Free Math Worksheets

- Identify where the given side lies relative to the two given angles. - Match the corresponding parts of the two triangles. - Which criterion uses two angles and the included side?

1. In \(\triangle ABC\), side \(c\) lies between angles \(\alpha\) and \(\beta\). 2. In \(\triangle DEF\), side \(f\) lies between angles \(\delta\) and \(\epsilon\). 3. The corresponding information matches: \(c = f = 5.4\,\text{cm}\), \(\alpha = \delta = 42^\circ\), and \(\beta = \epsilon = 76^\circ\). 4. Therefore, the triangles are congruent by ASA.

Yes. The triangles are congruent by ASA.

- What does the perimeter formula for a rhombus tell you? - Can a four-bar model change shape without changing side lengths? - Which angle or diagonal measurement would prevent that flexing?

1. The perimeter of a rhombus is \(P = 4s\), where \(s\) is the side length. Equal perimeters therefore guarantee only equal side lengths. 2. A rhombus can flex while keeping all four side lengths fixed, so its interior angles and diagonal lengths can change. 3. For example, a square with side length \(5\,\text{cm}\) and a non-square rhombus with side length \(5\,\text{cm}\) have the same perimeter but are not congruent. 4. One additional interior angle or one diagonal length, together with the side length, determines the rhombus up to congruence. Therefore, Sophie is correct.

Sophie is correct. Equal perimeters determine equal side lengths but not equal angles or diagonal lengths. An interior angle or a diagonal length is also needed to determine the rhombus’s shape.

- Compare the vertex names of the two triangles. - Match the equal legs and equal diagonals. - Which criterion uses three pairs of congruent sides?

1. Both triangles contain side \(\overline{CD}\). 2. In \(\triangle ADC\), the three sides are \(AD = 5\,\text{cm}\), \(AC = 7\,\text{cm}\), and \(CD\). 3. In \(\triangle BCD\), the three sides are \(BC = 5\,\text{cm}\), \(BD = 7\,\text{cm}\), and \(CD\). 4. Since \(AD = BC\), \(AC = BD\), and \(CD\) is shared, the triangles are congruent by SSS.

a) The shared side is \(\overline{CD}\). b) The triangles are congruent by SSS because \(AD = BC\), \(AC = BD\), and \(CD = CD\).

- Subtract the two known side lengths from the perimeter. - Which congruence criterion uses all three side lengths? - Check that the three lengths can form a triangle.

1. Find the third side length: \(20\,\text{cm} - 6\,\text{cm} - 9\,\text{cm} = 5\,\text{cm}\). 2. Both triangles therefore have side lengths \(5\,\text{cm}\), \(6\,\text{cm}\), and \(9\,\text{cm}\). 3. Since all three pairs of corresponding side lengths are equal, the triangles are congruent by SSS.

Yes. Both triangles have side lengths \(5\,\text{cm}\), \(6\,\text{cm}\), and \(9\,\text{cm}\), so they are congruent by SSS.

- Begin with the given side. - Construct each given angle at the correct endpoint of that side. - Which congruence criterion uses two angles and their included side?

1. Draw \(\overline{AB}\) with \(AB = 7.5\,\text{cm}\). 2. At \(A\), draw a ray that forms a \(40^\circ\) angle with \(\overline{AB}\). 3. At \(B\), draw a ray that forms a \(65^\circ\) angle with \(\overline{BA}\) on the same side of \(\overline{AB}\). 4. Label the intersection of the two rays \(C\). This completes \(\triangle ABC\). 5. The side \(\overline{AB}\) and the two angles at its endpoints are fixed, so the triangle is uniquely determined by ASA.

a) Draw \(AB = 7.5\,\text{cm}\), construct a \(40^\circ\) ray at \(A\) and a \(65^\circ\) ray at \(B\), and label their intersection \(C\). b) Every correct construction is congruent by ASA.

- List the three sides of each triangle. - Identify the side that belongs to both triangles. - Which congruence criterion uses three pairs of sides?

1. The given opposite-side relationships are \(AB = CD\) and \(BC = DA\). 2. The triangles share side \(\overline{AC}\), so \(AC = CA\). 3. Thus, all three pairs of corresponding sides are congruent. Therefore, \(\triangle ABC \cong \triangle CDA\) by SSS.

\(\triangle ABC \cong \triangle CDA\) by SSS because \(AB = CD\), \(BC = DA\), and \(AC = CA\).

- Identify the marked side and angles in each triangle. - Check whether the marked angles are at the endpoints of the marked side. - Rotating or reflecting a triangle does not change congruence. - Which criterion uses two angles and their included side?

1. Triangle 1 has a \(5.0\,\text{cm}\) side with adjacent angles of \(40^\circ\) and \(30^\circ\). 2. Triangle 2 has a \(5.0\,\text{cm}\) side with adjacent angles of \(30^\circ\) and \(40^\circ\). 3. Triangle 3 has sides of \(5.0\,\text{cm}\) and \(3.5\,\text{cm}\) with an included angle of \(40^\circ\). These are not the same given measurements as in the other triangles. 4. Triangle 4 has a \(5.0\,\text{cm}\) side with adjacent angles of \(40^\circ\) and \(30^\circ\). 5. Therefore, triangles 1, 2, and 4 are congruent by ASA. Triangle 3 is not congruent to them.

Triangles 1, 2, and 4 are congruent by ASA.

- Compare all the given measurements. - Which congruence theorem applies when all three corresponding sides are equal? - For the angle-side-angle data, check that the angle measures match exactly.

1. Triangles (1) and (3) each have side lengths \(3\,\text{cm}\), \(5\,\text{cm}\), and \(6\,\text{cm}\). 2. Therefore, triangles (1) and (3) are congruent by SSS. 3. Triangles (2) and (4) each have a side of \(4.5\,\text{cm}\) and one angle of \(35^{\circ}\), but their other given angles are different: \(105^{\circ}\) and \(100^{\circ}\). They are not congruent. 4. Neither triangle (2) nor triangle (4) can be congruent to triangle (1) or (3), because the given side length \(4.5\,\text{cm}\) does not match any side length in those triangles.

Only triangles (1) and (3) are congruent, by SSS. No other pair is congruent.

- Identify the side shared by both triangles. - List the other marked equal side and angle. - Is the equal angle included between the two equal sides?

1. The markings give \(KN = KL\). 2. The triangles share side \(\overline{KM}\), so \(KM = KM\). 3. The marked angles satisfy \(\angle NKM = \angle LKM\), and each is included between the corresponding equal sides. 4. Therefore, \(\triangle KNM \cong \triangle KLM\) by SAS.

The triangles are congruent by SAS.

- Use the equal legs of the isosceles triangle. - State what an angle bisector does. - Identify the shared side.

1. The given equal sides are \(PQ = PR\). 2. Because \(\overline{PS}\) bisects \(\angle QPR\), \(\angle QPS = \angle RPS\). 3. The triangles share side \(\overline{PS}\), so \(PS = PS\). 4. Therefore, \(\triangle PQS \cong \triangle PRS\) by SAS.

\(\triangle PQS \cong \triangle PRS\) by SAS.

- Recall the side properties of a rectangle. - Compare the included angles at \(B\) and \(D\). - Look for a second valid criterion using the diagonal.

1. Opposite sides of a rectangle are congruent, so \(AB = CD\) and \(BC = DA\). 2. Angles \(\angle ABC\) and \(\angle CDA\) are both right angles. 3. Therefore, \(\triangle ABC \cong \triangle CDA\) by SAS. 4. Alternatively, the triangles also share \(\overline{AC}\), so SSS can be used.

The triangles are congruent by SAS. SSS is also valid.

- State what midpoint means. - Identify the shared side. - Check whether the equal angles are included between the equal sides.

1. Since \(M\) is the midpoint of \(\overline{AB}\), \(AM = BM\). 2. The triangles share side \(\overline{MC}\), so \(MC = MC\). 3. The included angles satisfy \(\angle AMC = \angle BMC = 90^\circ\). 4. Therefore, \(\triangle AMC \cong \triangle BMC\) by SAS.

The triangles are congruent by SAS.

- Let \(M\) be the midpoint of \(\overline{AB}\). What equal lengths and right angles follow from the perpendicular bisector? - Compare triangles \(AMC\) and \(BMC\).

1. Let \(M\) be the midpoint of \(\overline{AB}\), where the perpendicular bisector meets \(\overline{AB}\). Then \(AM=BM\), and \(\angle AMC\) and \(\angle BMC\) are right angles. 2. Segment \(CM\) is shared by triangles \(AMC\) and \(BMC\). 3. Therefore, \(\triangle AMC\cong\triangle BMC\) by SAS. 4. Corresponding sides are congruent, so \(AC=BC\). Thus, \(\triangle ABC\) is isosceles.

\(\triangle AMC\cong\triangle BMC\) by SAS, so \(AC=BC\). Therefore, \(\triangle ABC\) is isosceles.

- Identify the hypotenuse and a pair of congruent legs. - Which theorem proves two right triangles congruent from that information?

1. Both triangles are right triangles: \(\angle ABD = \angle ACD = 90^{\circ}\). 2. They share hypotenuse \(\overline{AD}\). 3. One pair of corresponding legs is congruent: \(\overline{AB} = \overline{CD}\). 4. Therefore, \(\triangle ABD \cong \triangle DCA\) by the Hypotenuse–Leg theorem.

\(\triangle ABD \cong \triangle DCA\) by HL.

- Find the third angle in each triangle. - Identify which angles are adjacent to or opposite the \(6\,\text{cm}\) side. - Which congruence criterion uses two angles and the included side? - Congruent data must appear in corresponding positions.

1. In Lucas’s triangle, side \(c\) lies between the \(50^\circ\) and \(70^\circ\) angles. The third angle is \(180^\circ - 50^\circ - 70^\circ = 60^\circ\). 2. Sophie’s triangle has the same included side length and the same two adjacent angle measures. Therefore, \(\triangle ABC \cong \triangle DEF\) by ASA. 3. In Jonah’s triangle, the third angle is also \(60^\circ\), but side \(g = 6\,\text{cm}\) is opposite the \(50^\circ\) angle. 4. In Lucas’s and Sophie’s triangles, the \(6\,\text{cm}\) side is opposite the \(60^\circ\) angle. Since the equal-length side is not in the corresponding position, Jonah’s triangle is not congruent to the other two.

Lucas’s and Sophie’s triangles are congruent by ASA. Jonah’s triangle is not congruent to either one because its \(6\,\text{cm}\) side is opposite a different angle.

- Convert the side lengths to the same unit. - Use the triangle angle sum to find each missing angle. - Identify the two angles adjacent to the given side in each triangle.

1. Convert the side length: \(4.2\,\text{cm} = 42\,\text{mm}\). 2. In \(\triangle ABC\), \(\alpha = 180^\circ - 100^\circ - 35^\circ = 45^\circ\). 3. In \(\triangle GHI\), \(\angle I = 180^\circ - 100^\circ - 45^\circ = 35^\circ\). 4. Side \(a = 42\,\text{mm}\) lies between the \(100^\circ\) and \(35^\circ\) angles. Side \(g = 42\,\text{mm}\) also lies between the \(100^\circ\) and \(35^\circ\) angles. 5. Therefore, the triangles are congruent by ASA.

Yes. After converting units and finding the missing angles, both triangles have a \(42\,\text{mm}\) side included between angles of \(100^\circ\) and \(35^\circ\), so they are congruent by ASA.

- Apply the triangle inequality to both triangles formed by the diagonal. - Equality in the triangle inequality gives collinear points, not a triangle. - For three given side lengths, use intersections of circles. - To make a convex quadrilateral, place the two remaining vertices on opposite sides of the diagonal.

1. For \(\triangle ABC\), the side lengths would be \(7\,\text{cm}\), \(5\,\text{cm}\), and \(10\,\text{cm}\). Since \(7 + 5 > 10\), this triangle can be formed. 2. For \(\triangle ADC\), the side lengths would be \(4\,\text{cm}\), \(6\,\text{cm}\), and \(10\,\text{cm}\). Since \(4 + 6 = 10\), the points would be collinear and no nondegenerate triangle would form. Therefore, a \(10\,\text{cm}\) diagonal is not possible for the quadrilateral. 3. For \(AC = 8\,\text{cm}\), draw \(\overline{AC}\). Locate \(B\) at an intersection of a circle centered at \(A\) with radius \(7\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). 4. On the opposite side of \(\overline{AC}\), locate \(D\) at an intersection of a circle centered at \(A\) with radius \(6\,\text{cm}\) and a circle centered at \(C\) with radius \(4\,\text{cm}\). Connect the vertices in order. 5. Each component triangle is determined by three side lengths, so SSS applies.

a) No. The lengths \(4\,\text{cm}\), \(6\,\text{cm}\), and \(10\,\text{cm}\) form a degenerate triangle because \(4 + 6 = 10\). b) Construct \(\triangle ABC\) and \(\triangle ADC\) on opposite sides of the \(8\,\text{cm}\) diagonal using intersecting circles. Each triangle is determined by SSS.

- First construct the triangle whose three side lengths are known. - How many intersections can two circles have when their radii and center distance satisfy the strict triangle inequalities? - Compare placing \(D\) on the same side of \(AC\) as \(B\) with placing it on the opposite side.

1. Construct \(\triangle ABC\) from side lengths \(6\,\text{cm}\), \(5\,\text{cm}\), and \(7\,\text{cm}\) by SSS. 2. To locate \(D\), draw a circle centered at \(A\) with radius \(3\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). 3. Since \(|5 - 3| < 7 < 5 + 3\), the circles intersect at two points on opposite sides of \(\overline{AC}\). 4. Choosing \(D\) on the side of \(AC\) opposite \(B\) produces a convex quadrilateral. Choosing \(D\) on the same side as \(B\) produces a concave quadrilateral because the two component triangles overlap.

There are two possible locations for \(D\), one on each side of \(AC\). One choice produces a convex quadrilateral, and the other produces a concave quadrilateral.

- Which information is already implied by the two side lengths? - When do two sides and an included angle determine a triangle? - Can the two sides and the diagonal form a triangle, and which criterion applies?

1. For a), the side lengths already determine the perimeter: \(P = 2(a+b) = 2(5+3) = 16\,\text{cm}\). This condition gives no angle information, so it is not sufficient. 2. For b), the two adjacent sides and their included angle determine \(\triangle ABD\) by SAS. The fourth vertex of the parallelogram is then fixed, so the condition is sufficient. 3. For c), \(\triangle ABD\) has side lengths \(5\,\text{cm}\), \(3\,\text{cm}\), and \(7\,\text{cm}\). Since \(5 + 3 > 7\), the triangle exists and is determined by SSS. The parallelogram is then determined up to reflection, so the condition is sufficient.

a) Not sufficient; the perimeter is already determined by the side lengths. b) Sufficient; the parallelogram is determined using SAS. c) Sufficient; the diagonal and two adjacent sides determine a triangle by SSS.

- What is true about opposite sides and opposite angles of a parallelogram? - How are adjacent angle measures related? - Identify the included angle between each pair of corresponding sides.

1. Opposite angles of a parallelogram are congruent, so \(\angle C = \angle A = 60^\circ\). 2. Adjacent angles of a parallelogram are supplementary, so \(\angle B = \angle D = 180^\circ - 60^\circ = 120^\circ\). 3. Opposite sides of a parallelogram are congruent: \(AB = CD = 5\,\text{cm}\) and \(BC = DA = 3\,\text{cm}\). 4. The equal \(120^\circ\) angles are included between these corresponding side pairs. Therefore, \(\triangle ABC \cong \triangle CDA\) by SAS.

a) \(\angle C = 60^\circ\) b) \(AB = CD\), \(BC = DA\), and \(\angle B = \angle D = 120^\circ\), so the triangles are congruent by SAS.

- Two equal angle measures make triangles similar, but what fixes their scale? - Is a side between the two angles the same corresponding side as a side opposite one of them? - Recall what the order of the letters in ASA and AAS describes.

1. Two angle measures determine the third angle, so any triangles with those angle measures are similar. 2. A side length fixes the scale only when the side has the same corresponding position in both triangles. 3. For example, in one triangle the given side could lie between the two stated angles, while in the other triangle the equal-length side could be opposite one of those angles. 4. Those equal-length sides would not be corresponding sides, so the triangles can have different sizes. Lucas's claim is false as stated. 5. The triangles are congruent by ASA or AAS only when the equal side is in the same corresponding position relative to the equal angles.

No. Two angles and one side determine a unique triangle only when the side corresponds to the same side in both triangles. With the proper correspondence, ASA or AAS proves congruence.

- Use the triangle angle-sum theorem. - Must the \(6\,\text{cm}\) side lie between the two stated angles? - List the angle that could be opposite the given side in each case.

1. The third angle is \(180^\circ - 45^\circ - 75^\circ = 60^\circ\). 2. The \(6\,\text{cm}\) side can be opposite the \(45^\circ\), \(60^\circ\), or \(75^\circ\) angle. 3. In each case, the two angles at the endpoints of the \(6\,\text{cm}\) side are fixed, so ASA determines one triangle. 4. The three triangles are not congruent because the same \(6\,\text{cm}\) side is opposite a different angle in each case. Therefore, Lucas is correct.

Lucas is correct. The third angle is \(60^\circ\), and the \(6\,\text{cm}\) side can be opposite the \(45^\circ\), \(60^\circ\), or \(75^\circ\) angle. These three placements produce three noncongruent triangles.

- Check the triangle inequality when three side lengths are given. - Ask whether angle measures alone determine a triangle's size. - For c), determine whether the given angle is included between the two given sides.

1. For a), the triangle inequality fails because \(5 + 3 = 8 < 9\). No triangle exists. 2. For b), three angles determine the shape but not the size. Infinitely many similar, noncongruent triangles have these angle measures. 3. For c), sides \(a\) and \(c\) meet at angle \(\beta\). Two sides and their included angle determine exactly one triangle by SAS.

a) No triangle exists because \(5 + 3 < 9\). b) The triangle is not uniquely determined; AAA determines similarity, not congruence. c) Exactly one triangle is determined by SAS.

- Identify the side lengths shared by both triangles. - Identify the angle at \(A\). - Decide whether the angle is included between the two known sides. - Recall why SSA is ambiguous.

1. Both triangles have \(AB = 8\,\text{mi}\), \(BC_1 = BC_2 = 5\,\text{mi}\), and the same \(30^{\circ}\) angle at \(A\). 2. These data give two sides and a nonincluded angle. This is the ambiguous SSA case, which can produce two different triangles. 3. Therefore, \(ABC_1\) and \(ABC_2\) are not congruent.

a) \(AB = 8\,\text{mi}\), \(BC_1 = BC_2 = 5\,\text{mi}\), and \(\angle A = 30^{\circ}\). b) The data form the ambiguous SSA case, which does not determine a unique triangle.

- Divide the quadrilateral into two triangles. - Identify three measurements that determine the first triangle. - Decide which two additional measurements locate the fourth vertex. - Use one of the standard triangle congruence criteria.

1. One valid choice is \(a\), \(b\), \(c\), \(d\), and \(e\). 2. Construct \(\triangle ABC\) from side lengths \(a\), \(b\), and \(e\). It is unique up to reflection by SSS. 3. On the side of \(\overline{AC}\) opposite \(B\), construct \(\triangle ACD\) from side lengths \(c\), \(d\), and \(e\). It is also unique by SSS. 4. Therefore, the convex quadrilateral \(ABCD\) is unique up to reflection.

a) One valid choice is \(a\), \(b\), \(c\), \(d\), and \(e\). b) Construct \(\triangle ABC\) from \(a\), \(b\), and \(e\), then construct \(\triangle ACD\) on the opposite side of \(\overline{AC}\) from \(c\), \(d\), and \(e\). c) Both triangles are determined by SSS, so the convex quadrilateral is unique up to reflection.

- Compare triangles \(ADB\) and \(CDB\). - Identify two pairs of congruent sides and the included angles. - Use the converse of the Perpendicular Bisector Theorem after proving \(AB=CB\).

1. In triangles \(ADB\) and \(CDB\), \(AD=CD\) is given, \(DB=DB\) by the reflexive property, and \(\angle ADB=\angle BDC\) because \(\overline{BD}\) bisects \(\angle ADC\). 2. Therefore, \(\triangle ADB\cong\triangle CDB\) by SAS. 3. Corresponding sides are congruent, so \(AB=CB\). 4. Since \(B\) is equidistant from the endpoints of \(\overline{AC}\), the converse of the Perpendicular Bisector Theorem shows that \(B\) lies on the perpendicular bisector of \(\overline{AC}\).

\(\triangle ADB\cong\triangle CDB\) by SAS, so \(AB=CB\). Therefore, \(B\) lies on the perpendicular bisector of \(\overline{AC}\).

- Compare triangles \(AED\) and \(CED\). - Use the midpoint, the given congruent sides, and the shared segment to prove the triangles congruent. - Then connect line \(ED\) to the Perpendicular Bisector Theorem.

1. Since \(D\) is the midpoint of \(\overline{AC}\), \(AD=DC\). Also, \(AE=CE\) is given and \(ED=ED\) by the reflexive property. 2. Therefore, \(\triangle AED\cong\triangle CED\) by SSS. 3. Corresponding angles \(\angle ADE\) and \(\angle EDC\) are congruent. They form a linear pair, so each measures \(90^\circ\). Thus, \(\overline{ED}\) is perpendicular to \(\overline{AC}\) at its midpoint and lies on the perpendicular bisector of \(\overline{AC}\). 4. Because \(B\) lies on line \(ED\), the Perpendicular Bisector Theorem gives \(AB=CB\). Therefore, \(\triangle ABC\) is isosceles.

\(\triangle AED\cong\triangle CED\) by SSS, so line \(ED\) is the perpendicular bisector of \(\overline{AC}\). Since \(B\) lies on that line, \(AB=CB\), and \(\triangle ABC\) is isosceles.

- List the three side lengths of each triangle. - Include the shared diagonal. - Use the order of the congruence statement to match corresponding vertices.

1. The triangles have \(AB = CD = 5\,\text{cm}\) and \(BC = DA = 3\,\text{cm}\). 2. They share diagonal \(\overline{AC}\), so \(AC = CA = 6\,\text{cm}\). 3. Therefore, \(\triangle ABC \cong \triangle CDA\) by SSS. 4. The vertex correspondence is \(A \leftrightarrow C\), \(B \leftrightarrow D\), and \(C \leftrightarrow A\). 5. Thus, \(\angle BAC\) corresponds to \(\angle DCA\), and \(\angle BCA\) corresponds to \(\angle DAC\).

a) The triangles are congruent by SSS. b) \(\angle BAC\) corresponds to \(\angle DCA\), and \(\angle BCA\) corresponds to \(\angle DAC\).

- Think of a four-sided figure that can flex while its side lengths remain fixed. - How do a square and a non-square rhombus differ? - Divide the quadrilateral with a diagonal and count the information needed for each triangle.

1. A square with four side lengths of \(5\,\text{cm}\) has four right angles. A non-square rhombus can also have four side lengths of \(5\,\text{cm}\), but it has acute and obtuse angles. 2. The square and rhombus have the same four side lengths but are not congruent. Therefore, four side lengths do not determine a quadrilateral. 3. In general, five suitable independent measurements are needed. 4. Draw a diagonal to divide the quadrilateral into two triangles. Three suitable measurements determine the first triangle and therefore determine the diagonal. The second triangle shares that diagonal and needs two additional suitable measurements. Thus, the total is \(3 + 2 = 5\).

a) SSSS is not a quadrilateral congruence criterion because a \(5\,\text{cm}\) square and a non-square \(5\,\text{cm}\) rhombus have the same side lengths but different angles. b) An arbitrary quadrilateral generally requires five suitable independent measurements, found by determining two triangles that share a diagonal.

- First construct the triangle containing the given included angle. - How is the remaining vertex located using its distances from \(B\) and \(D\)? - What do zero, one, or two circle intersections mean geometrically? - Does a tangent intersection form a nondegenerate quadrilateral?

1. Sides \(a\) and \(d\) with included angle \(\alpha\) determine \(\triangle ABD\) by SAS. Therefore, vertices \(A\), \(B\), and \(D\), as well as diagonal \(BD\), are fixed up to congruence. 2. Point \(C\) must lie on a circle centered at \(B\) with radius \(b\) and on a circle centered at \(D\) with radius \(c\). 3. The circles may have no intersection, so no quadrilateral exists. If they have two intersections, they generally produce two different placements of \(C\), such as a convex placement and a concave or self-intersecting placement. 4. If the circles are tangent, their single intersection lies on line \(BD\), producing a degenerate figure rather than a proper quadrilateral. 5. Therefore, the five measurements do not always determine exactly one nondegenerate quadrilateral.

Lucas is incorrect. The first component triangle is fixed by SAS, but the two circles used to locate \(C\) may have no intersection or two intersections. A single tangent intersection is degenerate, so the data do not always determine exactly one proper quadrilateral.

- Build the quadrilateral one vertex at a time. - How does an angle determine a ray, and how does a length determine a point on that ray? - Why is the convex-orientation condition necessary?

1. Draw \(\overline{AB}\) with the given length. 2. At \(B\), construct the given angle \(\angle B\). Mark point \(C\) on the appropriate ray so that \(BC\) has the given length. 3. At \(C\), construct the given angle \(\angle C\), choosing the ray that places \(D\) in the specified convex orientation relative to \(A\) and \(\overline{BC}\). 4. Mark point \(D\) on that ray so that \(CD\) has the given length, then connect \(D\) to \(A\). 5. At each stage, a ray gives a unique direction and the specified side length gives a unique endpoint on that ray. With the convex orientation fixed, there is no second circle-intersection choice, so the quadrilateral is unique up to congruence.

a) Construct the chain \(A\)-\(B\)-\(C\)-\(D\) by alternating the given side lengths and included angles, then connect \(D\) to \(A\). b) The specified orientation fixes each angle ray, and each side length fixes one point on that ray. Therefore, the convex quadrilateral is uniquely determined up to congruence.

- Begin with the longer base. - Use the given angle and leg length to locate \(D\). - How does \(CD \parallel AB\) locate \(C\)? - In the parallelogram decomposition, find \(AE\) from the two base lengths.

1. Draw \(\overline{AB}\) with length \(7\,\text{cm}\). At \(A\), construct a \(70^\circ\) angle and mark \(D\) on its ray so that \(AD = 4\,\text{cm}\). 2. Through \(D\), draw a line parallel to \(AB\). On that line, in the same direction as \(A\) to \(B\), mark \(C\) so that \(DC = 3\,\text{cm}\). Connect \(B\) to \(C\). 3. If the line through \(D\) parallel to \(BC\) meets \(AB\) at \(E\), then \(EBCD\) is a parallelogram. Thus, \(EB = CD = 3\,\text{cm}\). 4. Therefore, \(AE = AB - EB = 7\,\text{cm} - 3\,\text{cm} = 4\,\text{cm}\). 5. In \(\triangle AED\), \(AE = 4\,\text{cm}\), \(AD = 4\,\text{cm}\), and the included angle is \(70^\circ\). The triangle is uniquely determined by SAS.

a) Draw \(AB\), construct the \(70^\circ\) angle at \(A\), mark \(AD = 4\,\text{cm}\), draw the parallel through \(D\), mark \(DC = 3\,\text{cm}\), and connect \(B\) to \(C\). b) SAS. The component triangle has \(AE = AD = 4\,\text{cm}\) with included angle \(70^\circ\).

- In a rectangle, what type of triangle is formed by two adjacent sides and a diagonal? - Which side must be longest in a right triangle? - How do the diagonals of a parallelogram intersect?

1. If \(ABCD\) were a rectangle, \(\angle ABC\) would be a right angle, so \(AC\) would be the hypotenuse of right triangle \(ABC\). 2. The hypotenuse must be the longest side of a right triangle. Here, \(AB = 6\,\text{cm}\) is longer than \(AC = 5\,\text{cm}\), so \(AC\) cannot be the hypotenuse. Therefore, the quadrilateral cannot be a rectangle. 3. If the figure is a parallelogram, triangle \(ABC\) is determined by side lengths \(6\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\) using SSS. 4. The diagonals of a parallelogram bisect each other, so \(D\) is the image of \(B\) under a \(180^\circ\) rotation about the midpoint of \(AC\). Thus, the parallelogram is uniquely determined up to congruence.

a) It cannot be a rectangle because \(AC\) would have to be the hypotenuse of \(\triangle ABC\), but \(AB = 6\,\text{cm}\) is longer than \(AC = 5\,\text{cm}\). b) Yes. Triangle \(ABC\) is determined by SSS, and the parallelogram condition fixes \(D\) as the half-turn image of \(B\) about the midpoint of \(AC\).

- Check whether the angle is included between the two given sides. - Compare each data set with SSS, SAS, ASA, and AAS. - Remember that SSA is not a general triangle congruence criterion.

1. In part a, one side and its two adjacent angles are given. The triangle is uniquely determined by ASA. 2. In part b, two sides and a nonincluded angle are given. This is the ambiguous SSA case, so the data can produce two noncongruent triangles and do not determine a unique triangle. 3. In part c, two sides and the included angle are given. The triangle is uniquely determined by SAS. 4. In part d, all three side lengths are given. The triangle is uniquely determined by SSS.

a) Yes, by ASA. b) No. SSA is ambiguous here and does not determine a unique triangle. c) Yes, by SAS. d) Yes, by SSS.

- Use the side and angle properties of an isosceles triangle. - Identify the shared side. - Decide whether the known angle is included between the known sides. - Use the fact that \(S\) is not the midpoint.

1. Since \(PQR\) is isosceles, \(PQ = PR\) and the base angles satisfy \(\angle PQS = \angle PRS\). The triangles also share side \(PS\). 2. The known angle is not included between the two known sides. The information is SSA, which is not a general congruence criterion. 3. Because \(S\) is not the midpoint of \(QR\), \(QS \ne RS\). These corresponding sides are unequal, confirming that the triangles are not congruent.

a) \(PQ = PR\), \(PS = PS\), and \(\angle PQS = \angle PRS\). b) The information is SSA, not a valid general congruence criterion. Also, \(QS \ne RS\), so the triangles are not congruent.