Free math worksheets

- What must the interior angles of a triangle total? - Add the three given measures and compare the result with that total.

1. The interior angles of any triangle must sum to \(180^\circ\). 2. The given measures sum to \(110^\circ + 45^\circ + 35^\circ = 190^\circ\). 3. Since \(190^\circ \ne 180^\circ\), no triangle can have all three given angle measures.

No. The given 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.

- Identify which vertices are connected by sides \(a\) and \(c\). - Locate angle \(\beta\) before deciding the order of the construction. - Ask whether the given angle lies between the two 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 \(BC=4.5\,\text{cm}\), then connect \(A\) to \(C\). 5. The triangle is uniquely determined by SAS.

1. The planning sketch must show \(AB=6\,\text{cm}\), \(BC=4.5\,\text{cm}\), 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.

- 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 \(5\,\text{cm}\).

- 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.

- Think about a compass-and-straightedge construction of a perpendicular through a point on a line. - The two given legs meet at the right-angle vertex. - Use the fixed leg lengths only after the perpendicular direction is established.

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. Connect \(A\) to \(B\).

Construct an exact perpendicular to \(AC\) through \(C\), mark \(B\) on that perpendicular so that \(BC=3\,\text{cm}\), and connect \(A\) to \(B\). With \(AC=5\,\text{cm}\), this gives the required right triangle without a protractor.

- Plot the points before deciding the direction of each ray. - Compare the horizontal and vertical coordinate changes from \(S\). - Identify horizontal, vertical, and equal-change diagonal directions.

1. Ray \(\overrightarrow{SA}\) is horizontal and points to the right. 2. Ray \(\overrightarrow{SB}\) is vertical and points upward, so it is perpendicular to \(\overrightarrow{SA}\). The angle is \(90^\circ\). 3. From \(S\) to \(C\), the horizontal and vertical changes are both \(4\), so \(\overrightarrow{SC}\) follows a \(45^\circ\) diagonal. The smaller angle is \(45^\circ\). 4. Ray \(\overrightarrow{SD}\) is vertical and points downward, so it is perpendicular to \(\overrightarrow{SA}\). The smaller angle is \(90^\circ\).

a) \(90^\circ\) b) \(45^\circ\) c) \(90^\circ\)

- Why are four fixed side lengths enough for two triangles but not automatically for one quadrilateral? - Think of the sides as rigid bars joined by hinges. - What added segment would divide the quadrilateral into two SSS triangles?

1. The same four side lengths can occur with different angle measures, so the quadrilaterals can 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 stay fixed. 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 then fixed by SSS. 5. For a convex quadrilateral, the two triangles lie on opposite sides of \(\overline{AC}\), so the quadrilateral is unique up to reflection.

Four side lengths alone do not determine a quadrilateral because its 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.

- What does each circle tell you about the distance from the third vertex to one endpoint of the base? - Compare the three side lengths of the triangle above the base with the triangle below it. - Decide whether reflection changes congruence.

1. The two circles intersect at two points, one on each side of \(\overline{AB}\). 2. Each intersection point is \(4\,\text{cm}\) from \(A\) and \(5\,\text{cm}\) from \(B\), while \(AB=6\,\text{cm}\). 3. The two resulting triangles have the same three side lengths, so they are congruent by SSS. 4. They are reflections across \(\overline{AB}\), so there is only one noncongruent triangle.

One noncongruent triangle. The two possible third-vertex locations produce reflected congruent copies by SSS.

- Compare the longest side with the sum of the other two. - Also compare the third side with the difference of the two known sides. - For the construction plan, think about how circles 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 \(c>2\), so \(2\,\text{cm}<c<8\,\text{cm}\). 2. Draw \(\overline{AB}\) with length \(7\,\text{cm}\). Draw a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(B\) with radius \(3\,\text{cm}\). 3. Label either intersection point \(C\), then connect \(A\) and \(B\) to \(C\). The reflected constructions are congruent by SSS.

1. \(2\,\text{cm}<c<8\,\text{cm}\) 2. Draw a \(7\,\text{cm}\) base and locate the third vertex at an intersection of circles of radii \(5\,\text{cm}\) and \(3\,\text{cm}\) centered at the base endpoints; connect that point to both endpoints.

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

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; their rays meet at the third vertex. 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. - For the possible case, think about how two circles can locate a point at two fixed distances.

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 \(4\,\text{cm}\). 4. Their intersection can be used as vertex \(C\); connect \(C\) to \(A\) and \(B\). This is an SSS construction.

a) No, because \(2.5+3<6\). b) Yes. Use 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 the base endpoints.

- 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.

- 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 \(60^\circ\) 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 \(C\).

- 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 \((180^\circ-70^\circ)\div2=55^\circ\). 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 \(C\).

- 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 \(90^\circ\).

- First use the relationship between \(\alpha\) and \(\gamma\). - Use the triangle angle sum to find the remaining angle. - Which side is between the two angles that can be constructed at the endpoints?

1. \(\gamma=2\cdot42^\circ=84^\circ\). 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. The side \(c=AB\) lies between the two constructed angles, so the triangle is uniquely determined by ASA.

1. \(\beta=54^\circ\) 2. Draw \(AB=7.2\,\text{cm}\), construct a \(42^\circ\) ray at \(A\) and a \(54^\circ\) ray at \(B\), and use their intersection as \(C\). 3. ASA

- Identify the two given sides and the vertex of the given angle. - Decide whether the given angle lies between the two given sides. - Use that relationship to identify the congruence criterion.

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. Connect \(B\) to \(C\). 5. The two given sides and their included angle determine the triangle uniquely by SAS.

Draw \(AB=4.0\,\text{cm}\), construct a \(105^\circ\) ray at \(A\), mark \(C\) on the ray so that \(AC=5.5\,\text{cm}\), and connect \(B\) to \(C\). 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 \(x=8\,\text{cm}\). 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 \(B\).

- 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 \(15\,\text{in.}\).

- Plot the points and draw the first ray before adding the second rays. - Opposite rays form a straight angle. - Compare horizontal, vertical, and equal-change diagonal directions.

1. Ray \(\overrightarrow{SA}\) is vertical and points upward. 2. The opposite ray points vertically downward, so the two rays form a straight angle. Thus, \(\beta=180^\circ\). 3. From \(S\) to \(Q\), the change is \(4\) units left and \(4\) units up. This equal-change diagonal makes a \(45^\circ\) angle with the vertical ray, so \(\beta=45^\circ\). 4. Ray \(\overrightarrow{SP}\) points horizontally left. A horizontal ray and a vertical ray are perpendicular, so \(\beta=90^\circ\).

a) \(180^\circ\) b) \(45^\circ\) c) \(90^\circ\)

- Plot each named point and draw the rays before calculating an angle. - Equal horizontal and vertical changes indicate a \(45^\circ\) diagonal direction. - Perpendicular rays form a right angle, and opposite rays form a straight angle.

1. Ray \(\overrightarrow{SA}\) is horizontal and points to the right. 2. From \(S\) to \(P\), the change is \(2\) units left and \(2\) units up. This northwest diagonal has direction \(135^\circ\) from the positive horizontal direction, so \(\gamma=135^\circ\). 3. A downward ray perpendicular to \(\overrightarrow{SA}\) is vertical. Therefore, \(\gamma=90^\circ\). It follows the line \(x=2\) and meets the x-axis at \((2, 0)\). 4. Ray \(\overrightarrow{SQ}\) points horizontally left, exactly opposite \(\overrightarrow{SA}\). Therefore, \(\gamma=180^\circ\).

a) \(135^\circ\) b) \(90^\circ\); the ray meets the x-axis at \((2, 0)\). c) \(180^\circ\)

- 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.

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

- Use the added diagonal to split the quadrilateral into two triangles. - Ask what congruence information is known for each triangle. - The convexity condition determines how the two triangles must be placed relative to the diagonal.

1. Diagonal \(AC\) divides the quadrilateral into \(\triangle ABC\) and \(\triangle ACD\). 2. Triangle \(ABC\) has fixed side lengths \(5\,\text{cm}\), \(4\,\text{cm}\), and \(6\,\text{cm}\), so it is determined by SSS. 3. Triangle \(ACD\) has fixed side lengths \(6\,\text{cm}\), \(6\,\text{cm}\), and \(3\,\text{cm}\), so it is also determined by SSS. 4. For a convex quadrilateral, the two triangles must lie on opposite sides of \(\overline{AC}\). The only alternative is the reflected copy of the whole quadrilateral.

The diagonal splits the quadrilateral into two SSS-determined triangles. Requiring the quadrilateral to be convex fixes them on opposite sides of \(AC\), so the quadrilateral is unique 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.

- Use the triangle inequality: the sum of any two sides must be greater than the third side. - Write all three inequalities. - Solve each inequality for \(x\). - Find the overlap of the solution sets before choosing the least integer.

1. Apply the triangle inequality to each pair of sides. 2. From \(x + 12 > 3x - 2\), obtain \(14 > 2x\), so \(x < 7\). 3. From \(x + (3x - 2) > 12\), obtain \(4x - 2 > 12\), so \(x > \frac{7}{2}\). 4. From \(12 + (3x - 2) > x\), obtain \(10 + 3x > x\), so \(x > -5\). 5. Side lengths must also be positive. Those restrictions are weaker than \(x > \frac{7}{2}\). 6. Combining all conditions gives \(\frac{7}{2} < x < 7\). The least integer in this interval is \(4\).

The range is \(\frac{7}{2} < x < 7\), and the least possible integer value is \(4\).

- Write one triangle inequality for each pair of sides. - Solve all three inequalities and find their common solution. - Check that every side length is positive. - Count the integers strictly between the two bounds.

1. Apply the triangle inequality to each pair of sides. 2. From \(2x + (x + 10) > 30\), obtain \(3x + 10 > 30\), so \(x > \frac{20}{3}\). 3. From \(2x + 30 > x + 10\), obtain \(x > -20\). 4. From \((x + 10) + 30 > 2x\), obtain \(x < 40\). 5. The positivity requirements for the side lengths are weaker than \(x > \frac{20}{3}\). Therefore, \(\frac{20}{3} < x < 40\). 6. The integer values are \(7, 8, \ldots, 39\). Their number is \(39 - 7 + 1 = 33\).

The range is \(\frac{20}{3} < x < 40\). There are \(33\) possible integer values of \(x\).

- 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 \((8, 8)\).