case of our isosceles right triangle, the legs are of equal length, so let's denote the length of each leg as [tex]\( l \)[/tex]. Therefore, the equation becomes:
[tex]\[ l^2 + l^2 = c^2 \][/tex]
[tex]\[ 2l^2 = c^2 \][/tex]
[tex]\[ c = \sqrt{2l^2} \][/tex]
[tex]\[ c = l\sqrt{2} \][/tex]
From this, we can see that the hypotenuse ([tex]\( c \)[/tex]) is [tex]\(\sqrt{2}\)[/tex] times the length of each leg ([tex]\( l \)[/tex]).
Now let's evaluate each statement:
A. Each leg is [tex]\(\sqrt{3}\)[/tex] times as long as the hypotenuse.
- This is incorrect because our derived relationship shows that the hypotenuse is [tex]\(\sqrt{2}\)[/tex] times the leg length, not the other way around, and [tex]\(\sqrt{3}\)[/tex] is not involved.
B. Each leg is [tex]\(\sqrt{2}\)[/tex] times as long as the hypotenuse.
- This is incorrect for the same reason as above. The relationship is the hypotenuse equals leg times [tex]\(\sqrt{2}\)[/tex].
C. The hypotenuse is [tex]\(\sqrt{3}\)[/tex] times as long as either leg.
- This is incorrect because the correct multiplier for the hypotenuse in terms of leg length in an isosceles right triangle is [tex]\(\sqrt{2}\)[/tex], not [tex]\(\sqrt{3}\)[/tex].
D. The hypotenuse is [tex]\(\sqrt{2}\)[/tex] times as long as either leg.
- This is correct because, based on our derived formulas using the Pythagorean theorem, the hypotenuse ([tex]\( c \)[/tex]) is indeed [tex]\(\sqrt{2}\)[/tex] times the length of each leg ([tex]\( l \)[/tex]).
Therefore, the correct answer is:
D. The hypotenuse is [tex]\(\sqrt{2}\)[/tex] times as long as either leg.
