Free Math Worksheets

- Check whether the sum of the two shorter sides is greater than the longest side. - If that sum were less than or equal to the longest side, the segments could not form a triangle.

1. Add the two shorter side lengths: \(5\,\text{cm} + 12\,\text{cm} = 17\,\text{cm}\). 2. Since \(17\,\text{cm} > 13\,\text{cm}\), the sum of the two shorter sides is greater than the longest side. 3. Therefore, the side lengths satisfy the triangle inequality, and the triangle exists.

Yes. The triangle exists because \(5\,\text{cm} + 12\,\text{cm} > 13\,\text{cm}\).

- Identify which vertices are connected by sides \(a\) and \(c\). - Locate angle \(\beta\). - Use a compass to transfer a fixed length onto a ray. - Determine whether the angle is included between the given sides.

1. In standard triangle notation, \(c = AB\), \(a = BC\), and \(\beta\) is the angle at \(B\). Thus, the given angle is included between the two given sides. 2. Draw \(\overline{AB}\) with \(AB = 6\,\text{cm}\). 3. At \(B\), construct a ray that forms a \(40^\circ\) angle with \(\overline{BA}\). 4. Use a compass to mark point \(C\) on the ray so that \(BC = 4.5\,\text{cm}\). 5. Draw \(\overline{AC}\). The triangle is uniquely determined by SAS.

1. The sketch shows \(AB = 6\,\text{cm}\), \(BC = 4.5\,\text{cm}\), and \(\angle ABC = 40^\circ\). 2. Draw \(AB\), construct the \(40^\circ\) ray at \(B\), mark \(C\) so that \(BC = 4.5\,\text{cm}\), and connect \(A\) to \(C\). 3. SAS

- 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.

- Recall that a median connects a vertex to the midpoint of the opposite side. - First construct the midpoint of the given side. - Locate \(C\) using its known distances from \(B\) and the 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. Label either intersection point \(C\). Draw \(\overline{AC}\) and \(\overline{BC}\). 6. The other intersection produces a reflected, congruent triangle.

Draw the \(8\,\text{cm}\) side and 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 \(5\,\text{cm}\).

- 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.

The parallelogram is impossible because its 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.

- Construct a perpendicular through \(C\) using equal compass arcs. - The two legs meet at the vertex of the right angle. - A perpendicular bisector construction can create an exact \(90^\circ\) angle.

1. Draw \(\overline{AC}\) with \(AC = 5\,\text{cm}\), and extend the line through \(C\). 2. Mark two points on the line that are the same distance from \(C\), one on each side of \(C\). 3. Construct the perpendicular bisector of the segment joining those two points. This line passes through \(C\) and is perpendicular to \(\overline{AC}\). 4. On the perpendicular line, mark \(B\) so that \(BC = 3\,\text{cm}\). 5. Draw \(\overline{AB}\). The completed triangle has a right angle at \(C\).

The construction produces \(\triangle ABC\) with \(AC = 5\,\text{cm}\), \(BC = 3\,\text{cm}\), and \(\angle C = 90^\circ\).

- Recall what the distance between two parallel lines represents. - Ask whether any connecting segment can be shorter than that distance.

1. The distance between parallel lines is the length of the shortest segment connecting them. 2. Because \(k\) and \(l\) are \(6\,\text{cm}\) apart, every segment joining a point on \(k\) to a point on \(l\) must be at least \(6\,\text{cm}\) long. 3. Therefore, \(\overline{CD}\) cannot be \(5\,\text{cm}\) long.

No. Every segment connecting the two lines must be at least \(6\,\text{cm}\) long.

- Think about what “exactly halfway” means for the two distances. - Consider the line parallel to both given lines inside the strip between them. - Divide the total distance into two equal parts.

1. The points equidistant from two parallel lines lie on a third line parallel to both and exactly halfway between them. 2. This middle parallel line divides the \(5\,\text{cm}\) distance into two equal distances. 3. Therefore, each point on the middle line is \(5\,\text{cm} \div 2 = 2.5\,\text{cm}\) from \(a\).

a) The points lie on a line parallel to \(a\) and \(b\), exactly halfway between them. b) Each point is \(2.5\,\text{cm}\) from \(a\).

- Compare the angle measures suggested by the two drawings. - Imagine the sides as rigid bars joined by hinges. - Which added segment would divide the quadrilateral into two triangles? - Which congruence criterion applies when all three sides of a triangle are known?

1. The diagrams have the same four side lengths but different angle measures, so the quadrilaterals have different shapes. 2. Unlike a triangle, a quadrilateral with four fixed side lengths is not rigid. Its vertices can move like hinged joints while the side lengths remain unchanged. 3. One useful additional measurement is the length of diagonal \(AC\). 4. Diagonal \(AC\) divides the quadrilateral into \(\triangle ABC\) and \(\triangle ACD\). Each triangle is determined by three side lengths, so each is fixed by SSS. 5. For a convex quadrilateral, the two triangles lie on opposite sides of \(\overline{AC}\). Therefore, the quadrilateral is unique up to reflection.

Four side lengths alone do not determine a quadrilateral because the angles can change. Adding the length of diagonal \(AC\), together with the requirement that the quadrilateral be convex, determines both component triangles by SSS and fixes the quadrilateral up to reflection.

- Compare the longest side with the sum of the other two. - Also compare the third side with the difference of the two known sides. - Use circles to locate a point at two fixed distances.

1. The triangle inequalities require \(3 + 5 > c\) and \(3 + c > 5\). These simplify to \(c < 8\) and \(c > 2\). 2. Therefore, \(2\,\text{cm} < c < 8\,\text{cm}\). 3. For \(c = 7\,\text{cm}\), draw \(\overline{AB}\) with length \(7\,\text{cm}\). 4. Draw a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(B\) with radius \(3\,\text{cm}\). 5. Label either intersection point \(C\), then draw \(\overline{AC}\) and \(\overline{BC}\). The two reflected constructions are congruent by SSS.

1. \(2\,\text{cm} < c < 8\,\text{cm}\) 2. Draw a \(7\,\text{cm}\) base and intersect circles of radii \(5\,\text{cm}\) and \(3\,\text{cm}\) centered at its endpoints.

- Use the triangle angle-sum theorem. - Subtract the two known angles from \(180^\circ\). - Identify the side between \(\alpha\) and \(\beta\).

1. The given angles have sum \(75^\circ + 115^\circ = 190^\circ\). Since the interior angles of a triangle sum to \(180^\circ\), no such triangle exists. 2. With \(\gamma = 60^\circ\), \(\beta = 180^\circ - 75^\circ - 60^\circ = 45^\circ\). 3. Draw \(\overline{AB}\) with \(AB = 5\,\text{cm}\). At \(A\), construct a \(75^\circ\) ray. At \(B\), construct a \(45^\circ\) ray on the same side of \(\overline{AB}\). Their intersection is \(C\). 4. The revised triangle is uniquely determined by ASA.

1. \(190^\circ\); the angle sum exceeds \(180^\circ\), so the triangle is impossible. 2. \(\beta = 45^\circ\) 3. Construct the two endpoint angles on the \(5\,\text{cm}\) side. The criterion is ASA.

- 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 \(4 + 4 < 10\).

- Use the triangle angle-sum theorem. - The third angle must have a positive measure. - When two angles are known, calculate the third before constructing.

1. Since the interior angles must sum to \(180^\circ\), \(45^\circ + \beta < 180^\circ\). Thus, any value with \(0^\circ < \beta < 135^\circ\) works; for example, \(\beta = 60^\circ\). 2. If \(\beta \ge 135^\circ\), then \(\alpha + \beta \ge 180^\circ\), leaving no positive measure for \(\gamma\). 3. If \(\gamma = 60^\circ\), then \(\beta = 180^\circ - 45^\circ - 60^\circ = 75^\circ\). 4. The side \(c\) lies between angles \(\alpha\) and \(\beta\), so the triangle is uniquely determined by ASA.

a) For example, \(\beta = 60^\circ\). Any \(0^\circ < \beta < 135^\circ\) works. b) At \(135^\circ\) or greater, \(\alpha + \beta \ge 180^\circ\), so no triangle exists. c) Yes. \(\beta = 75^\circ\), and the triangle is determined by ASA.

- Compare the longest side with the sum of the other two. - Use two circles for an SSS construction. - Measure or calculate each interior angle and check that the sum is \(180^\circ\).

1. For a), \(2.5 + 3 = 5.5 < 6\), so no triangle can be constructed. 2. For b), the two shorter sides satisfy \(4 + 5 > 7\), so a triangle can be constructed. 3. Draw \(AB = 7\,\text{cm}\). Draw a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(B\) with radius \(4\,\text{cm}\). Their intersection is \(C\). This is an SSS construction. 4. The angle measures are approximately \(\alpha \approx 34^\circ\), \(\beta \approx 44^\circ\), and \(\gamma \approx 102^\circ\).

a) No, because \(2.5 + 3 < 6\). b) Yes, by SSS. The angles are approximately \(34^\circ\), \(44^\circ\), and \(102^\circ\).

- Identify the locus of points at a fixed perpendicular distance from a line. - Identify 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. Draw a line parallel to \(\overline{AB}\) at a perpendicular distance of \(3\,\text{cm}\). Any possible point \(C\) on that side of the base must lie on this line. 3. Draw a circle centered at \(B\) with radius \(5\,\text{cm}\). 4. The circle intersects the parallel line at two points, producing two triangles. 5. Repeating the construction on the other side of \(\overline{AB}\) gives reflections of those triangles, not additional congruence classes. Therefore, there are two noncongruent triangles.

There are two noncongruent triangles.

- What two special relationships hold between the diagonals of a rhombus? - How far is each diagonal endpoint from the intersection point? - Which construction creates a perpendicular line through a segment’s midpoint?

1. The diagonals of a rhombus bisect each other and are perpendicular. 2. Therefore, their intersection is the midpoint of both diagonals. The endpoints of the \(10\,\text{cm}\) diagonal are each \(5\,\text{cm}\) from the intersection, and the endpoints of the \(6\,\text{cm}\) diagonal are each \(3\,\text{cm}\) from it. 3. Draw a \(10\,\text{cm}\) segment \(\overline{AC}\). 4. Construct the perpendicular bisector of \(\overline{AC}\), meeting it at midpoint \(M\). 5. On the perpendicular bisector, mark points \(B\) and \(D\) on opposite sides of \(M\) so that \(MB = MD = 3\,\text{cm}\). 6. Connect \(A\), \(B\), \(C\), and \(D\) in order. The fixed perpendicular half-diagonals determine all four vertices uniquely up to rigid motion and reflection.

a) The diagonals of a rhombus are perpendicular bisectors of each other, so their lengths fix all four vertices relative to their intersection. b) Draw the \(10\,\text{cm}\) diagonal, construct its perpendicular bisector, mark \(3\,\text{cm}\) in each direction from the midpoint, and connect the four 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 \(60^\circ\) angle. On its 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 \(C\).

- What equal side pairs follow from the symmetry axis? - Which two triangles share diagonal \(BD\)? - Find the base angles of the isosceles triangle with vertex angle \(70^\circ\).

1. Because \(AC\) is the symmetry axis, \(AB = AD = 4\,\text{cm}\) and \(BC = CD\). 2. Construct isosceles triangle \(ABD\) with side lengths \(4\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\) by SSS. 3. In isosceles triangle \(BCD\), the vertex angle at \(C\) is \(70^\circ\). Each base angle is \((180^\circ - 70^\circ) \div 2 = 55^\circ\). 4. At \(B\) and \(D\), construct \(55^\circ\) angles on the side of \(BD\) opposite \(A\). Their rays intersect at \(C\). 5. Connect \(B\) to \(C\) and \(C\) to \(D\).

Construct \(\triangle ABD\) from side lengths \(4\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\). On the opposite side of \(BD\), construct \(55^\circ\) base angles at \(B\) and \(D\); their intersection is \(C\).

- Focus on one triangle formed by the diagonal. - Use the sum and difference forms of the triangle inequality. - Decide whether the endpoint values are allowed. - Check that the same triangle can be reflected to form the kite.

1. The diagonal divides the kite into two congruent triangles, each with side lengths \(5\,\text{cm}\), \(9\,\text{cm}\), and \(e\). 2. The triangle inequality requires \(|9 - 5| < e < 9 + 5\). 3. Therefore, \(4\,\text{cm} < e < 14\,\text{cm}\). 4. The endpoint values are excluded because they would produce degenerate triangles with collinear vertices.

\(4\,\text{cm} < e < 14\,\text{cm}\)

- 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) The cut can produce two triangles, or 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 \(90^\circ\).

- First double \(\alpha\) to find \(\gamma\). - Use the triangle angle-sum theorem. - Construct the two angles at the endpoints of side \(c\).

1. \(\gamma = 2\cdot 42^\circ = 84^\circ\). 2. \(\beta = 180^\circ - 42^\circ - 84^\circ = 54^\circ\). 3. Draw \(AB = 7.2\,\text{cm}\). Construct a \(42^\circ\) ray at \(A\) and a \(54^\circ\) ray at \(B\) on the same side of \(\overline{AB}\). Their intersection is \(C\). 4. Measuring the construction gives \(a \approx 4.8\,\text{cm}\) and \(b \approx 5.9\,\text{cm}\).

\(\beta = 54^\circ\), \(a \approx 4.8\,\text{cm}\), and \(b \approx 5.9\,\text{cm}\)

- Identify which angle lies between the two given sides. - Make a quick sketch before beginning the accurate construction. - Remember that \(105^\circ\) is an obtuse angle.

1. Draw \(\overline{AB}\) with \(AB = c = 4.0\,\text{cm}\). 2. At \(A\), construct an angle of \(105^\circ\). 3. On the new ray, mark \(C\) so that \(AC = b = 5.5\,\text{cm}\). 4. Draw \(\overline{BC}\). 5. Measuring the completed triangle gives \(a \approx 7.6\,\text{cm}\), \(\beta \approx 44.4^\circ\), and \(\gamma \approx 30.6^\circ\). 6. The two given sides and their included angle determine the triangle uniquely by SAS.

\(a \approx 7.6\,\text{cm}\), \(\beta \approx 44.4^\circ\), and \(\gamma \approx 30.6^\circ\). The congruence criterion is SAS.

- 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 \(2x = 16\) and \(x = 8\,\text{cm}\). 2. For b), \(2x + 8 = 20\), so \(2x = 12\) and \(x = 6\,\text{cm}\). 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) A \(12\,\text{cm}\) base is impossible because the other two sides would have a total length of only \(8\,\text{cm}\), violating the triangle inequality.

- Think about the line of symmetry of an isosceles triangle. - Locate the midpoint of the base before marking the altitude. - Make sure the altitude is perpendicular to the base.

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. Draw \(\overline{AC}\) and \(\overline{BC}\). 6. Measure \(\overline{AC}\) and \(\overline{BC}\) with a ruler. Each measures approximately \(5.0\,\text{cm}\).

The perpendicular bisector locates the midpoint of the base and the altitude line. The completed triangle has \(a \approx b \approx 5.0\,\text{cm}\).

- What happens to the number of intersections when lines are parallel? - For the greatest number, think about how many new intersections each added line can make. - Try the problem first with two or three lines and look for a pattern.

1. To get the greatest number, arrange the lines so that no two are parallel and no three pass through the same point. 2. The second line adds \(1\) intersection, the third adds \(2\), and the fourth adds \(3\). 3. The greatest possible number is \(1 + 2 + 3 = 6\). 4. To get the least number, make all four lines parallel to one another. 5. Parallel lines do not intersect, so the least possible number is \(0\).

a) The greatest possible number is \(6\) intersection points. b) The least possible number is \(0\) intersection points.

- How many existing lines can the new line cross? - Add the new intersections to the previous total. - Look at how the totals \(1, 3, 6, \ldots\) increase.

1. A new line creates the most new intersections when it crosses every line already drawn at a different point. 2. With four lines already drawn, a fifth line can add \(4\) new intersection points. 3. Five lines can therefore form \(6 + 4 = 10\) intersection points. 4. A sixth line can cross the five existing lines at \(5\) new points. 5. Six lines can therefore form \(10 + 5 = 15\) intersection points.

a) The fifth line can add \(4\) new intersection points. b) Five lines can form at most \(10\) intersection points. c) Six lines can form at most \(15\) intersection points.

- Determine how many lines can be parallel to \(g\) at the given distance. - Consider choosing the same line twice. - Consider choosing one line on each side of \(g\).

1. There are exactly two lines parallel to \(g\) at a perpendicular distance of \(3\,\text{cm}\): one on each side of \(g\). 2. If \(a\) and \(b\) are the same line, the distance between them is \(0\,\text{cm}\). 3. If \(a\) and \(b\) are on opposite sides of \(g\), their distance is \(3\,\text{cm} + 3\,\text{cm} = 6\,\text{cm}\).

The possible distances are \(0\,\text{cm}\) and \(6\,\text{cm}\).

- Consider whether \(A\) is between the lines or outside them. - Compare the two possible arrangements. - Decide whether to subtract or add the distances.

1. If \(A\) lies between the two parallel lines, its distance to \(t\) is \(8-3=5\,\text{cm}\). 2. If \(A\) lies outside the strip on the side of \(s\) away from \(t\), its distance to \(t\) is \(8+3=11\,\text{cm}\).

\(5\,\text{cm}\) or \(11\,\text{cm}\)

- Consider the two possible positions of the parallel lines relative to \(P\). - In one arrangement, \(P\) is between the lines. - In the other arrangement, both lines are on the same side of \(P\).

1. If \(P\) lies between the lines, their distance is \(2+6=8\,\text{cm}\). 2. If both lines lie on the same side of \(P\), their distance is the difference \(6-2=4\,\text{cm}\).

\(4\,\text{cm}\) or \(8\,\text{cm}\)

- The perpendicular segment gives the shortest distance to a line. - Compare each proposed length with \(4\,\text{cm}\). - Segments to other points on the line may be longer than the perpendicular segment.

1. The distance from a point to a line is the length of the shortest segment from the point to the line. Here, that minimum is \(4\,\text{cm}\). 2. \(PX=4\,\text{cm}\) is possible when \(\overline{PX}\) is perpendicular to \(g\). 3. \(PY=3.5\,\text{cm}\) is impossible because it is shorter than the minimum distance. 4. \(PZ=5\,\text{cm}\) is possible because a nonperpendicular segment to the line can be longer than the minimum distance.

\(X\): possible \(Y\): not possible \(Z\): possible

- A perpendicular segment between parallel lines represents their distance. - Compare a perpendicular connection with a slanted connection.

1. A perpendicular segment between parallel lines has length equal to the distance between the lines. Therefore, \(AC=3\,\text{cm}\). 2. A nonperpendicular segment connecting the same parallel lines is longer than the perpendicular distance. Therefore, \(BD>AC\), so \(BD>3\,\text{cm}\).

a) \(AC=3\,\text{cm}\) b) \(BD>AC\), so \(BD>3\,\text{cm}\).

- Begin with the side shared by the two given angles. - After drawing that side, determine how each angle and distance locates one new vertex. - Compare a construction above the base with one below it. - A diagonal can help you think of the quadrilateral as two triangles.

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. Draw \(\overline{AD}\). The four vertices are now 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 base \(\overline{BC}\), the two angles, and the two adjacent side lengths fix points \(A\) and \(D\). The only other placement is a reflected, congruent copy.

- 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 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 \(20\,\text{cm}\).

- 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.

- Parallel lines do not intersect. - Picture several lines meeting like spokes at one hub. - Count how many old lines the new line can intersect.

1. Yes. Two lines can be parallel, while the third line intersects both of them. This creates exactly two intersection points. 2. For four distinct lines to have exactly one intersection point, all four must pass through the same point. 3. The first three lines have \(3\) intersection points. The fourth line does not intersect the line parallel to it, but it can intersect each of the other two lines at a new point. 4. The greatest possible total is \(3 + 2 = 5\) intersection points.

a) Yes. Two lines can be parallel, and the third can intersect both. b) All four lines must pass through the same point. c) At most \(5\) intersection points.

- Consider whether the two points must lie on the same side of \(g\). - Place equally distant points on opposite sides and consider the line through them.

1. The points at a fixed distance from \(g\) lie on two lines parallel to \(g\), one on each side. 2. If \(A\) and \(B\) lie on the same one of those lines, then \(\overleftrightarrow{AB}\) is parallel to \(g\). 3. However, place \(A\) and \(B\) on opposite sides of \(g\) at equal perpendicular distances. Then \(\overleftrightarrow{AB}\) crosses \(g\), so it is not parallel to \(g\). 4. Therefore, the statement is false.

False. Points \(A\) and \(B\) can lie on opposite sides of \(g\) at equal distances from it. Then \(\overleftrightarrow{AB}\) intersects \(g\) instead of being parallel to it.

- Find the fourth angle from the other three. - Decide whether all five listed measurements are independent. - Imagine the base fixed while the opposite side slides parallel to itself. - Congruent figures must have 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 \((8, 8)\).