Question

consider triangle ghj.
diagram: right triangle ghj with right angle at h, jg = 10 (hypotenuse), gh = 5 (leg), vertexes j, g, h
what is the length of line segment hj?
○ 5 units
○ $5\sqrt{3}$ units
○ 10 units
○ $10\sqrt{3}$ units

Explanation:

Step1: Identify triangle type

Triangle \( GHJ \) is right - angled at \( H \), with \( GJ = 10 \) (hypotenuse) and \( GH = 5 \) (one leg).

Step2: Apply Pythagorean theorem

The Pythagorean theorem states that for a right - triangle with legs \( a,b \) and hypotenuse \( c \), \( a^{2}+b^{2}=c^{2} \). Let \( HJ = x \), \( GH = 5 \), \( GJ = 10 \). Then \( x^{2}+5^{2}=10^{2} \).

Step3: Solve for \( x \)

\( x^{2}=10^{2}-5^{2}=100 - 25=75 \). So \( x=\sqrt{75}=\sqrt{25\times3}=5\sqrt{3} \)? Wait, no, wait. Wait, maybe I made a mistake. Wait, if \( GJ = 10 \) is the hypotenuse, \( GH = 5 \), then \( HJ=\sqrt{GJ^{2}-GH^{2}}=\sqrt{100 - 25}=\sqrt{75}=5\sqrt{3} \)? Wait, no, wait, maybe the triangle is a 30 - 60 - 90 triangle. In a 30 - 60 - 90 triangle, the sides are in the ratio \( 1:\sqrt{3}:2 \). If the shorter leg (opposite 30 degrees) is \( 5 \), the hypotenuse is \( 10 \) (which is \( 2\times5 \)), then the longer leg (opposite 60 degrees) is \( 5\sqrt{3} \). Wait, but let's re - check the Pythagorean theorem. \( HJ^{2}+GH^{2}=GJ^{2} \), so \( HJ^{2}=GJ^{2}-GH^{2}=10^{2}-5^{2}=100 - 25 = 75 \), so \( HJ=\sqrt{75}=5\sqrt{3} \)? Wait, no, wait, the options have \( 5\sqrt{3} \) as an option. Wait, but let's see the triangle again. The right angle is at \( H \), so \( GH \) and \( HJ \) are legs, \( GJ \) is hypotenuse. So \( HJ=\sqrt{GJ^{2}-GH^{2}}=\sqrt{100 - 25}=\sqrt{75}=5\sqrt{3} \). Wait, but maybe I mixed up the sides. Wait, if \( GH = 5 \), \( GJ = 10 \), then the angle at \( G \) is 60 degrees, angle at \( J \) is 30 degrees. So the longer leg (HJ) should be \( 5\sqrt{3} \).

Answer:

\( 5\sqrt{3} \) units