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) \(UV=2AB\).

- Count how many equal copied intervals make each requested segment. - Keep the points in their given order along the ray. - For the midpoint, compare the number of copied intervals on each side of a candidate point.

1. Four consecutive copies form \(\overline{P_0P_4}\), so \(P_0P_4=4\cdot2.35=9.40\,\text{cm}\). 2. Two consecutive copies form \(\overline{P_1P_3}\), so \(P_1P_3=2\cdot2.35=4.70\,\text{cm}\). 3. Point \(P_2\) is two copied lengths from \(P_0\) and two copied lengths from \(P_4\). 4. Thus \(P_0P_2=P_2P_4=4.70\,\text{cm}\), so \(P_2\) is the midpoint of \(\overline{P_0P_4}\).

a) \(9.40\,\text{cm}\). b) \(4.70\,\text{cm}\). c) \(P_2\), because \(P_0P_2=P_2P_4=4.70\,\text{cm}\).

- Track the initial and terminal ray of each copied angle. - Decide whether the two copied regions overlap or sit side by side. - Compare the combined angle with a straight angle.

1. The two copied angles are adjacent and do not overlap, so their measures add. 2. The combined angle is \(38^\circ+67^\circ=105^\circ\). 3. A straight angle measures \(180^\circ\), so the remaining angle is \(180^\circ-105^\circ=75^\circ\). 4. Copying preserves each source angle measure, and the shared ray makes the copied angles adjacent parts of the combined angle.

a) \(105^\circ\). b) \(75^\circ\). c) Each copy preserves its source measure, and the copies are adjacent with no overlap.

- Track the location of each copy relative to the shared initial ray. - Combine the two nonoverlapping angle regions. - Compare the two angles on either side of \(\overrightarrow{PX}\).

1. Each copied angle has measure \(52^\circ\), so \(m\angle UPX=52^\circ\) and \(m\angle XPV=52^\circ\). 2. The copied angles lie on opposite sides of \(\overrightarrow{PX}\), so the smaller angle between \(\overrightarrow{PU}\) and \(\overrightarrow{PV}\) is their sum. 3. Therefore, \(m\angle UPV=52^\circ+52^\circ=104^\circ\). 4. Since \(\angle UPX\) and \(\angle XPV\) are congruent, \(\overrightarrow{PX}\) bisects \(\angle UPV\).

a) \(104^\circ\). b) \(\overrightarrow{PX}\) is the angle bisector of \(\angle UPV\), because it divides the angle into two \(52^\circ\) angles.

- Determine the order of the three points on the line. - Decide whether the copied segments occupy separate directions or overlap along one direction. - Compare the whole distance in each placement with the two source lengths.

1. Since \(U\) and \(V\) lie on opposite rays from \(P\), point \(P\) lies between them. 2. Therefore, \(UV=UP+PV=4.8+7.1=11.9\,\text{cm}\). 3. If the copies lie on the same ray, the shorter copied segment is contained in the longer one. 4. The endpoint distance is then the difference \(7.1-4.8=2.3\,\text{cm}\).

a) \(11.9\,\text{cm}\). b) The endpoints lie on opposite sides of \(P\), so \(P\) separates them and the lengths add. c) \(2.3\,\text{cm}\).

- 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.

- Express the total distance from \(P\) in each construction. - Compare the two sums. - Use the fact that direction as well as distance is fixed on a ray.

1. The first construction places its endpoint at distance \(a+b\) from \(P\). 2. The second construction places its endpoint at distance \(b+a\) from \(P\). 3. Since \(a+b=b+a\), the two endpoints are the same distance from \(P\). 4. A ray contains exactly one point at any specified positive distance from its endpoint, so the two constructions end at the same point.

Both endpoints are at distance \(a+b=b+a\) from \(P\) on the same ray, so they coincide.

- Make the two copied angles share both a vertex and an initial ray. - Place the smaller copy within the larger one. - Interpret the angle between the terminal rays as a difference.

1. The app copies the \(83^\circ\) angle at a new vertex using a chosen initial ray. 2. From the same vertex and initial ray, it copies the \(37^\circ\) angle inside the first copy. 3. The angle between the two terminal rays is the difference of the copied angle measures. 4. Its measure is \(83^\circ-37^\circ=46^\circ\).

The app copies both angles from the same initial ray, placing the \(37^\circ\) copy inside the \(83^\circ\) copy. The angle between their terminal rays measures \(46^\circ\).

- 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 preserves distance from its center. - The direction of the copied angle matters. - Once the target ray is fixed, transfer the radius \(OP\).

1. At \(O\), the app copies the \(75^\circ\) source angle counterclockwise from ray \(\overrightarrow{OP}\). Let the new terminal ray be \(r\). 2. It copies length \(OP\) from \(O\) along ray \(r\) to locate \(P'\). 3. The construction gives \(\angle POP'=75^\circ\) in the required direction and \(OP'=OP\), which are exactly the defining conditions of the rotation image. 4. A ray contains exactly one point at a specified positive distance from its endpoint, so \(P'\) is unique.

The app copies the \(75^\circ\) angle at \(O\) counterclockwise from \(\overrightarrow{OP}\), then copies length \(OP\) onto the new ray. The marked point is the unique rotation image \(P'\).

- Identify which copied segment is the whole and which is the part removed from it. - Use the order of \(P\), \(R\), and \(Q\) to relate the three lengths. - Consider when the shorter transfer can end before reaching \(P\).

1. The first transfer gives \(PQ=9.2\,\text{cm}\). 2. The second transfer gives \(QR=3.7\,\text{cm}\), with \(R\) between \(P\) and \(Q\). 3. By segment addition, \(PR+RQ=PQ\), so \(PR=PQ-RQ=9.2-3.7=5.5\,\text{cm}\). 4. The point lies between \(P\) and \(Q\) only when the second copied length does not exceed the first, so the required condition is \(CD\le AB\).

a) \(PR=5.5\,\text{cm}\). b) The first copied length forms the whole segment \(\overline{PQ}\), and the second copied length removes \(\overline{RQ}\) from that whole. c) The required condition is \(CD\le AB\).

- Use the triangle angle sum to account for the angle not directly copied. - Identify the position of the copied side relative to the two copied angles. - Consider how choosing one side of the base removes the reflection choice.

1. The third angle measure is \(180^\circ-47^\circ-68^\circ=65^\circ\). 2. The app copies one side and the two angles at its endpoints. 3. These data give two angles and the included side, so the copied triangle is congruent to the source triangle by ASA. 4. Placing both copied angles on the same chosen side of \(\overline{BC}\) selects one of the two reflected congruent placements.

a) \(65^\circ\). b) The copied side lies between the two copied angles, so ASA guarantees a congruent triangle. The side choice determines which reflected placement is used.

- 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 both centers. - Compare the complete side-length sets of the two triangles. - Use the center distance and the two radii to analyze circle intersections.

1. Points \(R\) and \(S\) lie on both circles, so \(PR=QR=PS=QS=6.5\,\text{cm}\). 2. Both triangles share the base length \(PQ=8\,\text{cm}\), so each has side lengths \(6.5\), \(6.5\), and \(8\). 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 centers are \(8\,\text{cm}\) apart, while the sum of the radii is \(6.5+6.5=13\,\text{cm}\). Since \(0<8<13\), the congruent circles intersect at two points.

a) Each triangle has legs \(6.5\,\text{cm}\) and base \(8\,\text{cm}\). b) They are congruent by SSS and are reflections of each other across line \(PQ\). c) Two intersections occur because the center distance \(8\) is less than the radius sum \(13\) and greater than the radius difference \(0\).

- 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. The student 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 \(\overrightarrow{PQ}\).

- Translate the two copying steps into angle equalities in triangle \(PQR\). - Relate equal angles in a triangle to their opposite sides. - Use the triangle angle sum for the remaining angle.

1. The construction gives \(m\angle RPQ=m\angle PQR=55^\circ\). 2. A triangle with two congruent angles has congruent opposite sides, so \(PR=QR\). 3. The triangle angle sum gives \(m\angle PRQ=180^\circ-55^\circ-55^\circ=70^\circ\).

a) \(PR=QR\) because the copied base angles are congruent. b) \(m\angle PRQ=70^\circ\).

- 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 the two 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.

The app intersects a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(C\) with radius \(4\,\text{cm}\), using the two intersections as \(B\) and \(D\). Then \(AB=AD=5\,\text{cm}\) and \(CB=CD=4\,\text{cm}\), so \(ABCD\) is a kite.

- 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. The app copies 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, it transfers 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 \(B\).

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

- Compare the longest source segment with the other two placed end to end. - Use a common ray so the comparison is exact without numerical measurement. - After verifying the inequality, use two fixed-distance loci from the base endpoints.

1. The app copies the \(4.5\,\text{cm}\) and \(6\,\text{cm}\) segments consecutively on one ray. Their combined length is \(10.5\,\text{cm}\). 2. It copies the \(9.8\,\text{cm}\) segment from the same endpoint on the same ray. Its endpoint lies before the \(10.5\,\text{cm}\) endpoint, so \(4.5+6>9.8\). 3. Since \(9.8\,\text{cm}\) is the longest length, this comparison verifies the only nonautomatic triangle inequality. The three lengths can form a triangle. 4. The app copies the \(9.8\,\text{cm}\) segment as base \(\overline{AB}\). 5. It intersects a circle centered at \(A\) with radius \(4.5\,\text{cm}\) and a circle centered at \(B\) with radius \(6\,\text{cm}\). Either intersection can serve as the third vertex.

The app shows that \(4.5\,\text{cm}+6\,\text{cm}=10.5\,\text{cm}\) is longer than \(9.8\,\text{cm}\), so the lengths form a triangle. It uses \(9.8\,\text{cm}\) as the base and intersects circles of radii \(4.5\,\text{cm}\) and \(6\,\text{cm}\) centered at the base endpoints.

- Copy the angle by reproducing an arc and the distance between its side intersections. - Extend the copied angle’s initial side in the opposite direction through the vertex. - Use the \(180^\circ\) sum of a linear pair.

1. The app draws an initial ray \(\overrightarrow{PX}\) and an arc centered at the source angle’s vertex that meets its sides at two points. 2. Without changing the compass width, it draws an arc centered at \(P\) that meets \(\overrightarrow{PX}\) at \(D\). 3. It transfers the distance between the two intersection points on the source arc from \(D\) to the arc centered at \(P\), locating \(E\), and draws ray \(\overrightarrow{PE}\). Then \(m\angle XPE=67^\circ\). 4. It extends the initial side through \(P\) in the opposite direction to form ray \(\overrightarrow{PZ}\). 5. Angles \(\angle XPE\) and \(\angle ZPE\) form a linear pair, so their measures sum to \(180^\circ\). 6. Therefore, \(m\angle ZPE=180^\circ-67^\circ=113^\circ\).

The app copies the source angle by transferring one arc and its chord distance to vertex \(P\), then extends the initial side through \(P\) to create a linear pair. The supplement measures \(113^\circ\).

- Create perpendicular rays at both endpoints on the same side of the segment. - Transfer the original side length to both new rays. - Use a parallelogram test before applying the square conditions.

1. The app copies the right angle at \(A\) with one side along \(\overrightarrow{AB}\), and at \(B\) with one side along \(\overrightarrow{BA}\), both opening to the chosen side of \(\overline{AB}\). 2. It copies 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 \(BC\perp AB\), \(AD\parallel BC\). 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. It has a right angle and all four sides congruent, so it is a square.

The app copies right angles at \(A\) and \(B\) on the same side of \(\overline{AB}\), marks \(AD=BC=AB\), and connects \(C\) to \(D\). The figure is a parallelogram with four congruent sides and a right angle, so it is a square.

- First combine exact copies of the two source angles. - Think about what operation produces an average after the measures are added. - Check the final measure against the arithmetic mean.

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

The app copies the two angles adjacently to form a \(130^\circ\) angle, then bisects it. Each half measures \(65^\circ\), the average of \(50^\circ\) and \(80^\circ\).

- Place the two copied segments so they form a right angle. - Connect their unshared endpoints. - Relate the new segment to the two perpendicular copied lengths.

1. The app copies the \(6\,\text{cm}\) segment as \(\overline{AB}\). 2. At \(B\), it draws a ray perpendicular to \(AB\) and copies the \(8\,\text{cm}\) segment onto that ray to locate \(C\). 3. It connects \(A\) to \(C\). Triangle \(ABC\) is right at \(B\). 4. By the Pythagorean theorem, \(AC=\sqrt{6^2+8^2}=\sqrt{100}=10\,\text{cm}\).

The app uses the \(6\,\text{cm}\) and \(8\,\text{cm}\) segments as perpendicular legs of a right triangle. The hypotenuse is the required \(10\,\text{cm}\) segment.

- Identify which side of the right triangle is the diameter of the circle. - Use the theorem about an angle subtended by a diameter. - Apply the Pythagorean theorem to isolate the unknown leg.

1. Point \(C\) lies on the circle with diameter \(\overline{AB}\), so \(\angle ACB=90^\circ\). 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, \(BC^2=AB^2-AC^2=13^2-5^2=144\). 4. Therefore, \(BC=\sqrt{144}=12\,\text{cm}\).

The constructed segment is \(BC=12\,\text{cm}\). Because \(C\) lies on the circle with diameter \(\overline{AB}\), triangle \(ACB\) is right, so \(BC=\sqrt{13^2-5^2}\).

- 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. The unchanged compass widths give \(BD=PY\), \(BE=PZ\), and the transfer gives \(DE=YZ\). 4. Therefore, \(\triangle BDE\cong\triangle PYZ\) by SSS, so their vertex angles \(\angle DBE\) and \(\angle YPZ\) are congruent. These are \(\angle ABC\) and \(\angle XPZ\). 5. If the target arc radius changes, then \(BD=PY\) 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 where each copied angle lies relative to the two copied sides. - Compare the information in each plan with a triangle-congruence condition. - Consider whether the third vertex could occupy more than one valid 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 uses two sides and an angle that is not included between those two sides. 4. This side-side-angle information can allow more than one noncongruent triangle, so Plan B does not guarantee a unique congruent reproduction.

a) Plan A. b) Plan A transfers two sides and their included angle, which determines the triangle by SAS. Plan B uses a nonincluded angle with two sides, so the data may be ambiguous.

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

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.

The app copies the \(8\,\text{cm}\) length as \(\overline{PQ}\), then intersects a circle centered at \(P\) with radius \(5\,\text{cm}\) and a circle centered at \(Q\) with radius \(6\,\text{cm}\). Since \(|6-5|<8<6+5\), the circles have two intersections. Either intersection is the third vertex. The two possible triangles are reflections across line \(PQ\) and are congruent by SSS.

- Use the copied angle to establish the two adjacent side directions. - Place the two copied side lengths on those rays. - Complete the figure by enforcing the same two lengths on the opposite sides.

1. The app copies the \(65^\circ\) angle at point \(A\) to create rays \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\). 2. It copies the \(6\,\text{cm}\) segment onto \(\overrightarrow{AB}\) to locate \(B\), and the \(4\,\text{cm}\) segment onto \(\overrightarrow{AD}\) to locate \(D\). 3. It draws a circle centered at \(B\) with radius \(4\,\text{cm}\) and a circle centered at \(D\) with radius \(6\,\text{cm}\). 4. The circles meet at \(A\) and one other point, labeled \(C\). 5. Then \(AB=CD=6\,\text{cm}\) and \(AD=BC=4\,\text{cm}\). A quadrilateral with both pairs of opposite sides congruent is a parallelogram.

The app copies the given angle at \(A\), marks \(AB=6\,\text{cm}\) and \(AD=4\,\text{cm}\), then intersects the circle centered at \(B\) with radius \(4\,\text{cm}\) and the circle centered at \(D\) with radius \(6\,\text{cm}\). The second intersection is \(C\), and both pairs of opposite sides are congruent, so the quadrilateral is a parallelogram.

- Add the ratio parts before dividing the perimeter. - Use equal compass steps on an auxiliary ray and parallel lines to create nine exact parts. - Check the triangle inequality before completing the SSS construction.

1. The ratio parts total \(2+3+4=9\). 2. From one endpoint of the \(18\,\text{cm}\) segment, the app draws an auxiliary ray, chooses a fixed step length, and copies it consecutively nine times along the ray. 3. It connects the ninth mark to the other endpoint of the \(18\,\text{cm}\) segment and draws parallel lines through the other eight marks to divide the source segment into nine congruent parts. 4. Each part has length \(18\,\text{cm}\div9=2\,\text{cm}\). The app copies \(2\), \(3\), and \(4\) parts to create segments of lengths \(4\,\text{cm}\), \(6\,\text{cm}\), and \(8\,\text{cm}\). 5. It uses one segment as a base and intersects circles centered at its endpoints with radii equal to the other two segment lengths. Either circle intersection completes the triangle. 6. The triangle inequality holds because \(4+6>8\). Therefore, the construction circles intersect and the triangle exists.

The app divides the perimeter segment into nine congruent parts by using equal steps on an auxiliary ray and parallel lines. Copying \(2\), \(3\), and \(4\) parts gives side lengths \(4\,\text{cm}\), \(6\,\text{cm}\), and \(8\,\text{cm}\). It then completes an SSS construction. Since \(4+6>8\), these lengths form a triangle.

- Treat the copied chord length as a fixed distance from \(A\). - Count the intersections of the two circles. - Use the triangle formed by the center and one constructed chord.

1. The app draws a circle centered at \(A\) with radius equal to the copied \(8\,\text{cm}\) 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 \(\overline{OA}\), giving two reflected chord positions. 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. Thus \(m\angle AOB\approx106^\circ\).

a) The app intersects the original circle with the circle centered at \(A\) of radius \(8\,\text{cm}\). b) The two intersection points give two reflected chords through \(A\). c) The minor central angle is approximately \(106^\circ\).

- Form each new length by placing two source copies end to end. - Use the longest new length as a convenient base. - Check the triangle inequalities before intersecting the two construction circles.

1. On an auxiliary ray, the app copies the \(3\,\text{cm}\) and \(4\,\text{cm}\) segments consecutively to create a \(7\,\text{cm}\) segment. 2. In the same way, it creates segments of lengths \(3+5=8\,\text{cm}\) and \(4+5=9\,\text{cm}\). 3. It copies the \(9\,\text{cm}\) segment as a base and draws a circle centered at one endpoint with radius \(7\,\text{cm}\) and a circle centered at the other endpoint with radius \(8\,\text{cm}\). 4. The circles intersect because \(7+8>9\), \(7+9>8\), and \(8+9>7\). Either intersection completes an SSS construction of the triangle.

The required side lengths are \(7\,\text{cm}\), \(8\,\text{cm}\), and \(9\,\text{cm}\). The app copies the source lengths consecutively to form those three sums, then uses an SSS construction. All triangle inequalities hold, so the construction is possible.

- Place the two source lengths consecutively on one line. - Use the combined segment as the diameter of a semicircle. - Look for the right-triangle relationship involving an altitude to the hypotenuse.

1. On a line, the app copies the \(4\,\text{cm}\) segment as \(\overline{AB}\), then the \(9\,\text{cm}\) segment as \(\overline{BC}\), with \(A\), \(B\), and \(C\) in that order. 2. It draws the circle with diameter \(\overline{AC}\). 3. Through \(B\), it draws the perpendicular to \(AC\) and labels one intersection with the circle as \(D\). 4. Triangle \(ADC\) is right because \(AC\) is a diameter. The altitude-to-hypotenuse theorem gives \(BD^2=AB\cdot BC\). 5. Thus \(BD^2=4\cdot9=36\), so \(BD=6\,\text{cm}\).

The app copies the two source segments end to end, draws the circle on their total length as a diameter, and intersects it with the perpendicular through their shared endpoint. The perpendicular segment has length \(6\,\text{cm}\), the geometric mean of \(4\,\text{cm}\) and \(9\,\text{cm}\).

- Use the triangle angle sum to determine the missing endpoint angle. - Copy the known side before placing the two endpoint rays. - Check which congruence condition uses two angles and a nonincluded side.

1. The third angle is \(\angle B=180^\circ-48^\circ-67^\circ=65^\circ\). 2. The app copies \(\overline{AB}\) to create \(\overline{A'B'}\) with length \(7\,\text{cm}\). 3. At \(A'\), it copies the \(48^\circ\) angle with one side along \(\overrightarrow{A'B'}\). 4. At \(B'\), it creates a \(65^\circ\) angle on the same side of \(\overline{A'B'}\), with one side along \(\overrightarrow{B'A'}\). This angle can be formed by subtracting exact copies of the \(48^\circ\) and \(67^\circ\) angles from a straight angle. 5. The two new rays meet at \(C'\). The original data determine the triangle by AAS, and the constructed endpoint angles with included side \(\overline{A'B'}\) establish congruence by ASA.

The app copies the \(7\,\text{cm}\) side and the \(48^\circ\) angle at one endpoint. At the other endpoint, it forms the remaining angle \(180^\circ-48^\circ-67^\circ=65^\circ\). The original side and two nonincluded angles are AAS data; after the missing endpoint angle is found, the constructed triangle is congruent by ASA.

- 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.

- Start from the diagonal-bisection property of parallelograms. - Each full diagonal must be split equally across the intersection point. - The copied angle determines the angle between the two diagonal lines.

1. The app bisects the two source segments to obtain lengths \(5\,\text{cm}\) and \(3\,\text{cm}\). 2. At point \(O\), it copies the \(70^\circ\) angle to form two intersecting lines. 3. On opposite rays of the first line, it locates \(A\) and \(C\) so that \(OA=OC=5\,\text{cm}\). 4. On opposite rays of the second line, it locates \(B\) and \(D\) so that \(OB=OD=3\,\text{cm}\). 5. After the app connects \(A\), \(B\), \(C\), 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.

Place half of each diagonal on opposite rays of two lines meeting at \(70^\circ\): \(OA=OC=5\,\text{cm}\) and \(OB=OD=3\,\text{cm}\). The diagonals bisect each other at \(O\), so \(ABCD\) is a parallelogram.

- Treat the given diagonal as the common side of two SSS triangles. - Check the triangle inequalities for the triples \((6,5,8)\) and \((7,4,8)\). - To verify convexity, determine whether each new vertex projects to a point inside \(\overline{AC}\).

1. The app copies diagonal \(AC=8\,\text{cm}\). 2. It locates \(B\) 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. It locates \(D\) 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 inequalities \(6+5>8\), \(6+8>5\), and \(5+8>6\) show that triangle \(ABC\) exists. Likewise, \(7+4>8\), \(7+8>4\), and \(4+8>7\) show that triangle \(ADC\) exists. 5. Measured from \(A\) along \(AC\), the perpendicular projections of the two new vertices have positions \(\frac{6^2+8^2-5^2}{2\cdot8}=\frac{75}{16}\,\text{cm}\) and \(\frac{7^2+8^2-4^2}{2\cdot8}=\frac{97}{16}\,\text{cm}\). Both values lie strictly between \(0\) and \(8\). 6. Therefore, when \(B\) and \(D\) are selected on opposite sides of \(AC\), all four interior angles of \(ABCD\) are less than \(180^\circ\), so the quadrilateral is convex and has all five required lengths.

Use \(AC=8\,\text{cm}\) as the common side of two SSS triangles. The relevant circle pairs intersect because both triples satisfy the triangle inequalities. Choosing \(B\) and \(D\) on opposite sides of \(AC\), with both perpendicular projections inside \(\overline{AC}\), produces the required convex quadrilateral.

- A median ends at the midpoint of the opposite side. - The median-length condition gives a circle locus, while the angle condition gives a ray locus. - Compare \(AM\) with the circle's radius to determine the number of forward intersections.

1. The app copies \(AB=10\,\text{cm}\) and locates its midpoint \(M\). 2. At \(A\), it copies the \(40^\circ\) angle on the chosen side of \(AB\), creating the ray on which \(C\) must lie. 3. The condition \(CM=7\,\text{cm}\) places \(C\) on the circle centered at \(M\) with radius \(7\,\text{cm}\). 4. The app labels the intersection of this circle and the copied angle ray as \(C\). Because \(M\) is the midpoint of \(AB\), \(\overline{CM}\) is the required median. 5. Since \(AM=5\,\text{cm}<7\,\text{cm}\), point \(A\) lies inside the circle. A ray beginning at an interior point exits a circle at exactly one point, so the ray determines exactly one vertex \(C\) on the chosen side of \(AB\).

The app bisects \(\overline{AB}\) at \(M\), draws the \(40^\circ\) ray from \(A\), and intersects that ray with the circle centered at \(M\) of radius \(7\,\text{cm}\). Since \(AM=5\,\text{cm}<7\,\text{cm}\), the ray starts inside the circle and has exactly one forward intersection, giving a unique vertex \(C\) on the chosen side.

- Check that the two equal-radius circles have two intersections. - Use the perpendicular-bisector relationship created by the two circle intersections. - Apply the Pythagorean theorem to a right triangle with hypotenuse \(6\,\text{cm}\) and one leg \(5\,\text{cm}\).

1. The app copies diagonal \(AC=10\,\text{cm}\) and draws circles centered at \(A\) and \(C\), each with radius \(6\,\text{cm}\). 2. Since \(10<6+6\), the circles intersect at two points, \(B\) and \(D\). Therefore, \(AB=BC=CD=DA=6\,\text{cm}\), so \(ABCD\) is a rhombus. 3. The two circle intersections lie on the perpendicular bisector of \(\overline{AC}\). If \(O\) is the midpoint of \(AC\), then \(AO=5\,\text{cm}\), \(BO\perp AC\), and \(O\) is also the midpoint of \(BD\). 4. In right triangle \(AOB\), \(OB=\sqrt{AB^2-AO^2}=\sqrt{6^2-5^2}=\sqrt{11}\,\text{cm}\). 5. Therefore, \(BD=2OB=2\sqrt{11}\,\text{cm}\).

The other diagonal has length \(2\sqrt{11}\,\text{cm}\). The circle intersections lie on the perpendicular bisector of the \(10\,\text{cm}\) diagonal, so half of the other diagonal is \(\sqrt{6^2-5^2}=\sqrt{11}\,\text{cm}\).

- Start from the diagonal-bisection characterization of a parallelogram. - Use half of each required diagonal to locate their intersection. - A point reflection creates the opposite half of each diagonal.

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.

The app creates triangle \(AOB\) with side lengths \(5\,\text{cm}\), \(4\,\text{cm}\), and \(6\,\text{cm}\), then reflects \(A\) and \(B\) across \(O\). The resulting diagonals bisect each other and have lengths \(10\,\text{cm}\) and \(8\,\text{cm}\).

- Place the difference of the two base lengths on the longer base. - Use two circle intersections to enforce the two leg lengths and the shorter base length. - Look for an auxiliary quadrilateral with both pairs of opposite sides congruent.

1. The app copies the \(10\,\text{cm}\) segment to form base \(\overline{AB}\). 2. On \(\overline{AB}\), it locates point \(E\) so that \(EB=6\,\text{cm}\). Then \(AE=4\,\text{cm}\). 3. It locates point \(D\) 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. It locates point \(C\), 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.

The app copies \(AB=10\,\text{cm}\), marks \(EB=6\,\text{cm}\), locates \(D\) from \(AD=5\,\text{cm}\) and \(DE=4\,\text{cm}\), then locates \(C\) from \(DC=6\,\text{cm}\) and \(BC=4\,\text{cm}\). The auxiliary quadrilateral \(DCEB\) is a parallelogram, so \(CD\parallel AB\).

- After the base and angle are copied, identify the locus required by \(AC=7\,\text{cm}\). - Compare the circle radius with the perpendicular distance from \(A\) to the angle ray. - Use the two different distances from \(B\) to the intersection points to show the triangles are noncongruent.

1. The app copies \(AB=10\,\text{cm}\) and places at \(B\) the ray that forms a \(35^\circ\) angle with \(\overrightarrow{BA}\). 2. Point \(C\) must also 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 \(10\sin 35^\circ\approx5.74\,\text{cm}\), which is less than \(7\,\text{cm}\). Therefore, the line intersects the circle twice. 4. Along the ray, the distance from \(B\) to the perpendicular foot is \(10\cos 35^\circ\approx8.19\,\text{cm}\), and the two intersection distances from that foot are \(\sqrt{7^2-(10\sin 35^\circ)^2}\approx4.01\,\text{cm}\). Thus both intersections lie in the forward direction of the ray. 5. The resulting values are \(BC_1\approx4.18\,\text{cm}\) and \(BC_2\approx12.20\,\text{cm}\). Both triangles satisfy the given side, side, and nonincluded-angle conditions, but their third side lengths differ, so the triangles are not congruent.

The copied angle ray intersects the circle centered at \(A\) with radius \(7\,\text{cm}\) at two forward points, \(C_1\) and \(C_2\). They give \(BC_1\approx4.18\,\text{cm}\) and \(BC_2\approx12.20\,\text{cm}\), so the two valid SSA triangles are noncongruent.

- Begin with the square whose side is the copied source length. - Use the midpoint of one side and the opposite vertex to determine the circle radius exactly. - Express the longer side as the midpoint segment plus the constructed radius.

1. The app creates square \(ABCD\) with side length \(4\,\text{cm}\) and locates the midpoint \(M\) of \(\overline{AB}\). 2. It draws the circle centered at \(M\) through \(C\). The circle meets ray \(\overrightarrow{AB}\) beyond \(B\) at \(E\), and the app completes rectangle \(AEFD\). 3. In right triangle \(MBC\), \(MB=2\,\text{cm}\) and \(BC=4\,\text{cm}\), so \(MC=\sqrt{2^2+4^2}=2\sqrt{5}\,\text{cm}\). 4. Since \(ME=MC\), \(AE=AM+ME=2+2\sqrt{5}=2(1+\sqrt{5})\,\text{cm}\). 5. Therefore, \(\frac{AE}{AD}=\frac{2(1+\sqrt{5})}{4}=\frac{1+\sqrt{5}}{2}\).

The rectangle has side lengths \(4\,\text{cm}\) and \(2(1+\sqrt{5})\,\text{cm}\). Therefore, the ratio of its longer side to its shorter side is \(\frac{1+\sqrt{5}}{2}\).