in the triangles, (overline{gk} cong overline{pn}) and (overline{hg} cong overline{mp}). which statement correctly compares the angles? (\bigcirc) angle g is congruent to angle p. (\bigcirc) angle g is smaller than angle p. (\bigcirc) angle g is larger than angle p. (\bigcirc) angle g is congruent to angle n. (triangles: left triangle (hgk) with (hg) (marked congruent to (mp)), (gk) (marked congruent to (pn)), hypotenuse (hk = 32) cm. right triangle (mpn) with (mp) (marked congruent to (hg)), (pn) (marked congruent to (gk)), hypotenuse (mn = 40) cm.)

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Accepted Answer

# Explanation:
## Step1: Recall the relationship between sides and angles in a triangle
In a triangle, the larger the side opposite an angle, the larger the angle. Both triangles $ \triangle HGK $ and $ \triangle MPN $ are right - angled triangles (since $ \overline{HG}\perp\overline{GK} $ and $ \overline{MP}\perp\overline{PN} $) with $ \overline{GK}\cong\overline{PN} $ and $ \overline{HG}\cong\overline{MP} $. The hypotenuses are $ \overline{HK} = 32\space\text{cm} $ and $ \overline{MN}=40\space\text{cm} $.

## Step2: Analyze the angles opposite the hypotenuses
In $ \triangle HGK $, the angle opposite $ \overline{HK} $ is $ \angle G $. In $ \triangle MPN $, the angle opposite $ \overline{MN} $ is $ \angle P $. Since $ \overline{HK}=32\space\text{cm}<\overline{MN} = 40\space\text{cm} $, by the side - angle relationship in triangles (larger side opposite larger angle), the angle opposite the shorter side ($ \angle G $) is smaller than the angle opposite the longer side ($ \angle P $)? Wait, no, wait. Wait, in right - angled triangles, $ \angle G $ and $ \angle P $ are the right angles? No, wait, looking at the triangles: $ \overline{HG}$ and $ \overline{MP} $ are one pair of congruent sides, $ \overline{GK}$ and $ \overline{PN} $ are another pair of congruent sides. The hypotenuses are $ HK = 32 $ and $ MN=40 $. So in $ \triangle HGK $, angle at $ G $ is between $ HG $ and $ GK $, and in $ \triangle MPN $, angle at $ P $ is between $ MP $ and $ PN $. Wait, actually, let's consider the non - right angles. Wait, no, the triangles are right - angled at $ G $ and $ P $? Wait, the markings: $ \overline{HG}$ and $ \overline{MP} $ have one mark, $ \overline{GK}$ and $ \overline{PN} $ have two marks. So $ \triangle HGK\cong\triangle MPN $ in terms of two sides, but the hypotenuses are different. Wait, no, the hypotenuse of $ \triangle HGK $ is $ HK = 32 $, hypotenuse of $ \triangle MPN $ is $ MN = 40 $. So, in a triangle, if two sides are equal ( $ HG = MP $, $ GK=PN $) and the third side (hypotenuse) of the first triangle is shorter than the third side (hypotenuse) of the second triangle, then the angle opposite the shorter hypotenuse (angle at $ H $ in $ \triangle HGK $) is smaller than the angle opposite the longer hypotenuse (angle at $ M $ in $ \triangle MPN $). But we are interested in angle $ G $ and angle $ P $. Wait, angle $ G $ and angle $ P $: since $ \overline{HG}\perp\overline{GK} $ and $ \overline{MP}\perp\overline{PN} $, angles $ G $ and $ P $ are right angles? No, that can't be. Wait, no, the markings: the sides $ HG $ and $ MP $ have one tick, $ GK $ and $ PN $ have two ticks. So $ HG = MP $, $ GK=PN $. Then, by the Hinge Theorem (SAS Inequality Theorem), if two sides of one triangle are congruent to two sides of another triangle, but the included angle is different, the longer third side is opposite the larger included angle. Wait, the included angle between $ HG $ and $ GK $ is $ \angle G $, and the included angle between $ MP $ and $ PN $ is $ \angle P $. The third sides are $ HK = 32 $ and $ MN = 40 $. Since $ HK<MN $, by the Hinge Theorem, the included angle $ \angle G $ is smaller than the included angle $ \angle P $? Wait, no, 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. So since $ HK = 32<MN = 40 $, and $ HG = MP $, $ GK = PN $, then $ \angle G<\angle P $? Wait, no, wait: Let's denote $ \triangle ABC $ and $ \triangle DEF $ with $ AB = DE $, $ BC=EF $, and $ AC<DF $. Then $ \angle B<\angle E $. So in our case, $ HG = MP $ (AB = DE), $ GK = PN $ (BC = EF), $ HK = AC = 32 $, $ MN=DF = 40 $. So $ \angle G=\angle B $, $ \angle P=\angle E $. Since $ AC < DF $, then $ \angle G<\angle P $. Wait, but let's check the answer options. The options are: 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.

Wait, maybe I made a mistake. Let's re - examine the triangles. The triangles are $ \triangle HGK $ and $ \triangle MPN $. $ \overline{HG}\cong\overline{MP} $, $ \overline{GK}\cong\overline{PN} $. The hypotenuses: $ HK = 32 $, $ MN = 40 $. So in $ \triangle HGK $, angle at $ G $: since $ HG $ and $ GK $ are the legs, angle $ G $ is between them. In $ \triangle MPN $, angle at $ P $ is between $ MP $ and $ PN $. By the Hinge Theorem, since $ HK<MN $, the included angle $ \angle G $ is smaller than the included angle $ \angle P $. So the correct statement is "Angle G is smaller than angle P".

# Answer:
Angle G is smaller than angle P.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the relationship between sides and angles in a triangle

Step2: Analyze the angles opposite the hypotenuses

Answer: