in the triangles, $ \overline { c k } \cong \overline { f n } $ and $ \overline { h c } \cong \overline { m p } $.
which statement correctly compares the angles?
angle g is congruent to angle p.
angle g is smaller than angle p.
angle g is larger than angle p.
angle g is congruent to angle n.

Question

Accepted Answer

# Explanation:
## Step1: Recall the 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 third side of the first triangle is shorter than the third side of the second triangle, then the included angle of the first triangle is smaller than the included angle of the second triangle.

## Step2: Identify the congruent sides and the third sides
We know that $\overline{GK} \cong \overline{PN}$ (one pair of congruent sides) and $\overline{HG} \cong \overline{MP}$ (another pair of congruent sides). The third sides are $\overline{HK} = 32$ cm and $\overline{MN}=40$ cm. Since $32<40$, the included angle of the triangle with the shorter third side ($\angle G$ in $\triangle HGK$) is smaller than the included angle of the triangle with the longer third side ($\angle P$ in $\triangle MPN$)? Wait, no—wait, let's check the included angles. Wait, actually, in $\triangle HGK$, the sides around $\angle G$ are $HG$ and $GK$, and in $\triangle MPN$, the sides around $\angle P$ are $MP$ and $PN$. Since $HG \cong MP$ and $GK \cong PN$, and the third side $HK = 32$ cm, $MN = 40$ cm. So by the Hinge Theorem, since $HK < MN$, the angle opposite? Wait, no, the included angle. Wait, actually, the included angle for the two sides $HG$ and $GK$ is $\angle G$, and the included angle for $MP$ and $PN$ is $\angle P$. Wait, no—wait, $HG$ and $GK$ meet at $G$, so $\angle G$ is between $HG$ and $GK$. $MP$ and $PN$ meet at $P$, so $\angle P$ is between $MP$ and $PN$. Since $HG \cong MP$, $GK \cong PN$, and $HK < MN$, then by the Hinge Theorem, $\angle G < \angle P$? Wait, no—wait, the Hinge Theorem says that if two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is smaller than the included angle of the second, then the third side of the first is shorter than the third side of the second. Wait, we have the third sides: $HK = 32$, $MN = 40$, so $HK < MN$. Therefore, the included angle of the first triangle ($\angle G$) is smaller than the included angle of the second triangle ($\angle P$)? Wait, no—wait, the Hinge Theorem is: If $AB \cong DE$, $AC \cong DF$, and $BC < EF$, then $\angle A < \angle D$. So in our case, $HG \cong MP$ (let's say $HG = MP$), $GK \cong PN$ ( $GK = PN$ ), and $HK = 32$, $MN = 40$, so $HK < MN$. Therefore, the included angle $\angle G$ (between $HG$ and $GK$) is less than the included angle $\angle P$ (between $MP$ and $PN$)? Wait, no—wait, actually, the included angle for $HK$ is $\angle H$ and $\angle M$? Wait, maybe I mixed up the sides. Let's label the triangles properly.

In $\triangle HGK$: sides $HG$, $GK$, $HK = 32$ cm.

In $\triangle MPN$: sides $MP$, $PN$, $MN = 40$ cm.

Given $GK \cong PN$ and $HG \cong MP$. So two sides are congruent: $HG \cong MP$, $GK \cong PN$. The third sides: $HK = 32$, $MN = 40$. So $HK < MN$. By the Hinge Theorem, the angle opposite the shorter third side? Wait, no, the included angle. Wait, the Hinge Theorem is about the included angle. Wait, actually, the angle between the two congruent sides. So $HG$ and $GK$ form $\angle G$, and $MP$ and $PN$ form $\angle P$. Since $HG \cong MP$ and $GK \cong PN$, and the third side $HK < MN$, then the included angle $\angle G$ is smaller than the included angle $\angle P$. Wait, but let's think about right triangles? Wait, are these right triangles? The triangles have a right angle? Let's check the diagram: $G$ and $P$ look like right angles? Wait, the triangles are drawn with $G$ and $P$ as right angles? Wait, if $\angle G$ and $\angle P$ are right angles, but no, the sides: $HK$ is 32, $MN$ is 40. Wait, maybe I made a mistake. Wait, let's re-express:

If $\triangle HGK$ and $\triangle MPN$ have $GK = PN$, $HG = MP$, and $HK = 32$, $MN = 40$. Then by the Hinge Theorem, since $HK < MN$, the angle opposite? No, the included angle. Wait, the Hinge Theorem states that if two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is smaller than the included angle of the second, then the third side of the first is shorter than the third side of the second. Conversely, if the third side of the first is shorter than the third side of the second, then the included angle of the first is smaller than the included angle of the second.

So here, two sides are congruent ($HG \cong MP$, $GK \cong PN$), third side $HK = 32 < MN = 40$, so the included angle $\angle G$ (between $HG$ and $GK$) is smaller than the included angle $\angle P$ (between $MP$ and $PN$). Wait, but let's check the angles. Wait, maybe the triangles are right triangles? Wait, the diagram shows $G$ and $P$ as right angles? If $\angle G$ and $\angle P$ are right angles, then they would be congruent, but the hypotenuses are 32 and 40, which are different. So they can't both be right angles. Wait, maybe the triangles are not right triangles. Let's go back to the Hinge Theorem.

So, applying the Hinge Theorem:

- $HG \cong MP$ (given)
- $GK \cong PN$ (given)
- $HK = 32$ cm, $MN = 40$ cm, so $HK < MN$

Therefore, by the Hinge Theorem, the included angle $\angle G$ (between $HG$ and $GK$) is smaller than the included angle $\angle P$ (between $MP$ and $PN$). Wait, but let's check the options. The options are:

- Angle G is congruent to angle P. → No, because the third sides are different.
- Angle G is smaller than angle P. → This matches the Hinge Theorem result.
- Angle G is larger than angle P. → Opposite of what we found.
- Angle G is congruent to angle N. → No, because there's no reason for that.

Wait, but let's confirm the Hinge Theorem. The Hinge Theorem (SAS Inequality) says that if $AB \cong DE$, $AC \cong DF$, and $BC < EF$, then $\angle A < \angle D$ (where $\angle A$ is between $AB$ and $AC$, $\angle D$ is between $DE$ and $DF$). So in our case, $AB = HG$, $AC = GK$, $DE = MP$, $DF = PN$, $BC = HK$, $EF = MN$. So since $BC < EF$, $\angle A = \angle G < \angle D = \angle P$. So angle G is smaller than angle P.

# Answer:
Angle G is smaller than angle P. (The option: Angle G is smaller than angle P.)

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the Hinge Theorem

Step2: Identify the congruent sides and the third sides

Answer: