Free math worksheets

- Use the defining property of every point on the circle. - Compare how many directions extend from \(P\) on a line and on a ray. - Use the order \(U\)-\(P\)-\(V\) to relate the lengths.

1. Both \(U\) and \(V\) lie on the circle centered at \(P\) with radius \(AB\), so \(PU=AB\) and \(PV=AB\). 2. A full line extends in both directions from \(P\), so it meets the circle once on each side. A ray extends in only one direction, so it contains only one of those intersections. 3. Since \(P\) lies between \(U\) and \(V\), \(UV=UP+PV=AB+AB=2AB\).

a) Both are valid because \(PU=PV=AB\). b) One valid endpoint. c) \(UV=2AB\).

- The compass opening should stay fixed after it is matched to the source segment. - Each new endpoint becomes the center for the next transfer. - Compare how many copied intervals lie on each side of \(P_2\).

1. Set the compass opening to \(AB\) and do not change it. 2. Center the compass at \(P_0\) and mark \(P_1\) on the ray. Then center it successively at \(P_1\), \(P_2\), and \(P_3\) to mark \(P_2\), \(P_3\), and \(P_4\). 3. The four consecutive copied segments each have length \(2.35\,\text{cm}\), so \(P_0P_4=4\cdot2.35=9.40\,\text{cm}\). 4. Each of \(P_0P_2\) and \(P_2P_4\) consists of two copied segments, so both equal \(4.70\,\text{cm}\). Therefore, \(P_2\) is the midpoint.

Keep the compass set to \(AB\). Center successively at \(P_0,P_1,P_2,P_3\) to mark the next point on the ray each time. Then \(P_0P_4=9.40\,\text{cm}\), and \(P_2\) is the midpoint because \(P_0P_2=P_2P_4=4.70\,\text{cm}\).

- Look at the order of the rays and identify the two adjacent angle regions. - Combine the adjacent parts to find the larger angle. - Use the straight angle formed by the opposite rays for part b).

1. Angles \(\angle XPU\) and \(\angle UPV\) are adjacent and do not overlap, so their measures add. 2. \(m\angle XPV=38^\circ+67^\circ=105^\circ\). 3. Since \(\overrightarrow{PX}\) and \(\overrightarrow{PW}\) are opposite rays, \(m\angle XPW=180^\circ\). 4. Therefore, \(m\angle VPW=180^\circ-105^\circ=75^\circ\).

a) \(105^\circ\). b) \(75^\circ\). c) The two given angles are adjacent, share only ray \(\overrightarrow{PU}\), and do not overlap, so their measures add.

- Make the copied angle and the comparison angle share a vertex and initial ray. - Focus on the order of the two terminal rays. - Consider separately the inside, coincident, and outside cases.

1. The app places a copy of \(\angle A\) with the same vertex and initial ray as \(\angle B\), opening into the same region. 2. If the copied terminal ray lies inside \(\angle B\), then \(m\angle A<m\angle B\). 3. If the copied terminal ray coincides with the terminal ray of \(\angle B\), then \(m\angle A=m\angle B\). 4. If the copied terminal ray lies beyond the terminal ray of \(\angle B\), then \(m\angle A>m\angle B\).

With a common vertex and initial ray, a copied terminal ray inside, coincident with, or beyond the terminal ray of \(\angle B\) shows that \(m\angle A\) is less than, equal to, or greater than \(m\angle B\), respectively.

- Identify which angle is larger. - The smaller angle lies inside the larger angle and shares its initial ray. - Think about which operation gives the part of the larger angle that remains.

1. The larger angle is \(83^\circ\) and the inner angle is \(37^\circ\). 2. The angle between their terminal rays is the difference: \(83^\circ-37^\circ=46^\circ\).

\(46^\circ\)

- Use one common starting point and direction for both transfers. - Compare the positions of the two copied endpoints. - Consider equality and both possible endpoint orders.

1. The app transfers length \(AB\) from \(P\) onto the ray, locating point \(X\). 2. It transfers length \(CD\) from \(P\) onto the same ray, locating point \(Y\). 3. If \(X=Y\), then \(AB=CD\). 4. If \(X\) lies between \(P\) and \(Y\), then \(AB<CD\). 5. If \(Y\) lies between \(P\) and \(X\), then \(AB>CD\).

With both lengths copied from the same point on the same ray, coincident endpoints mean equal lengths, and the endpoint closer to \(P\) represents the shorter segment.

- A rotation fixes its center and preserves distance from that center. - The direction of the angle matters as well as its size. - Once the correct ray is fixed, ask how many points on that ray can be the required distance from \(O\).

1. From ray \(\overrightarrow{OP}\), construct a ray at \(O\) that is \(75^\circ\) counterclockwise from \(\overrightarrow{OP}\). 2. On the new ray, locate \(P'\) so that \(OP'=OP\). 3. These two conditions, \(m\angle POP'=75^\circ\) counterclockwise and \(OP'=OP\), are exactly the defining conditions for the rotation image. 4. A ray contains exactly one point at a specified positive distance from its endpoint, so \(P'\) is unique.

Construct the ray \(75^\circ\) counterclockwise from \(\overrightarrow{OP}\), then place \(P'\) on it so that \(OP'=OP\). These conditions determine one point, so the rotation image is unique.

- The source and target arcs already have the same radius. - Ask what second side-length condition will make the two construction triangles congruent.

1. The first pair of arcs has already transferred the source radius to the target vertex. 2. The remaining information that fixes the angle is the chord between the two source-arc intersection points. 3. Therefore, set the compass opening to \(DE\), then use that opening from \(X\) on the target arc.

Copy the distance \(DE\).

- First create a copy of the longer source segment on the target ray. - For the subtraction step, think about which endpoint should be the center of the second arc. - The second copied segment must fit inside the first copied segment for the requested point placement.

1. Set the compass opening to \(AB\). With center \(P\), mark its intersection \(Q\) with ray \(\overrightarrow{PX}\), so \(PQ=AB\). 2. Reset the compass opening to \(CD\). With center \(Q\), draw an arc that meets \(\overline{PQ}\) at \(R\). 3. Then \(RQ=CD\), so \(PR=PQ-RQ=AB-CD\). 4. Numerically, \(PR=9.2-3.7=5.5\,\text{cm}\). 5. The required intersection lies on \(\overline{PQ}\) when \(CD\le AB\).

First copy \(AB\) from \(P\) to locate \(Q\) on \(\overrightarrow{PX}\). Then set the compass to \(CD\), center it at \(Q\), and use the intersection of that arc with \(\overline{PQ}\) as \(R\). This gives \(PR=AB-CD=5.5\,\text{cm}\). The construction has \(R\in\overline{PQ}\) when \(CD\le AB\).

- A standard angle copy transfers more than just one compass opening. - First match the source and target arc radii; then consider the chord cut off by the source angle. - The two copies are placed on opposite sides of the same initial ray.

1. Draw an arc centered at the source vertex so that it meets both sides of the source angle. Preserve that arc radius. 2. With the same radius, draw target arcs centered at \(P\) on the two sides of \(\overrightarrow{PX}\). 3. Measure with the compass the chord between the two source-arc intersections. Transfer that chord from the point where each target arc meets \(\overrightarrow{PX}\), choosing opposite sides of the ray, and draw \(\overrightarrow{PU}\) and \(\overrightarrow{PV}\) through the transferred points. 4. Each copied angle is \(52^\circ\), so the smaller angle between the terminal rays is \(52^\circ+52^\circ=104^\circ\). 5. Ray \(\overrightarrow{PX}\) divides \(\angle UPV\) into two congruent angles, so it is the angle bisector.

Transfer the source arc radius and then the source chord between the arc intersections to construct the two copies on opposite sides of \(\overrightarrow{PX}\). The smaller angle is \(104^\circ\), and \(\overrightarrow{PX}\) is its angle bisector.

- Every copied segment must start from the specified point \(P\). - The compass must match the appropriate source segment before each transfer. - Decide whether \(P\) lies between the endpoints before choosing addition or subtraction.

1. Set the compass to the first source segment. Center it at \(P\) and mark its intersection with one ray as \(U\). 2. Reset the compass to the second source segment. Center it at \(P\) and mark its intersection with the opposite ray as \(V\). 3. Since \(P\) lies between \(U\) and \(V\), \(UV=PU+PV=4.8+7.1=11.9\,\text{cm}\). 4. For same-ray placement, center both transfers at \(P\) on one ray. The endpoints are then separated by \(7.1-4.8=2.3\,\text{cm}\).

Copy \(4.8\,\text{cm}\) from \(P\) onto one ray, reset the compass to \(7.1\,\text{cm}\), and copy that length from \(P\) onto the opposite ray. Then \(UV=11.9\,\text{cm}\). On the same ray, both copies start at \(P\), and their endpoints are \(2.3\,\text{cm}\) apart.

- Account for all three angles of a triangle. - Identify where the known side lies relative to the two known angles. - Match the information to a triangle-congruence criterion.

1. The third angle measure is \(180^\circ-47^\circ-68^\circ=65^\circ\). 2. The known side lies between the two known angles. 3. Therefore, the two triangles are congruent by ASA.

a) \(65^\circ\). b) The congruent side lies between the two congruent angles, so ASA guarantees congruence.

- Translate each point's membership on a circle into a distance equality. - Track whether the compass opening changes between the two circles. - Compare the three side lengths of the resulting triangle.

1. Point \(Q\) lies on the circle centered at \(P\) with radius \(AB\), so \(PQ=AB\). 2. Point \(R\) lies on the circle centered at \(P\) with the same radius, so \(PR=AB\). 3. Point \(R\) also lies on the circle centered at \(Q\) with radius \(AB\), so \(QR=AB\). 4. Thus \(PQ=PR=QR\), so \(\triangle PQR\) is equilateral.

All three sides are copies of \(\overline{AB}\): \(PQ=PR=QR=AB\). Therefore, \(\triangle PQR\) is equilateral.

- Translate each circle intersection into distances from the two centers. - Compare the complete side-length sets of the two triangles. - For two circles, compare the center distance with the sum and difference of the radii.

1. Since \(R\) and \(S\) lie on both circles, \(PR=QR=PS=QS=6.5\,\text{cm}\). 2. Both triangles also share the side \(PQ=8\,\text{cm}\). 3. Therefore, \(\triangle PRQ\cong\triangle PSQ\) by SSS. 4. The two intersection points lie on opposite sides of line \(PQ\), so the triangles are reflections of each other across line \(PQ\). 5. The center distance is \(8\,\text{cm}\), the radius sum is \(13\,\text{cm}\), and the radius difference is \(0\). Since \(0<8<13\), the circles intersect at two points.

a) The triangles are congruent by SSS. b) They are reflections of each other across line \(PQ\). c) Two intersections occur because \(0<8<13\).

- Compare the radii and chord lengths in the source and target constructions. - Think about the congruence of the triangles formed by the arc centers and intersection points. - Use the stated direction of rotation to select the required side.

1. The target arc has the same radius as the source arc. 2. The transferred chord has the same length as the chord cut off by the source angle. 3. For either intersection point, the two radii and the transferred chord form a triangle congruent to the source construction triangle by SSS. 4. Therefore, both \(\angle QPR\) and \(\angle QPS\) have the same measure as the source angle, but they lie on opposite sides of the target ray. 5. Amina must choose the intersection reached by rotating counterclockwise from \(\overrightarrow{PQ}\).

Both intersections produce congruent construction triangles, so both angles are valid copies. To satisfy the orientation condition, choose the point on the counterclockwise side of \(\overrightarrow{PQ}\).

- Recall the two compass quantities used in a standard angle-copy construction. - The phrase “same side” determines which of the possible copied rays to use. - After the construction, use the two copied base angles to analyze the triangle.

1. For each copy, use the standard angle-copy construction: transfer the source arc radius to the new vertex, then transfer the chord between the two source-arc intersections to locate the terminal ray. 2. Choose the terminal rays at \(P\) and \(Q\) on the same side of \(\overline{PQ}\); their intersection is \(R\). 3. The copied base angles satisfy \(\angle RPQ=\angle PQR=55^\circ\). 4. Equal base angles in a triangle have opposite congruent sides, so \(PR=QR\). 5. The triangle angle sum gives \(m\angle PRQ=180^\circ-55^\circ-55^\circ=70^\circ\).

Each angle copy transfers the source arc radius and then the source chord. With the two copies opening to the same side of \(\overline{PQ}\), their rays meet at \(R\). The equal copied base angles give \(PR=QR\), and \(m\angle PRQ=70^\circ\).

- Think of each copied length as a fixed distance from one endpoint of \(\overline{AC}\). - Use loci that contain all points at each required distance. - Check what side equalities two common circle intersections automatically create.

1. Draw a circle centered at \(A\) with radius equal to the copied \(5\,\text{cm}\) segment. 2. Draw a circle centered at \(C\) with radius equal to the copied \(4\,\text{cm}\) segment. 3. The circles intersect at two points because \(|5-4|<6<5+4\). Label the two intersections \(B\) and \(D\). 4. By the circle radii, \(AB=AD=5\,\text{cm}\) and \(CB=CD=4\,\text{cm}\). 5. Thus quadrilateral \(ABCD\) has two distinct pairs of congruent consecutive sides and is a kite.

Intersect a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(C\) with radius \(4\,\text{cm}\). Use the two intersections as \(B\) and \(D\). The radius equalities give the two pairs of congruent consecutive sides required for a kite.

- Distinguish copying a distance from \(P\) from appending a new distance to an existing copied segment. - The first endpoint should become relevant to the placement of the second copied segment. - Once each construction is complete, compare the final distance from \(P\).

1. In the first construction, set the compass to \(a\), center it at \(P\), and mark the first endpoint on the ray. Then reset the compass to \(b\), center it at that new endpoint, and mark the final endpoint farther along the ray. 2. In the second construction, set the compass to \(b\), center it at \(P\), and mark the first endpoint. Reset to \(a\), center at that endpoint, and mark the final endpoint. 3. Centering the second transfer again at \(P\) would merely mark another distance from \(P\); it would not append the second length to the first segment. 4. The first final endpoint is at distance \(a+b\) from \(P\), while the second is at distance \(b+a\). 5. Since \(a+b=b+a\), the two points lie at the same distance from \(P\) on the same ray, so they coincide.

For order \(a\) then \(b\), center the \(a\)-transfer at \(P\) and the \(b\)-transfer at the first endpoint. Reverse the openings for order \(b\) then \(a\). The second transfer must start at the first endpoint to append the lengths. The final distances are \(a+b\) and \(b+a\), so the final endpoints coincide.

- Find the total angle measure after all five copies. - Relate the copied angles to central angles in the circle. - Use the relationship between equal central angles and their chords.

1. The total measure of five copied angles is \(5\cdot72^\circ=360^\circ\), so the fifth terminal ray coincides with the starting ray. 2. The five central angles \(\angle AOB\), \(\angle BOC\), \(\angle COD\), \(\angle DOE\), and \(\angle EOA\) are all \(72^\circ\). 3. Equal central angles in the same circle intercept congruent chords, so \(AB=BC=CD=DE=EA\). 4. The vertices are equally spaced on the circle, so \(ABCDE\) is a regular pentagon.

Five copies total \(360^\circ\), so the construction closes. The five equal central angles intercept five congruent chords, making \(ABCDE\) a regular pentagon.

- Reproduce the direction of \(\overline{AB}\) symmetrically across \(m\). - Preserve the distance from \(A\). - Use the properties of an isosceles triangle to verify reflection symmetry.

1. Copy the angle between ray \(\overrightarrow{AB}\) and line \(m\) to the opposite side of \(m\), using the ray of \(m\) from \(A\) as the common reference side. 2. On the copied ray, transfer length \(AB\) to locate point \(B''\), so \(AB''=AB\). 3. The construction gives congruent angles between \(m\) and \(\overline{AB}\), and between \(m\) and \(\overline{AB''}\), on opposite sides of \(m\). 4. Thus line \(m\) bisects \(\angle BAB''\), and \(\triangle ABB''\) is isosceles with \(AB=AB''\). 5. In an isosceles triangle, the vertex-angle bisector is the perpendicular bisector of the base. Therefore, \(m\) is the perpendicular bisector of \(\overline{BB''}\), so \(B''\) is the reflection of \(B\).

Copy the angle made by \(\overline{AB}\) and \(m\) to the opposite side of \(m\), then copy length \(AB\) on the new ray. The resulting point \(B''\) is the reflection because \(m\) is the perpendicular bisector of \(\overline{BB''}\).

- Start by comparing the longest segment with the other two placed end to end. - A common ray lets you compare the copied lengths without numerical measurement. - For the construction, think of each remaining side length as a fixed-distance condition from one endpoint of the base.

1. Copy the \(4.5\,\text{cm}\) and \(6\,\text{cm}\) segments consecutively on one ray. Their combined length is \(10.5\,\text{cm}\). 2. Copy the \(9.8\,\text{cm}\) segment from the same starting point on the same ray. Its endpoint lies before the \(10.5\,\text{cm}\) endpoint, so \(4.5+6>9.8\). 3. Because \(9.8\,\text{cm}\) is the longest length, the other two triangle inequalities hold automatically. The three lengths can form a triangle. 4. Copy the \(9.8\,\text{cm}\) segment as base \(\overline{AB}\). 5. Draw a circle centered at \(A\) with radius equal to the \(4.5\,\text{cm}\) source segment and a circle centered at \(B\) with radius equal to the \(6\,\text{cm}\) source segment. 6. Either circle intersection can be used as the third vertex \(C\); then connect \(C\) to \(A\) and \(B\).

Yes. The copied lengths show \(4.5\,\text{cm}+6\,\text{cm}=10.5\,\text{cm}>9.8\,\text{cm}\). Copy \(9.8\,\text{cm}\) as the base, then intersect circles centered at its endpoints with radii copied from the \(4.5\,\text{cm}\) and \(6\,\text{cm}\) source segments.

- First reproduce the source angle at \(P\) without measuring it. - Extend one side of the copied angle straight through the vertex. - What total angle measure is formed by a linear pair?

1. Draw an initial ray \(\overrightarrow{PX}\). Reproduce at \(P\) the source construction arc and the chord between its two side intersections, locating a ray \(\overrightarrow{PE}\) so that \(m\angle XPE=67^\circ\). 2. Extend the initial side through \(P\) in the opposite direction to form ray \(\overrightarrow{PZ}\). 3. Angles \(\angle XPE\) and \(\angle ZPE\) form a linear pair, so their measures sum to \(180^\circ\). 4. Therefore, \(m\angle ZPE=180^\circ-67^\circ=113^\circ\).

Copy the \(67^\circ\) angle at \(P\), then extend its initial side through \(P\) in the opposite direction. The resulting supplement measures \(113^\circ\).

- Begin by creating two rays perpendicular to \(\overline{AB}\) on the same side. - Transfer the given side length onto both perpendicular rays. - Before calling the quadrilateral a square, identify a theorem that first proves it is a parallelogram.

1. Copy the right angle at \(A\) with one side along \(\overrightarrow{AB}\), and copy it at \(B\) with one side along \(\overrightarrow{BA}\), both opening to the chosen side of \(\overline{AB}\). 2. Copy length \(AB\) onto the perpendicular ray at \(A\) to locate \(D\), and onto the perpendicular ray at \(B\) to locate \(C\). 3. Since \(AD\perp AB\) and \(BC\perp AB\), lines \(AD\) and \(BC\) are parallel. 4. Also, \(AD=BC=AB\). A quadrilateral with one pair of opposite sides both parallel and congruent is a parallelogram, so \(ABCD\) is a parallelogram. 5. Therefore \(CD=AB\). The parallelogram has four congruent sides and a right angle, so it is a square.

Copy right angles at \(A\) and \(B\) on the same side of \(\overline{AB}\), copy length \(AB\) along both new rays to locate \(D\) and \(C\), and join \(C\) to \(D\). The result is a parallelogram with four congruent sides and a right angle, so it is a square.

- Think about how to combine the two given angle measures using exact copies. - After combining them, what construction corresponds to dividing their sum by \(2\)? - Check that the final construction represents the arithmetic mean, not merely one of the original angles.

1. From a new vertex, copy the \(50^\circ\) angle. 2. Starting from the terminal ray of that copy, copy the \(80^\circ\) angle adjacently on the same side. The combined angle measures \(50^\circ+80^\circ=130^\circ\). 3. Bisect the \(130^\circ\) angle. 4. Each resulting angle measures \(\frac{130^\circ}{2}=65^\circ\), which equals \(\frac{50^\circ+80^\circ}{2}\).

Copy the two source angles adjacently to form a \(130^\circ\) angle, then bisect it. Each half measures \(65^\circ\), the average of \(50^\circ\) and \(80^\circ\).

- Think of the two given segments as the legs of a right triangle. - The desired segment joins the unshared endpoints of those legs. - Relate the three side lengths of the right triangle.

1. Place copies of the \(6\,\text{cm}\) and \(8\,\text{cm}\) segments as perpendicular legs meeting at one endpoint. 2. Join the two unshared endpoints. This new segment is the hypotenuse. 3. By the Pythagorean theorem, its length is \(\sqrt{6^2+8^2}=\sqrt{100}=10\,\text{cm}\).

\(10\,\text{cm}\). Place the \(6\,\text{cm}\) and \(8\,\text{cm}\) segments perpendicular to each other and connect their unshared endpoints.

- What does a diameter tell you about an inscribed angle that subtends it? - Identify the hypotenuse before choosing a right-triangle relation. - Check that the unknown side must be shorter than the diameter.

1. Because \(\overline{AB}\) is a diameter and \(C\) lies on the circle, \(\angle ACB=90^\circ\). 2. In right triangle \(ACB\), \(AB=13\,\text{cm}\) is the hypotenuse and \(AC=5\,\text{cm}\) is one leg. 3. By the Pythagorean theorem, \(BC^2=13^2-5^2=144\). 4. Therefore, \(BC=12\,\text{cm}\).

\(12\,\text{cm}\)

- The source arc must exist before its chord length can be copied. - The target arc must exist before a point can be marked on it. - Draw the final ray only after its second point has been located.

1. First create the source chord by drawing the source arc, so step B comes first. 2. Next transfer that arc radius to the target vertex, so step A comes second. 3. Then transfer the source chord length \(DE\) to locate \(Q\), so step D comes third. 4. Finally draw the second side of the copied angle through \(Q\), so step C is last.

B, A, D, C

- Identify the three side lengths used in the usual SSS justification. - Two of those lengths are already correct because the arc radius was preserved. - Determine which remaining length the changed compass opening affects.

1. Equal source and target arc radii correctly create two equal sides in the source and target construction triangles. 2. The third required equality is the chord between the two arc-intersection points. 3. Using a longer compass opening makes the target chord different from the source chord. 4. The two construction triangles are no longer guaranteed congruent by SSS, so the target angle is not guaranteed congruent to the source angle.

The chord-length condition fails: the target chord is not equal to the source chord. Without that third side equality, SSS does not guarantee equal angles.

- A compass circle can represent all points at a fixed distance from one endpoint of the diagonal. - Use the known diagonal as the common baseline for locating both remaining vertices. - Think about how two circle intersections lie relative to the line through their centers.

1. Draw \(\overline{AC}\) with length \(7\,\text{cm}\). 2. Draw a circle centered at \(A\) with radius \(6\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). Choose either intersection as \(B\). 3. Draw a circle centered at \(A\) with radius \(3\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). 4. Since \(|5-3|<7<5+3\), the second pair of circles has two intersections, one on each side of \(\overline{AC}\). These are the two possible locations for \(D\). 5. Once \(B\) is fixed, choose \(D\) on the opposite side of \(\overline{AC}\) from \(B\) to obtain the convex quadrilateral. The other location produces the concave configuration.

Construct \(B\) as an intersection of circles centered at \(A\) and \(C\) with radii \(6\,\text{cm}\) and \(5\,\text{cm}\). Construct \(D\) as an intersection of circles centered at \(A\) and \(C\) with radii \(3\,\text{cm}\) and \(5\,\text{cm}\). After \(A\), \(B\), and \(C\) are fixed, there are two possible locations for \(D\); the location on the opposite side of \(AC\) from \(B\) gives the convex quadrilateral.

- Build the figure one vertex at a time rather than trying to place all four vertices at once. - After \(C\) is placed, compare the two possible rays for \(CD\) with the side of line \(BC\) that contains \(A\). - Distinguish a local choice that affects convexity from reflecting the entire finished figure.

1. Draw \(\overline{AB}\) with the given length. 2. At \(B\), copy the given angle \(\angle B\). On the chosen ray, mark \(C\) so that \(BC\) has the given length. 3. At \(C\), copy the given angle \(\angle C\). Choose the ray that places \(D\) on the same side of line \(BC\) as \(A\), as required for the convex orientation. 4. On that ray, mark \(D\) so that \(CD\) has the given length, then connect \(D\) to \(A\). 5. Once \(A\), \(B\), and the side of \(\overline{AB}\) containing \(C\) are fixed, each copied angle fixes a ray and each copied length fixes one point on that ray. At \(C\), convexity selects the ray that keeps \(A\) and \(D\) on the same side of line \(BC\). Reflecting the entire construction across line \(AB\) gives a mirror image, but that image is congruent to the first construction. Thus the quadrilateral is unique up to congruence.

a) Copy \(AB\), then copy \(\angle B\), copy \(BC\) along that ray, copy \(\angle C\) on the side of line \(BC\) that keeps \(A\) and \(D\) on the same side, and copy \(CD\) along the new ray before joining \(D\) to \(A\). b) Each copied angle fixes a ray after its side is chosen, and each copied length fixes one point on that ray. Convexity fixes the local choice at \(C\); reflecting the whole construction gives only a congruent mirror image. Therefore the result is unique up to congruence.

- Compare the small triangle formed at the original vertex with the one formed at the new vertex. - Which distances are fixed because they come from unchanged compass widths? - Consider what information is lost when one compass width changes.

1. In the source figure, \(BD=BE\) because both are radii of the same arc. 2. In the target figure, \(PY=PZ\) because both are radii of the target arc. 3. With the compass width unchanged in step 2, \(BD=PY\) and \(BE=PZ\). Step 3 also gives \(DE=YZ\). 4. Therefore, \(\triangle BDE\cong\triangle PYZ\) by SSS, so \(\angle DBE\cong\angle YPZ\). These are \(\angle ABC\) and \(\angle XPZ\). 5. If the target arc radius changes, \(BD=PY\) and \(BE=PZ\) are no longer guaranteed. The two triangles need not be congruent, so the copied angle is not guaranteed to match.

a) The source and target triangles have three pairs of congruent sides, so they are congruent by SSS. Their vertex angles are therefore congruent. b) No. Changing the target arc radius removes two of the required side-length matches, so the procedure no longer guarantees equal angles.

- Identify which angle lies between the two known sides in each plan. - Compare the information in each plan with the standard triangle-congruence criteria. - Ask whether the third vertex could satisfy the second plan in more than one position.

1. Plan A copies two sides and the included angle between them: \(AB\), \(AC\), and \(\angle BAC\). 2. These data determine a congruent triangle by SAS, so Plan A guarantees a reproduction of \(\triangle ABC\), up to reflection. 3. Plan B fixes \(AB\), \(AC\), and \(\angle ABC\). The copied angle is not the included angle between the two known sides. 4. This SSA information can produce two different positions for the third vertex, so it does not guarantee a unique congruent triangle.

a) Plan A. b) Plan A uses SAS. Plan B uses SSA, which can be ambiguous and therefore does not guarantee a congruent reproduction.

- Choose one copied length to serve as a fixed base. - Think about the locus of points at each specified distance from the base endpoints. - Compare the side lengths in the two possible placements.

1. Copy the \(8\,\text{cm}\) segment to make base \(\overline{PQ}\). 2. Draw a circle centered at \(P\) with radius equal to the copied \(5\,\text{cm}\) segment. 3. Draw a circle centered at \(Q\) with radius equal to the copied \(6\,\text{cm}\) segment. 4. The center distance and radii satisfy \(|6-5|<8<6+5\), so the circles intersect at two points on opposite sides of line \(PQ\). 5. Either intersection can be chosen as vertex \(R\). 6. In either position, \(PR=5\,\text{cm}\), \(QR=6\,\text{cm}\), and \(PQ=8\,\text{cm}\). The two triangles are reflections across line \(PQ\) and are congruent by SSS.

Copy the \(8\,\text{cm}\) length as \(\overline{PQ}\), then intersect a circle centered at \(P\) with radius \(5\,\text{cm}\) and a circle centered at \(Q\) with radius \(6\,\text{cm}\). Since \(|6-5|<8<6+5\), there are two possible third-vertex positions. They give reflected, congruent triangles.

- First use the copied angle to fix the directions of the two adjacent sides. - After \(B\) and \(D\) are fixed, what loci enforce the two required opposite-side lengths? - If the two loci meet twice, compare the two candidates relative to line \(BD\) and the desired uncrossed quadrilateral.

1. Copy the \(65^\circ\) angle at \(A\) to create rays \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\). 2. Copy \(6\,\text{cm}\) onto \(\overrightarrow{AB}\) to locate \(B\), and copy \(4\,\text{cm}\) onto \(\overrightarrow{AD}\) to locate \(D\). 3. Draw a circle centered at \(B\) with radius \(4\,\text{cm}\) and a circle centered at \(D\) with radius \(6\,\text{cm}\). These circles have two intersections. 4. Choose as \(C\) the intersection on the side of line \(BD\) opposite \(A\). The other intersection gives a crossed configuration rather than the intended parallelogram. 5. By construction, \(AB=CD=6\,\text{cm}\) and \(AD=BC=4\,\text{cm}\). 6. A quadrilateral with both pairs of opposite sides congruent is a parallelogram, so \(ABCD\) is the required parallelogram.

Copy the angle and the two adjacent side lengths at \(A\). Then intersect the circle centered at \(B\) with radius \(4\,\text{cm}\) and the circle centered at \(D\) with radius \(6\,\text{cm}\). Choose the intersection opposite \(A\) across line \(BD\) as \(C\). This gives \(AB=CD\) and \(AD=BC\), so \(ABCD\) is a parallelogram.

- A point at the other end of the required chord must satisfy two fixed-distance conditions. - Think about where a point can be if it must lie on the original circle and also be a fixed distance from \(A\). - After constructing one chord, use the triangle formed by its endpoints and the circle center to find the central angle.

1. Draw a circle centered at \(A\) with radius equal to the copied \(8\,\text{cm}\) source segment. 2. This circle intersects the original circle at two points, \(B\) and \(C\). Therefore, \(AB=AC=8\,\text{cm}\), so \(\overline{AB}\) and \(\overline{AC}\) are the required chords. 3. The two intersections lie on opposite sides of \(\overline{OA}\), so the two chord positions are reflections of each other across line \(OA\). 4. In \(\triangle AOB\), \(OA=OB=5\,\text{cm}\) and \(AB=8\,\text{cm}\). 5. By the law of cosines, \(8^2=5^2+5^2-2(5)(5)\cos\angle AOB\), so \(\cos\angle AOB=-0.28\). 6. Therefore, \(m\angle AOB\approx106^\circ\).

a) Intersect the original circle with a circle centered at \(A\) whose radius is a copy of the \(8\,\text{cm}\) source segment. b) The two circle intersections give two reflected chord positions through \(A\). c) The minor central angle is approximately \(106^\circ\).

- Form each new side by placing the appropriate pair of source segments end to end. - Once the three new lengths are known, choose one as a base. - Check whether the two remaining fixed-distance loci can intersect.

1. Copy the \(3\,\text{cm}\) and \(4\,\text{cm}\) segments consecutively on an auxiliary ray to create a \(7\,\text{cm}\) segment. 2. In the same way, create segments of lengths \(3+5=8\,\text{cm}\) and \(4+5=9\,\text{cm}\). 3. Copy the \(9\,\text{cm}\) segment as a base. 4. From one endpoint, draw a circle with radius equal to the copied \(7\,\text{cm}\) segment; from the other endpoint, draw a circle with radius equal to the copied \(8\,\text{cm}\) segment. 5. Since \(7+8>9\), the two circles intersect. Either intersection completes an SSS construction of the triangle.

The required side lengths are \(7\,\text{cm}\), \(8\,\text{cm}\), and \(9\,\text{cm}\). Form those lengths by copying source segments end to end, then use an SSS circle-intersection construction. The triangle can be constructed because \(7+8>9\).

- Identify the right triangle created by the diameter. - Which segment is the altitude to that right triangle's hypotenuse? - Relate that altitude to the two pieces into which it divides the hypotenuse.

1. Because \(AC\) is a diameter, \(\angle ADC=90^\circ\). 2. Segment \(BD\) is the altitude from the right angle to hypotenuse \(AC\). 3. The altitude-to-hypotenuse theorem gives \(BD^2=AB\cdot BC\). 4. Thus \(BD^2=4\cdot9=36\), so \(BD=6\,\text{cm}\). 5. Therefore \(BD=\sqrt{AB\cdot BC}\), the geometric mean of the two given lengths.

\(BD=6\,\text{cm}\). The altitude-to-hypotenuse theorem gives \(BD^2=AB\cdot BC\), so \(BD=\sqrt{4\cdot9}=6\,\text{cm}\).

- Use the triangle angle sum before deciding how to place the triangle. - Identify whether the given side lies between the two originally given angles. - Distinguish the criterion supported by the original data from a criterion that becomes available after finding the third angle.

1. The triangle angle sum gives \(\angle B=180^\circ-48^\circ-67^\circ=65^\circ\). 2. Copy \(\overline{AB}\) as a \(7\,\text{cm}\) segment. 3. At \(A\), construct a \(48^\circ\) ray, and at \(B\), construct a \(65^\circ\) ray on the same side of \(\overline{AB}\). 4. The two rays meet at the third vertex. 5. The original information is AAS data: two angles and a nonincluded side. After the third angle is determined, the construction can also be checked as ASA using \(\angle A\), \(\angle B\), and included side \(AB\).

a) \(65^\circ\) b) The data determine a unique triangle up to congruence by AAS. Copy the \(7\,\text{cm}\) side, then construct the \(48^\circ\) and \(65^\circ\) endpoint rays on the same side of it; their intersection is the third vertex.

- Compare the three side lengths in the source triangle with the three side lengths in the target triangle. - Decide whether congruence is necessary or whether similarity is enough. - Check whether all lengths were changed by the same scale factor.

1. In the source figure, \(VC=VD=r\), and the third side of \(\triangle VCD\) is \(CD\). 2. In the target figure, the two radii are each \(2r\), and the transferred chord is \(2CD\). 3. Every side of the target triangle is twice the corresponding side of the source triangle. 4. Therefore, the two triangles are similar by SSS similarity with scale factor \(2\). 5. Corresponding vertex angles are congruent, so the angle at \(P\) is an exact copy of \(\angle AVB\).

Yes. Scaling both the arc radius and the transferred chord by the same factor produces an SSS-similar triangle, so the copied vertex angle is congruent to the source angle.

- Focus first on what must be true of the diagonals at their intersection in every parallelogram. - The full diagonal lengths must be distributed symmetrically about their intersection point. - After the four vertices are placed, identify the quadrilateral theorem that certifies a parallelogram.

1. Bisect the two source segments to obtain lengths \(5\,\text{cm}\) and \(3\,\text{cm}\). 2. At a point \(O\), construct two intersecting lines making an angle of \(70^\circ\). 3. On opposite rays of the first line, locate \(A\) and \(C\) so that \(OA=OC=5\,\text{cm}\). 4. On opposite rays of the second line, locate \(B\) and \(D\) so that \(OB=OD=3\,\text{cm}\). 5. Connect \(A\), \(B\), \(C\), and \(D\) in order. Point \(O\) is the midpoint of both diagonals \(\overline{AC}\) and \(\overline{BD}\). 6. A quadrilateral whose diagonals bisect each other is a parallelogram, so \(ABCD\) is a parallelogram.

Bisect each source diagonal length, place equal halves on opposite rays of two lines meeting at \(70^\circ\), and join the four endpoints in order. The diagonals bisect each other at \(O\), so the quadrilateral is a parallelogram.

- Translate the median condition into a condition involving the midpoint of the base. - The angle condition and median-length condition describe two different loci for \(C\). - To count the forward intersections, compare the distance from the ray's endpoint to the circle center with the radius.

1. Copy \(AB=10\,\text{cm}\) and construct its midpoint \(M\). 2. At \(A\), construct the \(40^\circ\) ray on the chosen side of \(AB\). Point \(C\) must lie on this ray. 3. The condition \(CM=7\,\text{cm}\) also places \(C\) on the circle centered at \(M\) with radius \(7\,\text{cm}\). 4. Since \(AM=5\,\text{cm}<7\,\text{cm}\), point \(A\) lies inside that circle. A ray beginning at an interior point exits the circle at exactly one point. 5. That unique forward intersection is \(C\). Because \(M\) is the midpoint of \(AB\), \(\overline{CM}\) is the required median.

Construct the midpoint \(M\) of \(\overline{AB}\), the \(40^\circ\) ray from \(A\), and the circle centered at \(M\) with radius \(7\,\text{cm}\). Their unique forward intersection is \(C\). Since \(AM=5\,\text{cm}<7\,\text{cm}\), the ray starts inside the circle and exits it exactly once.

- Recall how the diagonals of a rhombus intersect. - Work with half of each diagonal and one side of the rhombus. - After finding half of the unknown diagonal, account for the full length.

1. The diagonals of a rhombus bisect each other at right angles. 2. Half of the \(10\,\text{cm}\) diagonal has length \(5\,\text{cm}\). Let half of the other diagonal have length \(x\). 3. A side of the rhombus and the two half-diagonals form a right triangle, so \(x^2+5^2=6^2\). 4. Thus \(x=\sqrt{11}\,\text{cm}\). 5. The full second diagonal has length \(2x=2\sqrt{11}\,\text{cm}\).

\(2\sqrt{11}\,\text{cm}\)

- Start from a defining property of the diagonals of a parallelogram. - Think about what lengths from the diagonal intersection to the vertices would be needed. - A point reflection can create the opposite half of a segment once its midpoint is fixed.

1. Bisect the source diagonal lengths to obtain \(5\,\text{cm}\) and \(4\,\text{cm}\). 2. Copy \(AB=6\,\text{cm}\). 3. Construct point \(O\) as an intersection of the circle centered at \(A\) with radius \(5\,\text{cm}\) and the circle centered at \(B\) with radius \(4\,\text{cm}\). 4. Reflect \(A\) across \(O\) to point \(C\), and reflect \(B\) across \(O\) to point \(D\). 5. Then \(O\) is the midpoint of both \(\overline{AC}\) and \(\overline{BD}\), with \(AC=10\,\text{cm}\) and \(BD=8\,\text{cm}\). 6. A quadrilateral whose diagonals bisect each other is a parallelogram, so \(ABCD\) has all required measurements.

Create \(\triangle AOB\) with \(AB=6\,\text{cm}\), \(AO=5\,\text{cm}\), and \(BO=4\,\text{cm}\), then reflect \(A\) and \(B\) across \(O\). The resulting diagonals bisect each other and have lengths \(10\,\text{cm}\) and \(8\,\text{cm}\), so the quadrilateral is a parallelogram.

- First determine how many equal ratio parts make the whole perimeter. - How can a segment be divided into an exact number of equal parts without using a marked ruler? - After the three lengths are obtained, what condition tells you whether an SSS triangle can be formed?

1. The ratio parts total \(2+3+4=9\). 2. From one endpoint of the \(18\,\text{cm}\) segment, draw an auxiliary ray and copy one fixed step nine times along the ray. 3. Connect the ninth mark to the far endpoint of the \(18\,\text{cm}\) segment. Through the other eight marks, construct parallels to that connector. The parallels divide the perimeter segment into nine congruent parts. 4. Each part has length \(18\,\text{cm}\div9=2\,\text{cm}\). Copy \(2\), \(3\), and \(4\) parts to obtain lengths \(4\,\text{cm}\), \(6\,\text{cm}\), and \(8\,\text{cm}\). 5. Use one length as a base and intersect circles centered at its endpoints with radii equal to the other two lengths. 6. Since \(4+6>8\), the triangle inequality holds, so the construction produces a triangle.

The side lengths are \(4\,\text{cm}\), \(6\,\text{cm}\), and \(8\,\text{cm}\). Divide the perimeter segment into nine equal parts by an auxiliary-ray parallel construction, copy groups of \(2\), \(3\), and \(4\) parts, then complete an SSS triangle. The triangle exists because \(4+6>8\).

- Compare the two base lengths and think about how their difference can be represented on the longer base. - Look for a way to enforce two distances at once when locating a new vertex. - A useful auxiliary quadrilateral may turn the required parallelism into a known quadrilateral property.

1. Copy the \(10\,\text{cm}\) segment to form base \(\overline{AB}\). 2. On \(\overline{AB}\), locate \(E\) so that \(EB=6\,\text{cm}\). Then \(AE=4\,\text{cm}\). 3. Locate \(D\) as an intersection of a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(E\) with radius \(4\,\text{cm}\). 4. Locate \(C\), on the same side of \(\overline{AB}\) as \(D\), as an intersection of a circle centered at \(D\) with radius \(6\,\text{cm}\) and a circle centered at \(B\) with radius \(4\,\text{cm}\). 5. In quadrilateral \(DCEB\), \(DC=EB=6\,\text{cm}\) and \(CB=DE=4\,\text{cm}\). Both pairs of opposite sides are congruent, so \(DCEB\) is a parallelogram. 6. Therefore, \(DC\parallel EB\). Since \(\overline{EB}\) lies on \(\overline{AB}\), \(CD\parallel AB\), and \(ABCD\) has the required side lengths.

Copy \(AB=10\,\text{cm}\), mark \(EB=6\,\text{cm}\), locate \(D\) from the two required distances to \(A\) and \(E\), then locate \(C\) from the two required distances to \(D\) and \(B\). The auxiliary quadrilateral \(DCEB\) has both pairs of opposite sides congruent, so it is a parallelogram and \(CD\parallel AB\).

- After fixing the side and angle, describe the locus imposed by \(AC=7\,\text{cm}\). - Compare the circle radius with the perpendicular distance from \(A\) to the angle ray. - Two valid intersections need not give congruent triangles; compare the resulting third sides.

1. Fix \(AB=10\,\text{cm}\) and draw the ray from \(B\) that makes a \(35^\circ\) angle with \(\overrightarrow{BA}\). 2. Point \(C\) must lie on the circle centered at \(A\) with radius \(7\,\text{cm}\). 3. The perpendicular distance from \(A\) to the line containing the ray is \(10\sin35^\circ\approx5.74\,\text{cm}<7\,\text{cm}\), so the line meets the circle twice. 4. The distance from \(B\) to the perpendicular foot is \(10\cos35^\circ\approx8.19\,\text{cm}\). The two circle intersections are \(\sqrt{7^2-(10\sin35^\circ)^2}\approx4.01\,\text{cm}\) from that foot along the line. 5. Both intersections lie on the forward ray, giving \(BC\approx8.19-4.01=4.18\,\text{cm}\) or \(BC\approx8.19+4.01=12.20\,\text{cm}\). 6. Because the possible third-side lengths differ, the two triangles are not congruent.

The angle ray meets the circle centered at \(A\) with radius \(7\,\text{cm}\) at two forward points. The two possible values are \(BC\approx4.18\,\text{cm}\) and \(BC\approx12.20\,\text{cm}\), so the SSA data determine two noncongruent triangles.

- View the diagonal as dividing the quadrilateral into two triangles. - First decide whether each three-length set can form a triangle. - Convexity requires more than putting the two new vertices on opposite sides; consider where their perpendicular projections fall along the diagonal.

1. Copy \(AC=8\,\text{cm}\). 2. Locate \(B\) on one side of \(AC\) by intersecting the circle centered at \(A\) with radius \(6\,\text{cm}\) and the circle centered at \(C\) with radius \(5\,\text{cm}\). 3. Locate \(D\) on the opposite side of \(AC\) by intersecting the circle centered at \(A\) with radius \(7\,\text{cm}\) and the circle centered at \(C\) with radius \(4\,\text{cm}\). 4. The triples \((6,5,8)\) and \((7,4,8)\) each satisfy the triangle inequalities, so both pairs of circles intersect. 5. The perpendicular projections of \(B\) and \(D\) onto line \(AC\), measured from \(A\), are \(\frac{6^2+8^2-5^2}{2\cdot8}=\frac{75}{16}\,\text{cm}\) and \(\frac{7^2+8^2-4^2}{2\cdot8}=\frac{97}{16}\,\text{cm}\), respectively. 6. Both projection positions lie strictly between \(0\) and \(8\). Choosing \(B\) and \(D\) on opposite sides of \(AC\) therefore gives a convex quadrilateral with all five required lengths.

Use \(AC\) as the common side of two SSS constructions: locate \(B\) from radii \(6\,\text{cm}\) and \(5\,\text{cm}\), and locate \(D\) on the opposite side from radii \(7\,\text{cm}\) and \(4\,\text{cm}\). Both triangles exist, and both new vertices project inside \(\overline{AC}\), so the resulting quadrilateral is convex.

- Use the midpoint to determine \(MB\). - Find the circle radius from right triangle \(MBC\). - Express the longer side \(AE\) as \(AM+ME\) before forming the ratio.

1. Since \(M\) is the midpoint of \(\overline{AB}\), \(MB=2\,\text{cm}\), while \(BC=4\,\text{cm}\). 2. In right triangle \(MBC\), \(MC=\sqrt{2^2+4^2}=2\sqrt5\,\text{cm}\). 3. Because \(E\) lies on the circle centered at \(M\) through \(C\), \(ME=MC=2\sqrt5\,\text{cm}\). 4. Therefore, \(AE=AM+ME=2+2\sqrt5=2(1+\sqrt5)\,\text{cm}\). 5. The shorter side is \(AD=4\,\text{cm}\), so \(\frac{AE}{AD}=\frac{2(1+\sqrt5)}{4}=\frac{1+\sqrt5}{2}\).

\(\frac{AE}{AD}=\frac{1+\sqrt5}{2}\)