Question

consider the following figure. (note that the figure is not drawn to scale.) order the side lengths ef, fh, eg, fg, and hg from least to greatest. try again your answer is incorrect.

Explanation:

Step1: Analyze Triangle EFC

In $\triangle EFC$, angles are $57^\circ$ (at E), $65^\circ$ (at C), so the third angle at F: $180 - 57 - 65 = 58^\circ$. By the "larger angle opposite longer side" rule:
- Angle at E: $57^\circ$ (opposite $FC$)
- Angle at C: $65^\circ$ (opposite $EF$)
- Angle at F: $58^\circ$ (opposite $EC$)
So, $FC < EF < EC$ (since $57 < 58 < 65$). Wait, no—wait, angle opposite $EF$ is $\angle ECF = 65^\circ$, angle opposite $FC$ is $\angle FEC = 57^\circ$, angle opposite $EC$ is $\angle EFC = 58^\circ$. So side lengths: $FC < EC < EF$? Wait, no, larger angle → longer side. So $57^\circ$ (opposite $FC$) < $58^\circ$ (opposite $EC$) < $65^\circ$ (opposite $EF$) → $FC < EC < EF$.

Step2: Analyze Triangle FHC

In $\triangle FHC$, angles are $55^\circ$ (at F), $57^\circ$ (at C), so third angle at H: $180 - 55 - 57 = 68^\circ$. By the same rule:
- Angle at F: $55^\circ$ (opposite $HC$)
- Angle at C: $57^\circ$ (opposite $FH$)
- Angle at H: $68^\circ$ (opposite $FC$)
So, $HC < FH < FC$ (since $55 < 57 < 68$). Wait, angle opposite $FC$ is $\angle FHC = 68^\circ$, so $FC$ is opposite the largest angle, so $FC$ is the longest in $\triangle FHC$? Wait, no: angle at H is $68^\circ$ (opposite $FC$), angle at C is $57^\circ$ (opposite $FH$), angle at F is $55^\circ$ (opposite $HC$). So side lengths: $HC < FH < FC$ (because $55 < 57 < 68$ → opposite sides: $HC$ (55°), $FH$ (57°), $FC$ (68°)). So $HC < FH < FC$.

Step3: Combine with Triangle EFC

From $\triangle EFC$: $FC < EC < EF$ (wait, no, earlier mistake: in $\triangle EFC$, angles are $\angle E = 57^\circ$, $\angle C = 65^\circ$, $\angle F = 58^\circ$. So opposite sides:
- $\angle E$ (57°) → opposite $FC$
- $\angle C$ (65°) → opposite $EF$
- $\angle F$ (58°) → opposite $EC$
Thus, $FC$ (opposite 57°) < $EC$ (opposite 58°) < $EF$ (opposite 65°) → $FC < EC < EF$.

From $\triangle FHC$: $HC < FH < FC$ (since $HC$ opposite 55°, $FH$ opposite 57°, $FC$ opposite 68°). So combining $HC < FH < FC$ and $FC < EC < EF$, we get $HC < FH < FC < EC < EF$? Wait, but the sides to order are $EF$, $FH$, $EC$, $FC$, $HG$? Wait, wait, the problem says "EF, FH, EG, FG, and HG"—wait, maybe a typo? Wait, the figure: E, F, C, H. So sides: EF, FH, EC, FC, and HC (HG? Maybe typo, should be HC). Let's correct: assume HG is HC.

So sides: EF, FH, EC, FC, HC.

From $\triangle FHC$: $HC < FH < FC$
From $\triangle EFC$: $FC < EC < EF$
So combining: $HC < FH < FC < EC < EF$

Wait, but let's recheck angles:

In $\triangle EFC$:
- $\angle FEC = 57^\circ$, $\angle ECF = 65^\circ$, $\angle EFC = 180 - 57 - 65 = 58^\circ$
So side opposite $\angle FEC$ (57°) is $FC$ → $FC$ length.
Side opposite $\angle ECF$ (65°) is $EF$ → $EF$ length.
Side opposite $\angle EFC$ (58°) is $EC$ → $EC$ length.
So $FC < EC < EF$ (since 57° < 58° < 65° → opposite sides: $FC < EC < EF$).

In $\triangle FHC$:
- $\angle HFC = 55^\circ$, $\angle HCF = 57^\circ$, $\angle FHC = 180 - 55 - 57 = 68^\circ$
Side opposite $\angle HFC$ (55°) is $HC$ → $HC$ length.
Side opposite $\angle HCF$ (57°) is $FH$ → $FH$ length.
Side opposite $\angle FHC$ (68°) is $FC$ → $FC$ length.
So $HC < FH < FC$ (since 55° < 57° < 68° → opposite sides: $HC < FH < FC$).

Now, combine the two:
$HC < FH < FC$ (from $\triangle FHC$) and $FC < EC < EF$ (from $\triangle EFC$). So transitivity: $HC < FH < FC < EC < EF$.

Assuming $HG$ is $HC$ (maybe a typo), the order of $EF$, $FH$, $EC$, $FC$, $HG$ (HC) is: $HG < FH < FC < EC < EF$.

Wait, but the problem lists sides as EF, FH, EG, FG, HG. Wait, maybe EG is EC? FG is FC? Let's adjust:
- EG = EC (maybe typo, since E, C, G? No, figure has E, F, C, H. So maybe EG is EC, FG is FC, HG is HC.

So sides: EF, FH, EC (EG), FC (FG), HC (HG).

Thus, order from least to greatest: HG (HC) < FH < FG (FC) < EG (EC) < EF.

Step3: Confirm with Angle-Side Relationship

The key is in a triangle, the larger the angle, the longer the opposite side. So by calculating angles in each triangle, we determine the side lengths relative to each other, then combine the two triangles (sharing side FC) to get the overall order.

Answer:

$HG < FH < FG < EG < EF$ (assuming $HG=HC$, $FG=FC$, $EG=EC$; if labels differ, adjust, but based on angle analysis, the order is $HG$ (or $HC$) < $FH$ < $FC$ (or $FG$) < $EC$ (or $EG$) < $EF$)