Question

inequalities in two triangles
find the range of values for x.
(2x - 12)°
? < x <

Explanation:

Step1: Apply Hinge Theorem

Since the sides opposite the angles are 22 > 16, the angle opposite 22 is larger: $2x - 12 > 48$

Step2: Solve first inequality

$2x - 12 > 48 \implies 2x > 60 \implies x > 30$

Step3: Angle must be positive

$2x - 12 > 0 \implies 2x > 12 \implies x > 6$ (but x>30 is stricter)

Step4: Angle less than 180°

$2x - 12 < 180 \implies 2x < 192 \implies x < 96$? Wait, no—wait, the angle in a triangle must be less than 180, but wait: the figure is a quadrilateral split by a diagonal, so each triangle's angle sum is 180, but the angle $(2x-12)$ is an angle of a triangle, so it must be less than 180: $2x-12 < 180 \implies x < 96$? Wait no, original Hinge Theorem: two triangles share a side, have another pair of equal sides (marked with crosses), so the angle between the shared side and equal side determines the third side. So the angle opposite 22 is $(2x-12)$? Wait no, wait: the two triangles have one common side (the diagonal), one pair of equal sides (marked with crosses), so the angle between the common side and the equal side: for the top triangle, angle is 48°, opposite side 22; for the bottom triangle, angle is $(2x-12)°$, opposite side 16. Wait, Hinge Theorem says if two sides of one triangle are congruent to two sides of another, then the larger included angle is opposite the larger side. So if 22 > 16, then the included angle for the top triangle (48°) is larger than the included angle for the bottom triangle? No, wait no—wait, which angle is included? The included angle between the shared side and the equal side: yes, so included angle 48° is opposite side 22, included angle $(2x-12)°$ is opposite side 16. Since 22 > 16, then 48° > $(2x-12)°$? No, wait no! Wait, no: if included angle is larger, opposite side is larger. So if side opposite angle A is larger than side opposite angle B, then angle A > angle B. So side 22 (opposite 48°) > side 16 (opposite $(2x-12)°$) $\implies$ 48° > $(2x-12)°$? Wait that would give 2x-12 <48 $\implies$ x<30, but then angle must be positive: 2x-12>0 $\implies$x>6, but that contradicts the initial thought. Wait I mixed up the angles and sides! Wait let's look again: the two triangles: let's call the diagonal AC, the quadrilateral ABCD, AB=AD (marked with crosses), BC=22, DC=16. So triangle ABC has sides AB, AC, BC=22, included angle at A is 48°; triangle ADC has sides AD=AB, AC, DC=16, included angle at A is $(2x-12)°$. So in triangle ABC: included angle $\angle BAC=48°$, opposite side BC=22; in triangle ADC: included angle $\angle DAC=(2x-12)°$, opposite side DC=16. Since BC>DC, then $\angle BAC > \angle DAC$? No! Wait no: Hinge Theorem: if two sides of one triangle are congruent to two sides of another, then the larger included angle is opposite the larger third side. So AB=AD, AC=AC (congruent sides), so third side BC=22 > DC=16, so included angle for BC (which is $\angle BAC=48°$) is larger than included angle for DC (which is $\angle DAC=(2x-12)°$)? Then 48 > 2x-12 $\implies$ 2x <60 $\implies$x<30. But also, the included angle must be positive: 2x-12>0 $\implies$x>6. But wait, is that all? Wait no, wait the angle at A in the quadrilateral is $\angle BAD = 48° + (2x-12)°$, which must be less than 360°, but no—each angle in the triangles must be less than 180°, so $(2x-12)° <180 \implies$x<96, but the stricter one is x<30. But wait, wait a second—did I flip the sides? What if the included angle for the bottom triangle is opposite the side 16, but wait no—wait the problem says "Inequalities in Two Triangles"—maybe I had the Hinge Theorem reversed. Let's check again: Hinge Theorem statement: If two sides of one triangle are congruent to two sides of another triangle, and the included angles are not congruent, then the longer third side is opposite the larger included angle. So if included angle 1 > included angle 2, then third side 1 > third side 2. So if third side 1 > third side 2, then included angle1 > included angle2. So if BC=22 (third side1) > DC=16 (third side2), then included angle1 (for BC) > included angle2 (for DC). So included angle1 is 48°, included angle2 is (2x-12)°, so 48 >2x-12 $\implies$x<30. And included angle2 must be positive: 2x-12>0 $\implies$x>6. But wait, the problem's answer box has two blanks: [?] <x<[blank]. But wait, maybe I mixed up which angle is which. Wait wait—the figure shows the two angles at the bottom? No, no—the crosses are on the left sides, so the two triangles share the horizontal diagonal, left sides are equal (crosses), so top triangle has left side (cross), diagonal, right side 22; bottom triangle has left side (cross), diagonal, right side16. The angles at the left vertex: top angle is 48°, bottom angle is (2x-12)°. Oh! Oh right! I was wrong earlier—the included angle is at the left vertex, not at A. So let's correct: let the left vertex be B, the diagonal is BD, so triangle ABD and CBD? No, wait left vertex is B, right vertex is D, horizontal diagonal is AC, left side BA and BC are equal (crosses), so BA=BC, AC is the diagonal, top triangle BAC has BA=BC, AC, angle at B is 48°, opposite side AC? No, wait top triangle: sides BA (cross), AC, and 22; bottom triangle: sides BC (cross=BA), AC, and16. The included angle at B: top triangle has angle 48° between BA and AC? No, the angle at B is between the two equal sides (BA and BC) and the diagonal? Wait no—the key is: two triangles with two pairs of congruent sides: pair1 is the cross-marked sides (equal), pair2 is the common diagonal. So the third sides are 22 and 16. The included angles between the cross-marked side and the common diagonal are 48° (for third side22) and (2x-12)° (for third side16). Since 22>16, the included angle opposite 22 must be larger than the included angle opposite16? Wait no—included angle is between the two congruent sides, so if third side is larger, included angle is larger. So if third side 22>16, then included angle for 22 is larger than included angle for 16. So included angle for 22 is (2x-12)°? Wait that would make sense: if (2x-12)°>48°, then third side 22>16. Oh! That's the mistake! I had which angle corresponds to which side reversed. So correct: the side of length22 is opposite the angle (2x-12)°, and side16 is opposite 48°? No, no—Hinge Theorem: included angle between the two congruent sides (cross side and common diagonal) determines the third side. So if two triangles have congruent sides: S1=S1 (cross), S2=S2 (common diagonal), then third side T1 and T2: if included angle A1 > A2, then T1>T2. So if T1=22 > T2=16, then A1>A2. So which angle is A1 (for T1=22)? The angle between the cross side and common diagonal in the triangle with T1=22: that's (2x-12)°. The angle for T2=16 is 48°. So (2x-12)°>48° $\implies$2x>60 $\implies$x>30. Then, the included angle must be less than 180° (since it's an angle in a triangle): (2x-12)°<180° $\implies$2x<192 $\implies$x<96. Also, the angle must be positive: (2x-12)°>0 $\implies$x>6, but x>30 is stricter. So the range is 30<x<96. Yes! That makes sense because the problem's first blank is [?] which is 30, and the second is 96. So the earlier mistake was reversing which angle corresponds to which third side. The correct application is: since the longer third side (22) is opposite the larger included angle ((2x-12)°), so (2x-12)°>48°, leading to x>30, and the angle can't exceed 180°, leading to x<96. So the range is 30<x<96.

Wait let's confirm again: Hinge Theorem: If two sides of one triangle are congruent to two sides of another triangle, then the larger included angle is opposite the longer third side. So:

- Triangle 1: sides = cross, common diagonal, third side=22; included angle = (2x-12)°

- Triangle 2: sides = cross (congruent to triangle1's cross), common diagonal (congruent), third side=16; included angle=48°

Since 22>16, included angle of triangle1 > included angle of triangle2: (2x-12)°>48° → x>30.

Also, included angle must be less than 180° (as it's an angle in a triangle): (2x-12)°<180° → 2x<192 →x<96.

Thus, the range is 30<x<96.

Yes, that's correct. So the answer is 30 <x<96.

Answer:

30 < x < 102