Question

version of the hinge theorem
lls
e theorem (geometry) (geometry)
s math tutoring
multiple-choice question
which angle is larger?
f
they are the same size.
a
rewatch

Explanation:

Step1: Recall Hinge Theorem

The Hinge Theorem (or SAS Inequality Theorem) states that if two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second. The converse is also true: if two sides of one triangle are congruent to two sides of another triangle, and the third side of the first is longer than the third side of the second, then the included angle of the first is larger than the included angle of the second.

Step2: Analyze the Triangles (from the left diagram)

Looking at triangles \( \triangle ABC \) and \( \triangle DEF \):
- \( AC = DF = 8 \) (one pair of congruent sides)
- \( AB = 6 \), \( DE = 4 \); \( BC = 5 \), \( EF = 6 \)? Wait, no, let's re - check. Wait, in the left - most triangle \( \triangle ABC \): sides \( AB = 6 \), \( BC = 5 \), \( AC = 8 \). In the middle triangle \( \triangle DEF \): sides \( DE = 4 \), \( EF = 6 \), \( DF = 8 \). So two sides: \( AC = DF = 8 \), and we compare the other sides. Wait, actually, for the angles \( \angle A \) (in \( \triangle ABC \)) and \( \angle F \) (in \( \triangle DEF \))? Wait, no, maybe the triangles with sides: Let's look at the two triangles where two sides are equal. Wait, in the first two triangles ( \( \triangle ABC \) and \( \triangle DEF \)): \( AC = DF = 8 \), and \( AB = 6 \), \( DE = 4 \); \( BC = 5 \), \( EF = 6 \). Wait, maybe a better way: in the triangle with sides 6, 5, 8 ( \( \triangle ABC \)) and 4, 6, 8 ( \( \triangle DEF \)). The two sides that are equal: \( AC = DF = 8 \), and then the other sides: \( AB = 6 \), \( DE = 4 \); \( BC = 5 \), \( EF = 6 \). Wait, maybe the included angles: for \( \angle A \) (included between \( AB \) and \( AC \)) and \( \angle F \) (included between \( DF \) and \( EF \))? Wait, no, let's use the converse. Wait, in the two triangles, if we have two sides congruent: let's say in two triangles, side1 = side1, side2 = side2, and third side1 > third side2, then included angle1 > included angle2.

Wait, looking at the triangles: \( \triangle ABC \): sides \( AB = 6 \), \( AC = 8 \), \( BC = 5 \). \( \triangle DEF \): sides \( DE = 4 \), \( DF = 8 \), \( EF = 6 \). So \( AC = DF = 8 \), and \( AB = 6 \), \( DE = 4 \); \( BC = 5 \), \( EF = 6 \). Wait, maybe the triangle with the longer third side has the larger included angle. Wait, in \( \triangle ABC \), the sides adjacent to \( \angle A \) are \( AB = 6 \) and \( AC = 8 \), and the third side is \( BC = 5 \). In \( \triangle DEF \), the sides adjacent to \( \angle F \) are \( DF = 8 \) and \( EF = 6 \), and the third side is \( DE = 4 \). Wait, no, maybe I got the angles wrong. Wait, the question is about which angle is larger, \( \angle A \) or \( \angle F \)? Wait, no, the options are \( F \), "They are the same size", and \( A \). Wait, let's use the converse of the Hinge Theorem. If two sides of one triangle are congruent to two sides of another triangle, and the third side of the first is longer than the third side of the second, then the included angle of the first is larger than the included angle of the second.

Wait, in \( \triangle ABC \): sides \( AB = 6 \), \( AC = 8 \), \( BC = 5 \). In \( \triangle DEF \): sides \( DE = 4 \), \( DF = 8 \), \( EF = 6 \). So \( AC = DF = 8 \), \( AB = 6 \), \( EF = 6 \) (so \( AB = EF = 6 \)), and \( BC = 5 \), \( DE = 4 \). So two sides: \( AB = EF = 6 \), \( AC = DF = 8 \). Then the third sides: \( BC = 5 \), \( DE = 4 \). Since \( BC>DE \) (5 > 4), by the converse of the Hinge Theorem, the included angle of \( \triangle ABC \) ( \( \angle A \)) is larger than the included angle of \( \triangle DEF \) ( \( \angle F \))? Wait, no, the included angle for \( AB \) and \( AC \) is \( \angle A \), and for \( EF \) and \( DF \) is \( \angle F \). Since \( BC>DE \) (the third sides), the included angle \( \angle A \) (between \( AB \) and \( AC \)) is larger than \( \angle F \) (between \( EF \) and \( DF \))? Wait, no, maybe I mixed up the sides. Wait, let's re - assign: Let's take triangle 1: sides \( a = 6 \), \( b = 8 \), \( c = 5 \). Triangle 2: sides \( a'=6 \), \( b' = 8 \), \( c'=4 \). Wait, no, if two sides are equal ( \( a = a'=6 \), \( b = b' = 8 \)) and \( c>c' \) (5 > 4? No, 5 < 4? No, 5>4? Wait, 5 is greater than 4? Yes, 5 > 4. Wait, no, in triangle 1, third side is 5, in triangle 2, third side is 4. So if two sides (6 and 8) are equal, and third side 5>4, then the included angle between 6 and 8 in triangle 1 is larger than the included angle between 6 and 8 in triangle 2. Wait, but in the diagram, the angle at \( A \) is in triangle with sides 6, 8, 5, and angle at \( F \) is in triangle with sides 6, 8, 4? Wait, maybe I made a mistake. Alternatively, looking at the other set of triangles (the ones with sides 4, 11, x and 4, 11, y? Wait, no, the key is: when two sides are congruent, the longer third side implies a larger included angle.

Wait, in the first two triangles ( \( \triangle ABC \) and \( \triangle DEF \)): \( AC = DF = 8 \), \( AB = 6 \), \( DE = 4 \); \( BC = 5 \), \( EF = 6 \). Wait, \( AB = 6 \), \( DE = 4 \) (so \( AB>DE \)), \( BC = 5 \), \( EF = 6 \) (so \( BC<EF \)). This is getting confusing. Wait, the question is "Which angle is larger? F, They are the same size, A". From the Hinge Theorem converse: if two sides of one triangle are congruent to two sides of another triangle, and the third side of the first is longer than the third side of the second, then the included angle of the first is larger than the included angle of the second.

Looking at the triangles: Let's assume that in \( \triangle ABC \) and \( \triangle DEF \), \( AC = DF \), \( AB = DE \)? No, \( AB = 6 \), \( DE = 4 \). Wait, maybe the triangles are \( \triangle ABC \) with sides \( AB = 6 \), \( AC = 8 \), \( BC = 5 \) and \( \triangle DFE \) with sides \( DF = 8 \), \( EF = 6 \), \( DE = 4 \). So two sides: \( AC = DF = 8 \), \( BC = 5 \), \( EF = 6 \). Wait, \( BC = 5 \), \( EF = 6 \), so \( EF>BC \). Then, by the Hinge Theorem, the included angle for \( EF \) and \( DF \) ( \( \angle F \)) and the included angle for \( BC \) and \( AC \) ( \( \angle C \))? No, this is not right. Wait, maybe the correct approach is: In the two triangles where two sides are equal (let's say side1 and side2), and the third side of triangle A is longer than the third side of triangle F, then angle A is larger than angle F. Wait, from the diagram, the triangle with the longer third side will have the larger included angle. If we consider the two triangles, and we see that the side opposite angle A is shorter than the side opposite angle F? No, this is getting too convoluted. Wait, the answer options are F, same, A. From the Hinge Theorem, if we have two triangles with two sides equal, and the third side of triangle with angle A is shorter or longer? Wait, looking at the lengths: in \( \triangle ABC \): sides 6, 5, 8. In \( \triangle DEF \): sides 4, 6, 8. So the side opposite angle A is BC = 5, and the side opposite angle F is DE = 4? No, angle A is in \( \triangle ABC \), opposite side BC = 5. Angle F is in \( \triangle DEF \), opposite side DE = 4. Since 5>4, by the Law of Sines ( \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \) ), \( \frac{BC}{\sin A}=\frac{DE}{\sin F} \), and since \( BC>DE \), \( \sin A>\sin F \). Also, since both angles are acute (probably, as triangle sides are positive and sum to valid triangles), angle A is larger than angle F. Wait, no, wait: Law of Sines: \( \frac{BC}{\sin A}=\frac{AC}{\sin B} \), \( \frac{DE}{\sin F}=\frac{DF}{\sin E} \). Since \( AC = DF \), and \( BC>DE \), \( \sin A>\sin F \), so angle A is larger than angle F.

Step3: Conclusion

So angle A is larger than angle F.

Answer:

A. A