- What is the total of the three interior angles in any triangle? - Add all three proposed angles as one set. - Compare that sum with the required total.

1. The three interior angles of a triangle must total \(180^\circ\). 2. These three angles add to \(110^\circ+45^\circ+35^\circ=190^\circ\). 3. Since \(190^\circ\ne180^\circ\), no such triangle exists.

No. The three angles total \(190^\circ\), not \(180^\circ\).

- Identify the longest proposed side first. - Compare that length with the sum of the other two sides. - State the inequality explicitly before giving the conclusion.

1. The longest side is \(15\,\text{cm}\). 2. Add the other two side lengths: \(6\,\text{cm}+10\,\text{cm}=16\,\text{cm}\). 3. Since \(16\,\text{cm}>15\,\text{cm}\), the triangle inequality is satisfied, so the triangle can be formed.

Yes. Since \(6+10>15\), the side lengths can form a triangle.

- What does the labeled segment fix? - Which points lie on a ray from A? - What changes if the third vertex is farther from A?

1. The given segment fixes \(AB=5\,\text{cm}\) and the ray fixes the angle at A to \(40^\circ\). 2. Point C can be placed at different distances from A along the ray. Those choices make triangles of different shapes.

More than one triangle shape is possible because the length \(AC\) is not fixed.

- Which letter is the vertex in \(\angle BAC\)? - Which of its two sides is already drawn?

1. The angle's vertex is A, so center the protractor at A. 2. Align its zero mark with the existing ray from A toward B before marking \(60^\circ\).

Center at A; align the zero mark with ray \(AB\).

- Check the requested side against its label. - Which vertex is at the center of the \(45^\circ\) arc? - Does the task ask for that same vertex?

1. The drawn side AB is labeled \(5\,\text{cm}\), so the length condition matches. 2. The \(45^\circ\) label is at A, not B, so the angle condition at B is not met.

No. The side length matches, but the \(45^\circ\) angle is at A rather than B.

- Does one side tell you where the third vertex must be? - Could you change the triangle's height while keeping the base fixed?

1. The third vertex can be placed in many different locations away from the line through the segment. 2. Those positions make triangles of different shapes that all contain the required side.

More than one triangle; the third vertex is still free to move.

- Put the two shorter lengths end to end. - Could they reach across the longest side?

1. The two shorter sides would total \(3+4=7\,\text{cm}\). 2. Since \(7\,\text{cm}\) is shorter than the proposed \(8\,\text{cm}\) side, the ends cannot meet to close a triangle.

No. \(3+4<8\), so the two shorter sides cannot reach each other.

- Which vertex do the two named sides share? - Which measurements can you set down first with a ruler and a protractor? - After fixing the angle, what locates the last vertex?

1. The given angle is at \(B\), where the two known sides meet. 2. Draw \(AB=6\,\text{cm}\). At \(B\), draw a ray at \(40^\circ\) to \(BA\). Mark \(C\) on this ray at \(BC=4.5\,\text{cm}\) and connect \(A\) to \(C\). 3. Once the ray is fixed and the distance from \(B\) is fixed, there is only one possible position for \(C\) on the chosen side of \(AB\). The reflection makes the same triangle shape.

1. The sketch must show \(AB=6\,\text{cm}\), \(BC=4.5\,\text{cm}\), and \(\angle ABC=40^\circ\). 2. Draw \(AB\), make a \(40^\circ\) ray at \(B\), mark \(C\) at the stated distance from \(B\), and join \(A\) to \(C\). 3. The given angle fixes the ray and the second side length fixes the position of \(C\), so no different triangle shape is possible.

- Add the two shorter lengths in each set. - Their sum must be strictly greater than the longest length.

1. For a), the two shorter lengths have sum \(3 + 4 = 7\), and \(7 > 5\). A triangle can be constructed. 2. For b), \(2.5 + 3 = 5.5 < 6\). The two shorter sides cannot meet, so no triangle exists. 3. For c), \(5 + 5 = 10\). Equality produces a degenerate straight-line figure, not a triangle.

a) Yes, because \(3 + 4 > 5\). b) No, because \(2.5 + 3 < 6\). c) No, because \(5 + 5 = 10\), which is degenerate.

- Use the diagonal to identify a triangle inside the parallelogram. - Compare the diagonal with the sum of the two side lengths. - A triangle must satisfy a strict inequality.

1. Diagonal \(\overline{AC}\) would form \(\triangle ABC\) with side lengths \(7\,\text{cm}\), \(4\,\text{cm}\), and \(12\,\text{cm}\). 2. The two shorter sides have sum \(7+4=11\). 3. Since \(11<12\), the triangle inequality fails. 4. Therefore, \(\triangle ABC\) cannot exist, so the parallelogram cannot exist.

No. The diagonal would create a triangle with \(7+4<12\), which violates the triangle inequality.

- What does it mean for two figures to be congruent? - Consider cuts through the midpoints of opposite sides. - How does cutting from a vertex affect the number of sides in each piece? - Sketch several possible straight cuts.

1. For a), cut along either diagonal of the rectangle. The diagonal divides the rectangle into two congruent right triangles. 2. For b), a cut through the midpoints of the \(6\,\text{cm}\) sides and parallel to the \(10\,\text{cm}\) sides forms two \(10\,\text{cm} \times 3\,\text{cm}\) rectangles. 3. A cut through the midpoints of the \(10\,\text{cm}\) sides and parallel to the \(6\,\text{cm}\) sides forms two \(5\,\text{cm} \times 6\,\text{cm}\) rectangles. 4. For c), begin the cut at a vertex and end it at an interior point of either nonadjacent side. The corner piece is a triangle, and the other piece is a trapezoid.

a) Cut along a diagonal. b) The new rectangles can be \(10\,\text{cm} \times 3\,\text{cm}\) or \(5\,\text{cm} \times 6\,\text{cm}\), using a midpoint cut parallel to a pair of sides. c) Yes. Connect a vertex to an interior point of a side that does not contain that vertex.

- What does congruent mean? - Which lines of symmetry does a square have? - What happens if all cuts are parallel to one side? - What regions form when both diagonals are drawn?

1. Squares: Draw the horizontal and vertical midlines. This forms four squares with side length \(4\,\text{cm}\). 2. Triangles: Draw both diagonals. This forms four congruent isosceles right triangles. 3. Rectangles: Draw three parallel lines, each \(2\,\text{cm}\) apart and parallel to one side. This forms four \(8\,\text{cm} \times 2\,\text{cm}\) rectangles.

1. Four \(4\,\text{cm}\) squares, using the two midlines. 2. Four congruent isosceles right triangles, using both diagonals. 3. Four \(8\,\text{cm} \times 2\,\text{cm}\) rectangles, using three equally spaced parallel lines.

- Use the fact that all four sides of a rhombus are congruent. - Begin with the given diagonal and locate points that are \(4.5\,\text{cm}\) from both endpoints. - Which lines divide a rhombus into mirror-image halves?

1. Draw \(\overline{AC}=7\,\text{cm}\) as one diagonal. 2. Draw an arc centered at \(A\) with radius \(4.5\,\text{cm}\) and an arc centered at \(C\) with the same radius. 3. Label the two intersection points of the arcs \(B\) and \(D\). 4. Connect \(A\), \(B\), \(C\), and \(D\) in order. Each side is \(4.5\,\text{cm}\), so the figure is a rhombus. 5. This non-square rhombus has two lines of symmetry, one along each diagonal.

Draw the \(7\,\text{cm}\) diagonal and use arcs of radius \(4.5\,\text{cm}\) from its endpoints to locate the other two vertices. The rhombus has two lines of symmetry: its diagonals.

- At which vertex must the right angle be? - Which side can you draw first with a ruler? - How can a protractor help you draw the other side at a right angle?

1. Draw \(AC=5\,\text{cm}\). 2. Use a protractor to draw a ray from \(C\) perpendicular to \(AC\). 3. Mark \(B\) on the ray \(3\,\text{cm}\) from \(C\). 4. Join \(A\) and \(B\). The two measured sides meet at a right angle at \(C\), as required.

Draw \(AC=5\,\text{cm}\), make a \(90^\circ\) ray at \(C\), mark \(B\) \(3\,\text{cm}\) along the ray, and join \(A\) to \(B\).

- Apply the triangle inequality to each side. - Use it to find both a lower and an upper bound for \(x\).

1. The lower bound is the difference of the known side lengths: \(x > 10 - 6 = 4\,\text{cm}\). 2. The upper bound is their sum: \(x < 10 + 6 = 16\,\text{cm}\). 3. Therefore, \(4\,\text{cm} < x < 16\,\text{cm}\).

The third side length must satisfy \(4\,\text{cm} < x < 16\,\text{cm}\).

- What distance from each base endpoint is represented by its circle? - Compare the three side lengths if you choose the intersection above or below the base. - Can one triangle be flipped onto the other?

1. The circles meet at two points on opposite sides of \(AB\). 2. Either intersection is \(4\,\text{cm}\) from \(A\) and \(5\,\text{cm}\) from \(B\). The base \(AB\) measures \(6\,\text{cm}\). 3. The two triangles are reflections across the base. One can be flipped onto the other without changing its size or shape. 4. Therefore, there is just one triangle shape and size, even though there are two possible vertex positions.

One. The two possible triangles are reflected copies with the same three side lengths.

- How many places could be exactly \(4\,\text{cm}\) from one point? - What must be special about the angle between the two given sides? - Which drawing tool lets you check that angle?

1. A point \(4\,\text{cm}\) from the end of the first side could lie in many directions. 2. Devon must use a protractor or square corner to draw a ray perpendicular to the first side at that endpoint. 3. Mark the second vertex \(4\,\text{cm}\) along that ray and join the free vertices.

No. Draw a ray at \(90^\circ\) to the first side at its endpoint, then mark \(4\,\text{cm}\) along the ray.

- What angle is formed by the left-pointing ray and \(AB\)? - Which part of that straight angle does Sam's mark measure? - Where should the protractor start when measuring from \(AB\)?

1. The wrong starting ray points opposite \(AB\), so the two rays form a straight angle of \(180^\circ\). 2. The marked ray makes \(180^\circ-145^\circ=35^\circ\) with \(AB\). 3. Sam must align the protractor's starting direction with \(AB\), not the opposite ray.

\(35^\circ\). Sam measured from the ray pointing left instead of ray \(AB\).

- What does every point on the circle have in common? - Which points also lie on the angled ray? - Would either intersection make the same side lengths but a different triangle?

1. Every point on the circle is exactly \(5\,\text{cm}\) from B. 2. The circle crosses the ray from A in two places, and both lie in the ray's forward direction. 3. Either intersection can be C, giving two triangles with the stated sides and angle.

Two positions, at the two intersections of the circle and the ray.

- Compare how many places \(C\) could be in each case. - What can change while both named side lengths stay the same? - Does each instruction allow triangles with different angles at \(A\)?

1. In a), C can move along a circle centered at A while staying \(4\,\text{cm}\) from A, giving different angles at A. 2. In b), the \(60^\circ\) angle fixes the ray AC, and its \(4\,\text{cm}\) length fixes C on that ray. 3. Thus b) fixes the shape, up to turning or reflecting the whole triangle.

b) fixes the triangle's shape. The \(60^\circ\) angle fixes the direction of \(AC\), and its \(4\,\text{cm}\) length fixes \(C\) on that ray. In a), the direction of \(AC\) can change, so different triangles are possible.

- Which points are the required distance from A? - Which points are the required distance from B? - Where do both conditions hold at once?

1. A possible C must lie on both circles, so it must be at an intersection. 2. The circles cross at two points, one on either side of AB. 3. Either intersection can be used for C.

Two locations: one above AB and one below AB.

- Compare one named side at a time. - Does the longest side have the requested length? - Are the shorter lengths attached to the right pairs of vertices?

1. Segment AB is correctly labeled \(5\,\text{cm}\). 2. The drawing labels AC as \(3\,\text{cm}\) and BC as \(4\,\text{cm}\), the opposite of what was requested. 3. The student switched the assignments of the two shorter side lengths.

AC and BC are switched; AC should be \(4\,\text{cm}\) and BC should be \(3\,\text{cm}\).

- Do the two base lengths tell you how far apart to place the parallel lines? - Could the shorter base slide along its line without changing its length? - Which other measurement could make a drawing more specific?

1. Draw a \(7\,\text{cm}\) base and a parallel \(4\,\text{cm}\) base above it. 2. You can place the shorter base nearer or farther from the longer one, or shift it to one side, while keeping their lengths and direction fixed. 3. Joining the ends makes different trapezoids, so the two base lengths do not fix one shape.

No. The parallel bases can be different distances apart or shifted sideways, giving different trapezoids.

- How many rays make the specified angle with AB at A on one chosen side? - Do the two rays from A and B share more than one intersection point?

1. The \(45^\circ\) condition determines one ray starting at A. 2. The \(65^\circ\) condition determines a ray starting at B on the same side of AB. 3. The two rays meet at one point, which fixes C on that side of AB.

No. The rays from A and B meet at one fixed point C on the chosen side of AB.

- How much total length is left for the other two sides? - Can that sum be divided in more than one way? - Which of your choices also let the sides reach to form a triangle?

1. Since AB is \(5\,\text{cm}\), the other two sides must total \(13-5=8\,\text{cm}\). 2. One choice is \(AC=4\,\text{cm}\) and \(BC=4\,\text{cm}\); another is \(AC=3\,\text{cm}\) and \(BC=5\,\text{cm}\). 3. Both pass the triangle inequality. Their side lengths differ, so the perimeter and one side do not fix the shape.

No. For example, \((AC,BC)=(4,4)\,\text{cm}\) and \((AC,BC)=(3,5)\,\text{cm}\) both work and give different shapes.

- Compare the longest side with the sum of the other two. - Also compare the third side with the difference of the two known sides. - What points are a fixed distance from an endpoint?

1. The third side must be shorter than the sum and longer than the difference of the given sides: \(5-3<AB<5+3\), so \(2\,\text{cm}<AB<8\,\text{cm}\). 2. Draw \(AB=7\,\text{cm}\). Locate a point \(C\) that is \(5\,\text{cm}\) from \(A\) and \(3\,\text{cm}\) from \(B\), for example by intersecting two circles with those radii. 3. Connect \(C\) to both endpoints. A point on the other side of \(AB\) makes the reflected copy of the same triangle.

1. \(2\,\text{cm}<AB<8\,\text{cm}\). 2. Draw the \(7\,\text{cm}\) base, locate a point \(5\,\text{cm}\) from one endpoint and \(3\,\text{cm}\) from the other, and join it to both endpoints.

- Use the triangle angle sum to test the original data. - For the revised data, find the missing angle before planning the construction. - Which given side lies between the two angles used in the construction?

1. The first two angles total \(75^\circ+115^\circ=190^\circ\), already more than the \(180^\circ\) total for a triangle. No triangle is possible. 2. For the new triangle, \(\angle B=180^\circ-75^\circ-60^\circ=45^\circ\). 3. Draw \(AB=5\,\text{cm}\). Make a \(75^\circ\) ray at \(A\) and a \(45^\circ\) ray at \(B\), both on the same side of \(AB\). Their meeting point is \(C\). 4. With the base and its two endpoint rays fixed, the rays meet at only one point on that side. The other side gives a reflection, not a new shape.

1. No. The given angles total \(190^\circ\), more than \(180^\circ\). 2. \(\angle B=45^\circ\). 3. Draw the base and the two endpoint rays with a protractor, and use their intersection as \(C\). The ray intersection fixes the triangle's shape up to reflection.

- Use both the sum and difference forms of the triangle inequality. - Remember that the inequalities are strict. - Then choose the first and last whole numbers in the interval.

1. The triangle inequality gives \(|12 - 8| < c < 12 + 8\), so \(4 < c < 20\). 2. Since \(c\) is a whole number, its smallest possible value is \(5\,\text{cm}\), and its largest possible value is \(19\,\text{cm}\). 3. If \(c = 20\,\text{cm}\), then \(8 + 12 = 20\). The three points would be collinear, producing a degenerate figure rather than a triangle.

1. Smallest: \(5\,\text{cm}\); largest: \(19\,\text{cm}\) 2. At \(c = 20\,\text{cm}\), the two shorter sides sum exactly to the longest side, so no triangle is formed.

- Consider whether the known side is the base or one of the equal legs. - Write a perimeter equation for each case. - Check each result with the triangle inequality.

1. Case 1: The \(4\,\text{cm}\) side is the base. If each equal leg has length \(x\), then \(2x + 4 = 18\), so \(x = 7\). The side lengths are \(7\,\text{cm}\), \(7\,\text{cm}\), and \(4\,\text{cm}\). 2. These lengths satisfy the triangle inequality because \(7 + 4 > 7\). 3. Case 2: The \(4\,\text{cm}\) side is a leg, so both legs are \(4\,\text{cm}\). The base would be \(18 - 4 - 4 = 10\,\text{cm}\). 4. These lengths do not form a triangle because \(4 + 4 < 10\). 5. Therefore, the only possible side lengths are \(7\,\text{cm}\), \(7\,\text{cm}\), and \(4\,\text{cm}\).

The only possible side lengths are \(7\,\text{cm}\), \(7\,\text{cm}\), and \(4\,\text{cm}\). The alternative \(4\), \(4\), and \(10\) fails because \(4 + 4 < 10\).

- What total must the three interior angles have? - A triangle's third angle must be greater than zero. - If two angles are known, can you use them to draw rays from opposite ends of the known side?

1. The angles of a triangle total \(180^\circ\), so \(0^\circ<B<135^\circ\). For example, \(B=60^\circ\) works. 2. If \(B\) measures \(135^\circ\) or more, the first two angles total at least \(180^\circ\), leaving no positive angle at \(C\). 3. If \(C=60^\circ\), then \(B=180^\circ-45^\circ-60^\circ=75^\circ\). 4. The fixed length \(AB\) and the angles at \(A\) and \(B\) determine two rays that meet at just one point \(C\) on the chosen side of \(AB\). A reflected copy has the same shape and size.

a) For example, \(60^\circ\). Any measure greater than \(0^\circ\) and less than \(135^\circ\) works. b) The remaining angle would be \(0^\circ\) or negative, so no triangle exists. c) Yes. Angle \(B=75^\circ\); the fixed side and two angles locate the third vertex.

- How must the two shorter side lengths compare with the longest? - How can you represent all points at a fixed distance from one endpoint? - What does an intersection of the two sets of possible points mean?

1. In a), the shorter lengths total \(2.5+3=5.5\,\text{cm}\), less than the longest side of \(6\,\text{cm}\). No triangle is possible. 2. In b), \(4+5>7\), so the three lengths can form a triangle. 3. Draw a \(7\,\text{cm}\) segment \(AB\). Locate a point \(5\,\text{cm}\) from \(A\) and \(4\,\text{cm}\) from \(B\) by intersecting two circles of those radii. 4. Use either intersection as \(C\), then join \(AC\) and \(BC\).

a) No; \(2.5+3<6\). b) Yes; draw a \(7\,\text{cm}\) base and locate the third vertex at an intersection of circles with radii \(5\,\text{cm}\) and \(4\,\text{cm}\) centered at its endpoints.

- A median connects a vertex to the midpoint of the opposite side. - First determine the midpoint of the given side. - Then locate \(C\) from its known distances to \(B\) and to that midpoint.

1. Draw \(\overline{AB}\) with \(AB=8\,\text{cm}\). 2. Construct the midpoint \(M\) of \(\overline{AB}\), so \(AM=MB=4\,\text{cm}\). 3. Draw a circle centered at \(B\) with radius \(6\,\text{cm}\). 4. Draw a circle centered at \(M\) with radius \(5\,\text{cm}\). 5. Either intersection of the circles can be labeled \(C\); connect \(A\) and \(B\) to \(C\). The two choices are reflected, congruent triangles.

Draw the \(8\,\text{cm}\) side and construct its midpoint. Point \(C\) is an intersection of a circle centered at \(B\) with radius \(6\,\text{cm}\) and a circle centered at the midpoint with radius \(5\,\text{cm}\).

- Describe the locus of points a fixed perpendicular distance from a line. - Describe the locus of points a fixed distance from \(B\). - Count intersections, then account for reflected copies.

1. Draw \(\overline{AB}\) with \(AB=7.5\,\text{cm}\). 2. A point \(C\) with altitude \(3\,\text{cm}\) from \(AB\) must lie on a line parallel to \(AB\) at perpendicular distance \(3\,\text{cm}\). 3. The condition \(a=BC=5\,\text{cm}\) places \(C\) on a circle centered at \(B\) with radius \(5\,\text{cm}\). 4. On one side of \(AB\), the circle intersects the parallel line at two points, giving two triangles. The corresponding points on the other side give reflected copies, not new congruence classes. 5. Therefore, there are two noncongruent triangles.

Locate \(C\) at intersections of the locus \(3\,\text{cm}\) from line \(AB\) and the circle centered at \(B\) with radius \(5\,\text{cm}\). There are two noncongruent triangles.

- Where do the diagonals of a rhombus cross? - What fraction of each diagonal lies on either side of the crossing point? - Which ruler-and-protractor steps give the four endpoints?

1. A rhombus's diagonals cross at their midpoints at right angles. 2. The diagonal endpoints are \(5\,\text{cm}\) and \(3\,\text{cm}\) from their common midpoint, respectively. Their positions are therefore fixed, apart from turning or reflecting the figure. 3. Draw segment \(AC\) of length \(10\,\text{cm}\). Mark its midpoint \(M\), \(5\,\text{cm}\) from each end. 4. Using a protractor, draw a line perpendicular to \(AC\) at \(M\). Mark \(B\) and \(D\) on opposite sides of \(M\), each \(3\,\text{cm}\) away. 5. Join \(A\), \(B\), \(C\), and \(D\) in order.

a) The diagonals bisect each other at right angles; the four half-diagonals have fixed lengths of \(5\,\text{cm}\) and \(3\,\text{cm}\). b) Draw a \(10\,\text{cm}\) segment, find its midpoint, draw a perpendicular there, mark \(3\,\text{cm}\) on each side, and join the endpoints.

- Begin with the side that has the most connected information. - Use the parallel-base condition to determine the line containing \(C\). - How does the diagonal length locate \(C\) on that line?

1. Draw \(\overline{AB}\) with length \(7.5\,\text{cm}\). 2. At \(A\), construct a \(60^\circ\) ray. On that ray, mark \(D\) so that \(AD=4.5\,\text{cm}\). 3. Through \(D\), draw a line parallel to \(AB\). 4. Draw a circle centered at \(A\) with radius \(8\,\text{cm}\). 5. Choose the intersection of the circle and the parallel line that lies to the right of \(D\); label it \(C\). This choice gives a non-self-intersecting trapezoid. 6. Connect \(B\) to \(C\).

Draw \(AB\), construct the \(60^\circ\) ray at \(A\), mark \(AD=4.5\,\text{cm}\), draw the parallel through \(D\), and locate \(C\) at the appropriate intersection with the circle centered at \(A\) of radius \(8\,\text{cm}\). Then connect \(B\) to \(C\).

- Which pairs of sides are equal in a kite with symmetry line \(AC\)? - What triangle can you make with the known diagonal and two equal sides? - How can the angle at \(C\) help you find the two angles along \(BD\)?

1. Symmetry across \(AC\) gives \(AB=AD=4\,\text{cm}\) and \(BC=CD\). 2. Draw \(BD=5\,\text{cm}\). Find \(A\) at a point \(4\,\text{cm}\) from both \(B\) and \(D\), using intersecting circles or measured distances. 3. Triangle \(BCD\) is isosceles. Its base angles are each \((180^\circ-70^\circ)\div2=55^\circ\). 4. On the other side of \(BD\) from \(A\), draw rays making \(55^\circ\) with \(BD\) at \(B\) and \(D\). Their intersection is \(C\). 5. Join \(A\), \(B\), \(C\), and \(D\).

Draw \(BD=5\,\text{cm}\). Locate \(A\) at distance \(4\,\text{cm}\) from each end. On the other side of \(BD\), draw \(55^\circ\) rays from \(B\) and \(D\); their meeting point is \(C\). Join the four vertices.

- The stated diagonal is not the kite’s symmetry diagonal; identify the side lengths in each triangle it creates. - Apply the triangle inequality separately to the \(5,5,e\) triangle and the \(9,9,e\) triangle. - The final range must satisfy both sets of inequalities.

1. The diagonal \(e\) divides the kite into two isosceles triangles. One has side lengths \(5\,\text{cm}\), \(5\,\text{cm}\), and \(e\); the other has side lengths \(9\,\text{cm}\), \(9\,\text{cm}\), and \(e\). 2. For the triangle with two \(5\,\text{cm}\) sides, the triangle inequality requires \(0 < e < 10\,\text{cm}\). 3. For the triangle with two \(9\,\text{cm}\) sides, the triangle inequality requires \(0 < e < 18\,\text{cm}\). 4. Both triangles must exist, so the common range is \(0 < e < 10\,\text{cm}\). The endpoints are excluded because \(e=0\) does not give a side and \(e=10\,\text{cm}\) makes the smaller isosceles triangle degenerate.

\(0 < e < 10\,\text{cm}\)

- Begin each procedure with one side and the given angle. - Compare how equal side lengths are arranged in a parallelogram and in a kite. - In a kite, the equal sides occur in adjacent pairs.

1. For the parallelogram, draw \(\overline{AB}=5\,\text{cm}\). At \(A\), construct a \(50^\circ\) angle and mark \(D\) so that \(\overline{AD}=3\,\text{cm}\). Draw a line through \(B\) parallel to \(\overline{AD}\) and a line through \(D\) parallel to \(\overline{AB}\). Their intersection is \(C\). 2. For the kite, draw \(\overline{AB}=5\,\text{cm}\). At \(B\), construct a \(50^\circ\) angle and mark \(C\) so that \(\overline{BC}=3\,\text{cm}\). Draw a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(C\) with radius \(3\,\text{cm}\). Their second intersection, other than \(B\), is \(D\). Connect \(C\) to \(D\) and \(D\) to \(A\). 3. The parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals. The kite has line symmetry across diagonal \(\overline{AC}\).

a) Draw adjacent sides of lengths \(5\,\text{cm}\) and \(3\,\text{cm}\) with an included angle of \(50^\circ\), then complete the parallelogram using parallel lines. b) Construct \(AB=AD=5\,\text{cm}\) and \(BC=CD=3\,\text{cm}\) with \(\angle ABC=50^\circ\). c) The parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals. The kite has line symmetry across \(\overline{AC}\).

- Begin with the triangle's three original vertices. - How many new boundary points does a straight cut create? - Compare a cut through a vertex with a cut that avoids every vertex. - Count the vertices of both pieces in each case.

1. If the cut connects a vertex to a point on the opposite side, it forms two triangles. 2. If the cut connects interior points of two sides and avoids the vertices, it forms one smaller triangle and one quadrilateral. The quadrilateral is a trapezoid only when the cut is parallel to the third side. 3. A proper straight cut creates two new boundary points. If the cut avoids the vertices, one piece has one original vertex and the two new points, for a total of three vertices; the other has two original vertices and the two new points, for a total of four vertices. 4. If the cut passes through a vertex, each piece has three vertices. Therefore, no possible one-cut case produces two quadrilaterals.

a) A cut from a vertex to the opposite side produces two triangles. A cut joining interior points of two sides produces one triangle and one quadrilateral. b) A cut avoiding the vertices gives pieces with three and four vertices. A cut through a vertex gives two three-vertex pieces. Therefore, two quadrilaterals are impossible.

- First divide the rectangle into two congruent smaller rectangles. - How does a diagonal divide a rectangle? - All four triangles must have the same two leg lengths. - Which midpoint cut halves the \(10\,\text{cm}\) dimension?

1. Draw the segment through the midpoints of the \(10\,\text{cm}\) sides, parallel to the \(4\,\text{cm}\) sides. This divides the original rectangle into two congruent \(5\,\text{cm} \times 4\,\text{cm}\) rectangles. 2. Draw one diagonal in each smaller rectangle. 3. Each diagonal divides its \(5\,\text{cm} \times 4\,\text{cm}\) rectangle into two congruent right triangles. Since the smaller rectangles are congruent, all four triangles are congruent. 4. The legs of each triangle are the sides of a smaller rectangle, so their lengths are \(5\,\text{cm}\) and \(4\,\text{cm}\).

Divide the rectangle into two \(5\,\text{cm} \times 4\,\text{cm}\) rectangles with a midpoint cut parallel to the \(4\,\text{cm}\) sides, then draw one diagonal in each smaller rectangle. Each right triangle has legs of \(5\,\text{cm}\) and \(4\,\text{cm}\).

- Which diagonal of a kite is bisected by the line of symmetry? - At what angle do the diagonals intersect? - Begin with the diagonal that is the line of symmetry.

1. Draw \(\overline{AC}=8\,\text{cm}\). 2. Mark point \(S\) on \(\overline{AC}\) so that \(AS=2\,\text{cm}\). 3. Construct a line through \(S\) perpendicular to \(\overline{AC}\). 4. Because \(\overline{AC}\) is the line of symmetry, it bisects \(\overline{BD}\). Mark \(B\) and \(D\) on the perpendicular line, on opposite sides of \(S\), so that \(SB=SD=2\,\text{cm}\). 5. Connect \(A\), \(B\), \(C\), and \(D\) in order.

Draw \(AC=8\,\text{cm}\), mark \(S\) so that \(AS=2\,\text{cm}\), construct the perpendicular through \(S\), and place \(B\) and \(D\) \(2\,\text{cm}\) from \(S\) on opposite sides.

- Recall the symmetry of a square. - Combine the diagonal properties of a rectangle and a rhombus. - Use the perpendicular bisector to locate the midpoint and direction of the second diagonal.

1. Draw the first diagonal \(\overline{AC}=6\,\text{cm}\). 2. Construct the perpendicular bisector of \(\overline{AC}\). Let \(M\) be the midpoint. 3. Because the diagonals are congruent and bisect each other, mark points \(B\) and \(D\) on the perpendicular bisector so that \(MB=MD=3\,\text{cm}\). 4. Connect \(A\), \(B\), \(C\), and \(D\) in order. 5. The diagonals of a square are congruent, bisect each other, and are perpendicular.

Construct the perpendicular bisector of a \(6\,\text{cm}\) diagonal and mark the endpoints of the second diagonal \(3\,\text{cm}\) from the midpoint. The diagonals are congruent, bisect each other, and intersect at \(90^\circ\).

- Use the given relationship to find angle \(C\). - What must all three triangle angles add to? - Once the base and both base angles are drawn, can the third vertex move freely?

1. Angle \(C=2\cdot42^\circ=84^\circ\). 2. Angle \(B=180^\circ-42^\circ-84^\circ=54^\circ\). 3. Draw \(AB=7.2\,\text{cm}\). Draw a \(42^\circ\) ray at \(A\) and a \(54^\circ\) ray at \(B\), on the same side of \(AB\). Their intersection gives \(C\). 4. The angles fix the directions of both rays, and they meet at one point. A drawing on the other side is only a reflected copy.

1. \(54^\circ\) 2. Draw \(AB=7.2\,\text{cm}\) with \(42^\circ\) at \(A\) and \(54^\circ\) at \(B\). Use the rays' intersection as \(C\). 3. Both rays start at fixed endpoints and have fixed directions, so they meet at one point (up to reflection).

- Which two segments start at \(A\)? - Where should you draw the given angle? - Can the point \(C\) be anywhere other than the measured point on the ray?

1. Draw \(AB=4.0\,\text{cm}\). 2. At \(A\), draw a ray forming \(105^\circ\) with \(AB\). 3. Measure \(5.5\,\text{cm}\) from \(A\) along that ray and mark \(C\). 4. Join \(B\) and \(C\). With the two sides and the angle between them fixed, \(C\) cannot move to another position on that ray. The opposite-side construction is a reflection.

Draw \(AB=4.0\,\text{cm}\), construct a \(105^\circ\) ray from \(A\), mark \(C\) on it \(5.5\,\text{cm}\) from \(A\), and connect \(B\) to \(C\). The fixed ray and distance locate \(C\), apart from reflection.

- Write the perimeter as the base plus two equal side lengths. - Keep the total perimeter fixed as the base changes. - Use the triangle inequality to test the proposed \(12\,\text{cm}\) base.

1. For a), let each congruent side have length \(x\). Then \(2x+4=20\), so \(x=8\,\text{cm}\). 2. For b), \(2x+8=20\), so \(x=6\,\text{cm}\). Each congruent side decreases by \(2\,\text{cm}\). 3. For c), a \(12\,\text{cm}\) base would leave only \(20-12=8\,\text{cm}\) for the other two sides combined. 4. The triangle inequality requires the sum of those two sides to be greater than the base, but \(8<12\). Therefore, no triangle is possible.

a) Each congruent side is \(8\,\text{cm}\). b) Each congruent side is \(6\,\text{cm}\), a decrease of \(2\,\text{cm}\). c) No. A \(12\,\text{cm}\) base would leave only \(8\,\text{cm}\) for the other two sides combined, violating the triangle inequality.

- Think about the line of symmetry of an isosceles triangle. - Locate the midpoint of the base before placing the vertex. - The altitude length tells you how far the vertex is from the base along that perpendicular line.

1. In an isosceles triangle, the altitude from the vertex to the base is also the perpendicular bisector of the base. It locates both the midpoint of the base and the line containing vertex \(C\). 2. Draw \(\overline{AB}\) with \(AB=6\,\text{cm}\). 3. Construct the perpendicular bisector of \(\overline{AB}\), and label its intersection with \(\overline{AB}\) as \(M\). 4. On the perpendicular bisector, mark \(C\) so that \(MC=4\,\text{cm}\). 5. Connect \(A\) and \(B\) to \(C\).

1. The perpendicular bisector locates the midpoint of the base and the altitude line. 2. Draw \(AB=6\,\text{cm}\), construct its perpendicular bisector, and mark \(C\) on that line \(4\,\text{cm}\) from the midpoint; then connect \(C\) to \(A\) and \(B\).

- Compute the three side lengths for several allowed values of \(x\). - The longest side can change as \(x\) changes; check which is longest each time. - Compare the longest length with the total of the other two.

1. A triangle is possible when the two shorter sides add to more than the longest side. 2. For \(x=1,2,3\), the lengths are \((1,1,12)\), \((2,4,12)\), and \((3,7,12)\). In each case, the two shorter lengths do not total more than \(12\). 3. For \(x=4,5,6\), the lengths are \((4,10,12)\), \((5,13,12)\), and \((6,16,12)\). In each case, the shorter sides total more than the longest. 4. For \(x=7\) or \(8\), the lengths are \((7,19,12)\) or \((8,22,12)\); the two shorter lengths total only \(19\) or \(20\), respectively, so no triangle is possible. 5. The possible values are \(4,5,6\), and the smallest is \(4\).

\(x=4,5,6\); smallest \(x=4\).

- Identify the longest side. - Compare the sum of the two shorter sides with the longest side. - After finding the least integer value of \(k\), add all three side lengths.

1. For positive \(k\), the longest side is \(k + 4\). The sum of the two shorter sides must be greater than the longest side. 2. Write \(k + (k + 2) > k + 4\). 3. Simplify: \(2k + 2 > k + 4\), so \(k > 2\). 4. The other two triangle inequalities are satisfied whenever \(k\) is positive. 5. The least integer greater than \(2\) is \(3\). 6. The perimeter is \(k + (k + 2) + (k + 4) = 3k + 6\). For \(k = 3\), the perimeter is \(3 \cdot 3 + 6 = 15\,\text{in.}\).

The triangle exists when \(k > 2\). If \(k\) is an integer, the least possible perimeter is \(15\,\text{in.}\).

- What must be true about the sum of the two shorter sides? - Compare \(2x+(x+10)\) with \(30\) for small and larger allowed integers. - Check that the other side lengths do not become too large at the upper end.

1. The lengths must be positive, and the two shorter lengths must add to more than the longest. 2. For \(x\) from \(1\) through \(6\), the first two lengths total \(2x+(x+10)=3x+10\), which is at most \(28\), less than \(30\). None of these values works. 3. For \(x\) from \(7\) through \(12\), the first two lengths total at least \(31>30\). Also, \(2x\) is at most \(24\), and \(x+10\) is at most \(22\), so neither exceeds the sum of the other two sides. 4. The possible values are \(7,8,9,10,11,12\), giving \(6\) values.

\(x=7,8,9,10,11,12\); \(6\) values.

- Begin with the side shared by the two given angles. - Determine how each angle and fixed distance locates one new vertex. - Compare placing the construction on opposite sides of the base.

1. Draw \(\overline{BC}\) with \(BC=3\,\text{cm}\). 2. At \(B\), construct a \(100^\circ\) angle and mark \(A\) on its ray so that \(BA=6\,\text{cm}\). 3. At \(C\), on the same side of \(\overline{BC}\), construct an \(80^\circ\) angle and mark \(D\) on its ray so that \(CD=4\,\text{cm}\). 4. Connect \(A\) to \(D\). The four vertices are fixed, and the resulting quadrilateral is simple. 5. Constructing the points on the other side of \(\overline{BC}\) produces only a reflected, congruent quadrilateral. Therefore, the quadrilateral is unique up to reflection.

Yes. The side \(BC\), the two adjacent angles, and the two adjacent side lengths fix points \(A\) and \(D\). The only other placement is a reflected, congruent copy.

- What can change in a four-sided frame with hinged corners? - Would the distance between two opposite vertices always stay the same? - How could one diagonal divide the quadrilateral into two triangles with known sides?

1. Leon's claim is false. Imagine the four sides joined at movable corners: the angles can change while the side lengths stay fixed. 2. For example, diagonal \(AC\) can have different lengths between \(2\,\text{cm}\) and \(8\,\text{cm}\). The triangles on its two sides can be formed with side lengths \(5,3,AC\) and \(4,4,AC\), respectively, producing different quadrilaterals. 3. Giving the length of \(AC\) fixes the three side lengths of each triangle. Each triangle then has a fixed shape. 4. For a convex quadrilateral, the two triangles must lie on opposite sides of \(AC\); only a reflected copy remains possible.

a) No. The corners can move, changing the angles even though all four sides keep their lengths. b) The length of diagonal \(AC\) is one possible extra measurement. It fixes the two triangles on either side.

- What are the two possible choices for the pair of congruent sides? - Can a triangle form when two side lengths have a sum equal to the third side length? - Find the perimeter only for the side-length set that forms a triangle.

1. The possible side-length sets are \(8\,\text{cm}, 8\,\text{cm}, 4\,\text{cm}\) and \(4\,\text{cm}, 4\,\text{cm}, 8\,\text{cm}\). 2. The set \(4\,\text{cm}, 4\,\text{cm}, 8\,\text{cm}\) does not form a triangle because \(4\,\text{cm} + 4\,\text{cm} = 8\,\text{cm}\), not greater than \(8\,\text{cm}\). 3. The set \(8\,\text{cm}, 8\,\text{cm}, 4\,\text{cm}\) satisfies the triangle inequality. 4. Its perimeter is \(8\,\text{cm} + 8\,\text{cm} + 4\,\text{cm} = 20\,\text{cm}\).

The perimeter is \(20\,\text{cm}\).

- What two triangles appear when you draw diagonal \(AC\)? - How many side lengths are known for each triangle? - On which sides of the diagonal must the triangles lie to make a convex quadrilateral?

1. Diagonal \(AC\) splits the quadrilateral into triangles \(ABC\) and \(ACD\). 2. Triangle \(ABC\) has all three sides fixed: \(5\,\text{cm}\), \(4\,\text{cm}\), and \(6\,\text{cm}\). Its shape cannot change while keeping these lengths. 3. Triangle \(ACD\) also has three fixed sides: \(6\,\text{cm}\), \(6\,\text{cm}\), and \(3\,\text{cm}\). Its shape is fixed too. 4. To form a convex quadrilateral, these triangles must be joined along \(AC\) on opposite sides. The other possible drawing is a reflection.

The known diagonal makes two triangles, each with three fixed side lengths. Joining them on opposite sides of that diagonal fixes one convex quadrilateral, up to reflection.

- Start with two perpendicular segments that cross. - For line symmetry, the endpoints on opposite sides of the symmetry line must be reflections of each other. - Consider what changes when each diagonal bisects the other.

1. Draw diagonal \(\overline{AC}=6\,\text{cm}\). Choose a point \(P\) on \(\overline{AC}\) that is not its midpoint, such as \(AP=2\,\text{cm}\) and \(PC=4\,\text{cm}\). 2. Draw a line through \(P\) perpendicular to \(\overline{AC}\). Place \(B\) and \(D\) on opposite sides of \(P\) so that \(PB=1\,\text{cm}\) and \(PD=3\,\text{cm}\). Then \(BD=4\,\text{cm}\). Connect \(A\), \(B\), \(C\), and \(D\) in order. Neither diagonal bisects the other, so the quadrilateral is not a kite. 3. To form a kite, one diagonal must be the perpendicular bisector of the other. That diagonal is the line of symmetry. 4. The kite is a rhombus when the diagonals bisect each other. Then the quadrilateral also has \(180^\circ\) rotational symmetry.

a) One example has \(AP=2\,\text{cm}\), \(PC=4\,\text{cm}\), \(PB=1\,\text{cm}\), and \(PD=3\,\text{cm}\), with \(\overline{AC}\perp\overline{BD}\). b) One diagonal must be the perpendicular bisector of the other. c) The diagonals must bisect each other.

- Compare placing equal-length sides opposite each other with placing them next to each other. - An isosceles trapezoid must have congruent legs. - Consider how changing the angles affects which symmetries remain.

1. Arrange the sticks as opposite pairs in the order \(4, 8, 4, 8\). With nonright angles, this forms a general parallelogram. With four right angles, it forms a rectangle. 2. Arrange the equal lengths as adjacent pairs in the order \(4, 4, 8, 8\). This forms a kite. 3. In an isosceles trapezoid, the legs must be congruent. If the two \(4\,\text{cm}\) sticks are the legs, the two bases are both \(8\,\text{cm}\). If the two \(8\,\text{cm}\) sticks are the legs, the two bases are both \(4\,\text{cm}\). In either case, one pair of opposite sides is both parallel and congruent, so the figure is a parallelogram. If the bases have different lengths, the remaining sticks are not congruent and cannot be the legs. 4. A general parallelogram has no lines of symmetry. A rectangle has two lines of symmetry through the midpoints of opposite sides. The kite has one line of symmetry through the vertices where each pair of congruent adjacent sides meets.

a) A general parallelogram, a rectangle, and a kite can be formed. b) Congruent legs leave two equal-length bases, which produces a parallelogram. Different-length bases leave unequal legs. c) General parallelogram: no lines of symmetry; rectangle: two lines through opposite side midpoints; kite: one diagonal line of symmetry.

- How many of the four angle measures are independent once their sum is known? - Fix the given side and imagine the opposite side sliding parallel to itself. - Congruent figures require equal corresponding side lengths, not only equal angles.

1. The interior angles of every quadrilateral have sum \(360^\circ\). Therefore, the four angle measurements are not independent. Once three angles are known, the fourth is determined by \(\delta=360^\circ-(\alpha+\beta+\gamma)\). 2. The data provide only four independent measurements: three angles and one side length. 3. Fixing \(\overline{AB}\) and the adjacent angles \(\alpha\) and \(\beta\) fixes the directions of the sides from \(A\) and \(B\). 4. A segment with the required direction for the opposite side can slide parallel to itself and intersect those two rays at different points. This changes the other three side lengths while preserving all four angles and \(AB=5\,\text{cm}\). 5. Therefore, infinitely many noncongruent quadrilaterals satisfy the measurements.

Lucas is incorrect. The fourth angle is determined by the other three because the angle sum is \(360^\circ\), so it is not an independent measurement. The opposite side can slide parallel to itself, producing infinitely many noncongruent quadrilaterals with the same four angles and the same \(5\,\text{cm}\) side.

- Decide whether \(\overline{AB}\) can be a side, a diagonal, or both. - A side-based square can lie on either side of \(\overline{AB}\). - Turn the horizontal and vertical change from \(A\) to \(B\) by \(90^\circ\) to get an equal-length perpendicular change.

1. Segment \(\overline{AB}\) can be a side of a square in two different orientations, one on each side of the segment. 2. Segment \(\overline{AB}\) can also be a diagonal of exactly one square. 3. Therefore, there are \(3\) possible squares. 4. From \(A\) to \(B\), move \(3\) units right and \(1\) unit up. Turning this movement \(90^\circ\) counterclockwise gives a movement of \(1\) unit left and \(3\) units up. 5. Apply that movement to both endpoints: from \(A\), reach \((1, 5)\); from \(B\), reach \((4, 6)\). These are the other two vertices of one square.

a) There are \(3\) different squares. b) One possible pair of additional vertices is \((4, 6)\) and \((1, 5)\).

- Any one of the three given points can be the shared endpoint of two adjacent sides. - Find a horizontal and vertical change between two points. - Apply the same change from the remaining point.

1. Reading the graph gives \(A(3, 3)\), \(B(7, 4)\), and \(C(4, 7)\). 2. First, treat \(A\) as the shared endpoint of two adjacent sides. From \(A\) to \(C\), move \(1\) unit right and \(4\) units up. Apply that movement to \(B\) to get \(D = (8, 8)\). 3. Next, treat \(B\) as the shared endpoint. From \(B\) to \(C\), move \(3\) units left and \(3\) units up. Apply that movement to \(A\) to get \(D = (0, 6)\). 4. Finally, treat \(C\) as the shared endpoint. From \(C\) to \(B\), move \(3\) units right and \(3\) units down. Apply that movement to \(A\) to get \(D = (6, 0)\). 5. Therefore, the three possible coordinates are \((8, 8)\), \((0, 6)\), and \((6, 0)\).

The three possible coordinates are \((6, 0)\), \((0, 6)\), and \((8, 8)\).