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The Hinge Theorem: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Match before comparing

Triangles ABC and DEF have $AB=DE$ .

They also have $AC=DF$ .

The included angles at A and D measure 38 and 72 degrees, respectively. Compare BC and EF, naming both matched pairs and explaining your conclusion.

The two tick families identify the two given side pairs.

Text description of this figure

Triangles ABC and DEF have matching single ticks on their first arms and matching double ticks on their second arms. Only stated givens are labeled. The second triangle is dashed.
1. Hint 1
  Locate the vertex where each pair of matched sides meets.
2. Hint 2
  Match the side opposite the larger included angle with the longer third side.
Answer

$BC<EF.$
The matching pairs are AB with DE, and AC with DF.

Full solution

AB matches DE, and AC matches DF.
The two pairs meet at A and D, so those are the included-angle vertices.

Angle D is larger than angle A.
The hinge theorem therefore makes its opposite side EF longer than BC.

The comparison is
$BC<EF.$

Answer

$BC<EF.$
The matching pairs are AB with DE, and AC with DF.

Key idea

The included angle is the one between the two matched arms.
Problem 2 Read the opening from the third side

Two triangles KLM and PQR have equal corresponding arms: KL and PQ both measure 8 units, and KM and PR both measure 6 units. Their third sides LM and QR measure 9 and 11 units, respectively.

Compare the included angles at K and P. Explain why this is a two-triangle comparison rather than an angle-side comparison within one triangle.

All labeled lengths are givens; the included angles are unknown.

Text description of this figure

Triangles KLM and PQR have matching single ticks on their first arms and matching double ticks on their second arms. Only stated givens are labeled. The second triangle is dashed.
1. Hint 1
  Check both arm pairs, then compare the two third sides.
2. Hint 2
  The longer third side points to the larger included angle under these hypotheses.
Answer

$m\angle K<m\angle P.$
The converse hinge theorem compares the included angles in the two triangles.

Full solution

The pairs KL with PQ and KM with PR match.
They surround angles K and P.

QR is longer than LM, since its length is 11 rather than 9.
The converse hinge theorem gives

$m\angle K<m\angle P.$

The third sides belong to different triangles, so ordering sides and opposite angles within one triangle does not directly establish this conclusion.

Answer

$m\angle K<m\angle P.$
The converse hinge theorem compares the included angles in the two triangles.

Key idea

Two equal corresponding side pairs connect the two triangles.
Problem 3 Build an opening from adjacent parts

Triangles PQR and PRS share PR and satisfy $PQ=PS$ . Ray PR lies strictly inside angle QPS.

Ray PT lies strictly inside angle QPR. The given angle measures are
$m\angle QPT=27^\circ.$
The next angle is
$m\angle TPR=45^\circ.$
The other triangle has
$m\angle RPS=49^\circ.$
Compare QR and RS. Explain which side supplies the second equal pair.

PQ and PS carry the given equal-side marks. Ray PT splits the opening QPR.

Text description of this figure

Triangles PQR and PRS share PR. Single ticks mark the given equal lengths PQ and PS. Ray PT is strictly inside angle QPR. The adjacent given angles QPT, TPR, and RPS measure 27, 45, and 49 degrees. The second triangle is dashed.
1. Hint 1
  Locate the two openings between the equal arms and the shared side.
2. Hint 2
  Angle QPR is made of the two adjacent angles separated by ray PT.
Answer

$QR>RS.$
The second equal pair is PR with itself.

Full solution

Ray PT splits angle QPR into the adjacent given parts:
$m\angle QPR=27^\circ+45^\circ.$

Thus
$m\angle QPR=72^\circ.$

The sides PQ and PS are equal, and PR is shared.
These matched pairs surround angles QPR and RPS.

The first opening is 72 degrees, greater than the other opening of 49 degrees.
The hinge theorem gives
$QR>RS.$

Answer

$QR>RS.$
The second equal pair is PR with itself.

Key idea

Adjacent angle parts can determine the included angle before the hinge comparison.
Problem 4 Complete an arm before matching it

In triangle GHI, point N lies strictly between G and H. All lengths are in the same unit.

The first piece has length
$GN=2u+3.$
The second piece has length
$NH=u+1.$
In triangle JKL, the proposed matching arm has length
$JK=u+20.$
The other arms GI and JL each measure 11 units. The included angles HGI and KJL measure 41 and 79 degrees, respectively.

Find the real value of u that makes GH equal to JK, then compare HI and KL. Show how the two pieces of GH enter your calculation.
1. Hint 1
  Use the point order to express the whole arm GH.
2. Hint 2
  Make that whole length equal to JK before comparing the included angles.
Answer

$u=8.$
The third-side comparison is
$HI<KL.$
GH and JK each measure 28 units.

Full solution

Because N is between G and H, segment addition gives
$GH=GN+NH.$

Substitute the expressions and collect terms:
$GH=3u+4.$

To match GH with JK, solve
$3u+4=u+20.$

Subtract u and then subtract 4:
$2u=16.$

Divide by 2:
$u=8.$

At this value, GN is 19 units and NH is 9 units.
Their positive sum makes GH 28 units, matching JK.

The other pair GI and JL is already equal, with both lengths 11.
The included angle at G is smaller than the one at J, so the hinge theorem gives
$HI<KL.$

The labeled pieces of GH are added before comparing the complete arms.

Answer

$u=8.$
The third-side comparison is
$HI<KL.$
GH and JK each measure 28 units.

Key idea

Compose a split arm and establish both equal pairs before using a hinge comparison.
Problem 5 Check Nora’s choice of angle

Triangles ABC and DEF satisfy $AB=DE$ .

They also satisfy $AC=DF$ .

Angle ABC measures 80 degrees, and angle DEF measures 60 degrees. Nora says, “The equal side pairs and the larger angle at B prove BC is longer than EF by the hinge theorem.”

Identify the error in her use of the theorem. Name the included angles she would need to compare for that argument. You do not need to find the actual third-side order.
1. Hint 1
  Find where AB and AC meet; do the same for DE and DF.
2. Hint 2
  Check whether those vertices are the ones used in Nora’s comparison.
Answer

Nora compared angles outside the matched side pairs. The included angles are BAC and EDF; their order is the needed hinge premise.

Full solution

AB and AC meet at A, while DE and DF meet at D.
Therefore the included angles are BAC and EDF.

Nora compares vertices B and E instead.
Those angles are not between the matched arms.

Her cited theorem therefore does not justify the claimed order.
An included-angle comparison at A and D is needed for that hinge argument.

The shaded openings at A and D lie between the matched arms. The given 80° and 60° angles are at B and E.

Answer

Nora compared angles outside the matched side pairs. The included angles are BAC and EDF; their order is the needed hinge premise.

Key idea

The angles compared directly by the hinge theorem must lie between the matched sides.
Problem 6 Keep the three possible orders separate

Each case concerns two nondegenerate triangles ABC and DEF with AB equal to DE and AC equal to DF. Treat the cases separately.

Case A:
$m\angle BAC<m\angle EDF.$
Compare BC and EF.

Case B:
$BC=EF.$
Compare the included angles.

Case C:
$BC>EF.$
Compare the included angles.

Give a reason in each case.
1. Hint 1
  Strict comparisons and equality use different named results.
2. Hint 2
  For equal third sides, count how many corresponding side pairs now match.
Answer

Case A:
$BC<EF.$
Case B:
$m\angle A=m\angle D.$
Case C:
$m\angle A>m\angle D.$

Full solution

Case A: The hinge theorem matches the smaller included angle to the shorter third side, so
$BC<EF.$

Case B: All three corresponding side pairs match.
SSS makes the triangles congruent, giving
$m\angle BAC=m\angle EDF.$

Case C: The converse hinge theorem applies to the longer third side and gives
$m\angle BAC>m\angle EDF.$

Answer

Case A:
$BC<EF.$
Case B:
$m\angle A=m\angle D.$
Case C:
$m\angle A>m\angle D.$

Key idea

Strict order uses the hinge theorems, while equality completes a congruence test.
Problem 7 Compare two folding frames

Two ideal folding frames each have straight rigid arms of lengths 12 and 20 centimeters, joined at one endpoint. Treat the arms and the straight segment between their free ends as a triangle.

The first frame opens to 45 degrees, and the second opens to 100 degrees. Let the free-end gaps be g and h centimeters, respectively. Compare g and h. Then use the triangle inequality to give the strict numerical upper bound for h obtained by summing the two fixed arm lengths. Explain both conclusions.
1. Hint 1
  The two frames keep the same pair of arm lengths.
2. Hint 2
  Use the included-angle comparison for the order, and the sum of the arms for an upper bound.
Answer

$\begin{aligned}g&<h.\\h&<32\text{ cm}.\end{aligned}$

Full solution

Both arm pairs match, and both openings are strictly between zero and 180 degrees.
They give nondegenerate triangles.

The wider opening is 100 degrees, so the hinge theorem gives

$g<h.$

The triangle inequality bounds h below the sum of the two arms.

$h<12+20.$

Thus h is less than 32 centimeters.
The bound is strict because a 100-degree opening is not flat.

The frames keep the same two arm lengths while their openings differ.

Answer

$\begin{aligned}g&<h.\\h&<32\text{ cm}.\end{aligned}$

Key idea

The hinge comparison orders two gaps while the triangle inequality bounds either gap.
Problem 8 Explain the equality case in both directions

Two nondegenerate triangles ABC and DEF have AB equal to DE and AC equal to DF. Prove that their included angles are equal exactly when their third sides are equal. State the congruence criterion used in each direction.
1. Hint 1
  Start by assuming equal included angles, then match the sides surrounding them.
2. Hint 2
  For the reverse direction, assume equal third sides and count all the matching side pairs.
Answer

Equal included angles give equal third sides by SAS. Equal third sides give equal included angles by SSS.

Full solution

First suppose the included angles BAC and EDF are equal.
The two given corresponding side pairs surround those angles, so SAS makes the triangles congruent.

Corresponding third sides match:
$BC=EF.$

For the reverse direction, suppose BC equals EF.
Together with the two given pairs, this supplies all three corresponding side pairs.
SSS makes the triangles congruent.

Corresponding included angles match:
$m\angle BAC=m\angle EDF.$

For the forward direction, the equal openings complete SAS; for the reverse, equal third sides complete SSS.

Answer

Equal included angles give equal third sides by SAS. Equal third sides give equal included angles by SSS.

Key idea

An “exactly when” statement needs a justified argument in both directions.
Problem 9 Repair the flat auxiliary case

In a hinge construction, A is off line BC. The points on that line occur in the order B, X, P, C. The copied sides AX and AC are equal, and AP bisects angle XAC.

SAS for triangles AXP and ACP has already established that PX equals PC. Explain why the strict triangle inequality cannot be applied to BPX. Then prove BC is longer than BX using the stated order.

The given order and the equality already established by SAS are shown.

Text description of this figure

A is above the line containing B, X, P, and C in that order. AX equals AC. Matching single arcs mark angles XAP and PAC, the two halves bisected by AP. The equality XP equals PC is given after SAS and has double ticks.
1. Hint 1
  Check whether B, P, and X are noncollinear.
2. Hint 2
  Write BC as the sum of the consecutive positive segments along B, X, P, C.
Answer

BPX is flat, so it is not a nondegenerate triangle. Nevertheless,
$BC>BX.$

Full solution

B, P, and X are collinear, so the strict triangle inequality for a nondegenerate triangle is unavailable.

The order B, X, P, C gives the segment sum

$BC=BX+XP+PC.$

Both XP and PC are positive, so adding them strictly increases BX.

$BC>BX.$

The equality of PX and PC is consistent with this argument, but the positive extra lengths already make the comparison strict.

Answer

BPX is flat, so it is not a nondegenerate triangle. Nevertheless,
$BC>BX.$

Key idea

An auxiliary figure can be flat even when the two original triangles are not.
Problem 10 Choose parameters that satisfy every condition

A designer wants two nondegenerate triangles ABC and DEF whose corresponding arm pairs match, with BC longer than EF. The proposed lengths, in units, are
$AB=2t+5.$ $DE=t+8.$
The second pair AC and DF each has length 9 units.

The first included angle, in degrees, is
$m\angle BAC=2s+18.$
The second included angle, in degrees, is
$m\angle EDF=5s-9.$
Find all real parameter pairs (t, s) that meet the requirements, with t positive. Explain what fails at each boundary of your s interval.
1. Hint 1
  Make both corresponding arm pairs equal before applying an angle comparison.
2. Hint 2
  Use the desired third-side order to compare the included angles.
3. Hint 3
  Keep both included angles strictly between zero and 180 degrees, then intersect all restrictions.
Answer

$\begin{aligned}t&=3.\\\frac95&<s<9.\end{aligned}$
At the lower s boundary the second opening is zero; at the upper boundary the openings and third sides are equal.

Full solution

The first equal pair requires
$2t+5=t+8.$

Subtract t, then subtract 5:
$t=3.$

Both first arms are then 11 units, and the second arms are both 9 units.
All are positive.

The converse hinge theorem turns the desired third-side order into an included-angle comparison.

$2s+18>5s-9.$

Subtract 2s and add 9:
$27>3s.$

Thus
$s<9.$

The first included angle is positive exactly when s exceeds negative 9, and it is below 180 degrees exactly when s is less than 81.

The second included angle must be positive.

$5s-9>0.$

This gives
$s>\frac95.$

Its upper angle condition gives

$s<\frac{189}{5}.$

Intersecting all restrictions leaves

$\frac95<s<9.$

For every s in this interval, the positive arms 11 and 9 can be joined at each permitted opening to form the two triangles.
The hinge theorem gives the requested strict side order.

At s equal to nine-fifths, the second opening is zero, so its triangle is flat.
At s equal to 9, both openings measure 36 degrees, so SAS makes the third sides equal.

One permitted choice has t of 3 and s of 5.
Its included angles measure 28 and 16 degrees.

One allowed choice has t of 3 and s of 5; the marked equal pairs and valid openings satisfy all requirements.

Answer

$\begin{aligned}t&=3.\\\frac95&<s<9.\end{aligned}$
At the lower s boundary the second opening is zero; at the upper boundary the openings and third sides are equal.

Key idea

A complete choice must satisfy equal-pair conditions, the strict order, and the angle domain together.

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