Question

an isosceles right triangle has leg lengths of 4 centimeters. what is the length of the altitude drawn from the right angle to the hypotenuse?
options: 2 cm, $2\sqrt{2}$ cm, 4 cm, $4\sqrt{2}$ cm

Explanation:

Step1: Identify triangle type

It's an isosceles right triangle, so both legs are equal. Given one leg \( a = 4 \, \text{cm} \), the other leg \( b = 4 \, \text{cm} \).

Step2: Find hypotenuse length

Using Pythagorean theorem \( c = \sqrt{a^2 + b^2} \), substitute \( a = 4 \), \( b = 4 \):
\( c = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \, \text{cm} \).

Step3: Relate altitude to hypotenuse

In an isosceles right triangle, the altitude to the hypotenuse (let’s call it \( h \)) can also be derived from area. Area \( A = \frac{1}{2} \times \text{leg} \times \text{leg} = \frac{1}{2} \times \text{hypotenuse} \times \text{altitude} \).
First, area from legs: \( A = \frac{1}{2} \times 4 \times 4 = 8 \, \text{cm}^2 \).
Then, set \( A = \frac{1}{2} \times 4\sqrt{2} \times h \). Solve for \( h \):
\( 8 = \frac{1}{2} \times 4\sqrt{2} \times h \)
\( 8 = 2\sqrt{2} \times h \)
\( h = \frac{8}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{cm} \). Wait, no—wait, in an isosceles right triangle, the altitude to hypotenuse is also the median and angle bisector, and in such a triangle, the altitude to hypotenuse is \( \frac{\text{leg}}{\sqrt{2}} \)? Wait, no, let's recheck. Wait, the legs are 4, hypotenuse is \( 4\sqrt{2} \). The area is \( \frac{1}{2} \times 4 \times 4 = 8 \). The area is also \( \frac{1}{2} \times \text{hypotenuse} \times \text{altitude} \), so \( 8 = \frac{1}{2} \times 4\sqrt{2} \times h \). Then \( h = \frac{16}{4\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \)? Wait, no, that can't be. Wait, no—wait, in an isosceles right triangle, the altitude to the hypotenuse is equal to half the hypotenuse? Wait, hypotenuse is \( 4\sqrt{2} \), half of that is \( 2\sqrt{2} \), which matches. But wait, another way: in an isosceles right triangle, the legs are equal, so the altitude to hypotenuse splits it into two smaller isosceles right triangles. So the altitude length should be equal to the length of the segments it divides the hypotenuse into. Wait, but let's confirm with the area. Legs are 4, so area is 8. Hypotenuse is \( 4\sqrt{2} \), so altitude \( h = \frac{2 \times \text{Area}}{\text{Hypotenuse}} = \frac{16}{4\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \). Wait, but the options include \( 2\sqrt{2} \, \text{cm} \). Wait, but wait—the problem says "an isosceles right triangle has leg lengths of 4 cm". Wait, maybe I misread: is the leg 4, or is the altitude related? Wait, no—the question is: "What is the length of the altitude drawn from the right angle to the hypotenuse?" So in an isosceles right triangle with leg length \( a \), the altitude to hypotenuse \( h \) is \( \frac{a}{\sqrt{2}} \) or \( \frac{\text{hypotenuse}}{2} \). Since hypotenuse \( c = a\sqrt{2} \), then \( h = \frac{c}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} \). For \( a = 4 \), \( h = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{cm} \). So the correct answer is \( 2\sqrt{2} \, \text{cm} \).

Answer:

\( 2\sqrt{2} \, \text{cm} \) (the second option: \( 2\sqrt{2} \, \text{cm} \))