Which is a true statement about an isosceles right triangle?

A. Each leg is [tex]\sqrt{3}[/tex] times as long as the hypotenuse.
B. The hypotenuse is [tex]\sqrt{2}[/tex] times as long as either leg.
C. The hypotenuse is [tex]\sqrt{3}[/tex] times as long as either leg.
D. Each leg is [tex]\sqrt{2}[/tex] times as long as the hypotenuse.



Answer :

Let's analyze the properties of an isosceles right triangle. An isosceles right triangle has two sides (legs) of equal length and the angle between these two sides is 90 degrees.

### Step-by-Step Solution:

1. Recalling the Definition:
In an isosceles right triangle, the two legs are equal in length, and the angles opposite these legs are each 45 degrees.

2. Using the Pythagorean Theorem:
For a right triangle, the Pythagorean theorem states that the square of the length of the hypotenuse \(c\) is equal to the sum of the squares of the lengths of the other two sides \(a\) and \(b\). Mathematically, this is \(c^2 = a^2 + b^2\).

3. Considering the Equal Sides:
In an isosceles right triangle where the equal sides (legs) are both \(a\), we have:
[tex]\[ c^2 = a^2 + a^2 \implies c^2 = 2a^2 \][/tex]

4. Solving for the Hypotenuse:
To find \(c\), the hypotenuse, we take the square root of both sides:
[tex]\[ c = \sqrt{2a^2} = a\sqrt{2} \][/tex]

5. Reviewing the Statements:
- A. Each leg is \(\sqrt{3}\) times as long as the hypotenuse.
This implies \(a = \sqrt{3}c\), which contradicts our derived equation.
- B. The hypotenuse is \(\sqrt{2}\) times as long as either leg.
This fits our derived equation \(c = a\sqrt{2}\).
- C. The hypotenuse is \(\sqrt{3}\) times as long as either leg.
This would imply \(c = a\sqrt{3}\), which contradicts our derived equation.
- D. Each leg is \(\sqrt{2}\) times as long as the hypotenuse.
This implies \(a = c\sqrt{2}\), which contradicts our derived equation.

6. Conclusion:
The correct statement is B. The hypotenuse is \(\sqrt{2}\) times as long as either leg.

Therefore, the true statement about an isosceles right triangle is:
B. The hypotenuse is [tex]\(\sqrt{2}\)[/tex] times as long as either leg.

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