Question

find eg.
f g
45°
21
e
write your answer in simplified, rationalized form. do not round.
eg =

Explanation:

Step1: Identify Triangle Type

Triangle \( EFG \) is a right - isosceles triangle (right - angled at \( E \), \( \angle G = 45^{\circ} \), so \( \angle F=45^{\circ} \), hence \( EF = EG \)).

Step2: Use Trigonometry or Properties

In right - triangle \( EFG \), \( \sin(45^{\circ})=\frac{EG}{FG} \)? Wait, no. Wait, \( \angle G = 45^{\circ} \), \( \angle E = 90^{\circ} \), \( EF = 21 \). Since it's a 45 - 45 - 90 triangle, the legs are equal. So \( EG = EF \)? Wait, no, wait. Wait, \( \tan(45^{\circ})=\frac{EF}{EG} \), but \( \tan(45^{\circ}) = 1 \), so \( EF=EG \)? Wait, no, let's correct. In right triangle \( EFG \), right - angled at \( E \), \( \angle G = 45^{\circ} \), so \( \angle F=45^{\circ} \), so the triangle is isosceles with \( EF = EG \)? Wait, no, \( EF \) is one leg, \( EG \) is the other leg. Wait, the side \( FG \) is the hypotenuse. Wait, \( \sin(\angle G)=\frac{EF}{FG} \), \( \cos(\angle G)=\frac{EG}{FG} \), but since \( \angle G = 45^{\circ} \), \( \sin(45^{\circ})=\cos(45^{\circ})=\frac{\sqrt{2}}{2} \). But also, in a 45 - 45 - 90 triangle, the legs are equal. So if \( EF = 21 \), then \( EG = 21 \)? Wait, no, that can't be. Wait, no, wait, the side labeled 21 is \( EF \), and we need to find \( EG \). Since \( \angle G = 45^{\circ} \), \( \angle E = 90^{\circ} \), so \( \tan(\angle G)=\frac{EF}{EG} \). Since \( \tan(45^{\circ}) = 1 \), then \( \frac{EF}{EG}=1 \), so \( EG = EF = 21 \)? Wait, no, that's not right. Wait, no, maybe I mixed up the angles. Wait, \( \angle G = 45^{\circ} \), so the angle at \( G \) is 45, right - angled at \( E \), so the side opposite \( \angle G \) is \( EF \), and the side adjacent to \( \angle G \) is \( EG \). So \( \tan(\angle G)=\frac{\text{opposite}}{\text{adjacent}}=\frac{EF}{EG} \). Since \( \tan(45^{\circ}) = 1 \), then \( EF = EG \). Since \( EF = 21 \), then \( EG = 21 \)? Wait, no, that's incorrect. Wait, no, wait, maybe it's a 45 - 45 - 90 triangle, so the legs are equal. So if one leg is 21, the other leg is also 21. So \( EG = 21 \)? Wait, but that seems too simple. Wait, no, maybe I made a mistake. Wait, let's use trigonometry correctly. \( \cos(\angle G)=\frac{EG}{FG} \), \( \sin(\angle G)=\frac{EF}{FG} \), but since \( \angle G = 45^{\circ} \), \( \cos(45^{\circ})=\sin(45^{\circ}) \), so \( EG = EF \). So if \( EF = 21 \), then \( EG = 21 \). Wait, but that's the case. Because in a 45 - 45 - 90 triangle, the two legs are equal in length. So since \( \angle G = 45^{\circ} \) and \( \angle E = 90^{\circ} \), \( \angle F = 45^{\circ} \), so \( EF = EG \). Given \( EF = 21 \), then \( EG = 21 \).

Wait, no, that's not correct. Wait, maybe the triangle is labeled differently. Let's re - examine the diagram. The triangle has vertices \( F \), \( G \), \( E \), right - angled at \( E \). So \( E \) is the right angle, so \( FE \) and \( GE \) are the legs, \( FG \) is the hypotenuse. \( \angle G = 45^{\circ} \), so in triangle \( EFG \), \( \angle E = 90^{\circ} \), \( \angle G = 45^{\circ} \), so \( \angle F=45^{\circ} \). Therefore, triangle \( EFG \) is an isosceles right triangle with \( FE = GE \). Since \( FE = 21 \), then \( GE = 21 \).

Answer:

\( 21 \)