Question

triangle fgh is an isosceles right triangle with a hypotenuse that measures 16 units. an altitude, \\(\overline{gj}\\), is drawn from the right angle to the hypotenuse. what is the length of \\(\overline{gj}\\)?
options:
6 units
2 units
4 units
8 units

Explanation:

Step1: Recall properties of isosceles right triangle

In an isosceles right triangle, the altitude to the hypotenuse bisects the hypotenuse and is equal to half the hypotenuse.

Step2: Calculate length of \( \overline{GJ} \)

Given hypotenuse length is 16 units. So the length of \( \overline{GJ} \) (altitude to hypotenuse) is \( \frac{16}{2}=8 \) units? Wait, no, wait. Wait, in an isosceles right triangle, the legs are equal. Let the legs be \( a \), then by Pythagoras \( a^2 + a^2=16^2 \), \( 2a^2 = 256 \), \( a^2 = 128 \), \( a = 8\sqrt{2} \). But the area of the triangle can be calculated in two ways: \( \frac{1}{2} \times \text{leg} \times \text{leg} \) and \( \frac{1}{2} \times \text{hypotenuse} \times \text{altitude to hypotenuse} \). So \( \frac{1}{2} \times a \times a=\frac{1}{2} \times 16 \times GJ \). Since \( a^2 = 128 \), then \( \frac{128}{2}=\frac{16 \times GJ}{2} \), so \( 128 = 16 \times GJ \), \( GJ = 8 \). Wait, but also, in an isosceles right triangle, the altitude to the hypotenuse is equal to half the hypotenuse? Wait, no, hypotenuse is 16, half of that is 8. Wait, but let's check again. Wait, maybe I made a mistake earlier. Wait, in a right triangle, the altitude to the hypotenuse is given by \( \frac{ab}{c} \), where \( a \) and \( b \) are legs, \( c \) hypotenuse. In isosceles right triangle, \( a = b \), so \( \frac{a^2}{c} \). But from Pythagoras, \( a^2=\frac{c^2}{2} \), so \( \frac{c^2/2}{c}=\frac{c}{2} \). So yes, altitude to hypotenuse is \( \frac{c}{2} \). So \( c = 16 \), so altitude is 8. So the length of \( \overline{GJ} \) is 8 units.

Answer:

8 units (corresponding to the option with 8 units)