Free math worksheets

- Recall the theorem about tangent segments from one external point. - Compare the two segments that begin at \(P\) and end at the tangency points. - Use the length marked in the diagram only after identifying the relationship.

1. Tangent segments drawn from the same external point to a circle are congruent. 2. The diagram gives \(PA=8.4\,\text{cm}\). Therefore, \(PA=PB\), so \(PB=8.4\,\text{cm}\).

\(PB=8.4\,\text{cm}\)

- Focus on the radius that ends at the point of tangency. - What angle does a tangent form with that radius? - The vertex of \(\angle OTP\) is the tangency point \(T\).

1. A tangent to a circle is perpendicular to the radius drawn to the point of tangency. 2. Therefore, \(\overline{OT}\perp\overline{PT}\), so \(m\angle OTP=90^\circ\).

\(90^\circ\)

- A radius to a point of tangency is perpendicular to the tangent line. - Identify the equation of the line through \(P\) and \(Q\). - For a vertical diameter, only the y-coordinate changes from the center.

1. The radius is the horizontal distance from \(M(3, 3)\) to the y-axis, so \(r = |3 - 0| = 3\). The point of tangency is \(B(0, 3)\). 2. The line through \(P(0, 0)\) and \(Q(7, 0)\) is the x-axis, \(y = 0\). 3. The perpendicular distance from \(M(3, 3)\) to the x-axis is \(3\). Since this distance equals the radius, \(w\) is tangent to the circle. 4. A vertical diameter lies on \(x = 3\). Moving \(3\) units above and below the center gives \(K(3, 6)\) and \(L(3, 0)\).

a) \(r = 3\) and \(B(0, 3)\). b) \(w\) is tangent because its distance from \(M\) is \(3\), equal to the radius. c) One choice is \(K(3, 6)\) and \(L(3, 0)\).

- Convert all lengths to the same unit. - Compare each perpendicular distance with the radius. - Relate the comparison to zero, one, or two intersection points.

1. Convert all measurements to centimeters: \(r = 4.5\,\text{cm}\), \(d_1 = 3.5\,\text{cm}\), \(d_2 = 4.5\,\text{cm}\), and \(d_3 = 5\,\text{cm}\). 2. Since \(d_1 < r\), line \(g_1\) intersects the circle twice and is a secant. 3. Since \(d_2 = r\), line \(g_2\) intersects the circle once and is a tangent. 4. Since \(d_3 > r\), line \(g_3\) does not intersect the circle and is exterior to the circle.

\(g_1\): secant because \(d_1 < r\). \(g_2\): tangent because \(d_2 = r\). \(g_3\): exterior line because \(d_3 > r\).

- Picture two parallel lines that just touch opposite sides of a circle. - Consider the radii from the center to the two points of tangency. - Relate the resulting segment to the circle’s diameter.

1. A tangent line is perpendicular to the radius drawn to its point of tangency. 2. Because the two tangents are parallel, the two perpendicular radii lie on the same line through the center and form a diameter. 3. Therefore, the distance between the tangent lines equals the diameter: \(2r = 2 \cdot 4.5\,\text{cm} = 9\,\text{cm}\).

The tangent lines are \(9\,\text{cm}\) apart. Their points of tangency are endpoints of a diameter, so the distance is \(2 \cdot 4.5\,\text{cm} = 9\,\text{cm}\).

- Find the distance between \(C\) and \(D\). - Recall the relationship between the center distance and radii for externally tangent circles. - Imagine increasing the second radius until the circles just touch.

1. The centers have the same x-coordinate, so \(CD = 7 - 1 = 6\,\text{cm}\). 2. For external tangency, the distance between the centers equals the sum of the radii: \(r_C + r_D = 6\,\text{cm}\). 3. Therefore, \(r_D = 6\,\text{cm} - 3\,\text{cm} = 3\,\text{cm}\).

\(r_D = 3\,\text{cm}\)

- Which segment is a radius to the point of tangency? - What theorem determines the angle between that radius and the tangent? - After establishing the right angle, use the two labeled lengths to relate them to \(MP\).

1. A radius drawn to a point of tangency is perpendicular to the tangent, so \(\triangle MTP\) is a right triangle. 2. The diagram shows legs \(MT=5\,\text{cm}\) and \(TP=12\,\text{cm}\). Apply the Pythagorean theorem: \(MP^2=5^2+12^2=169\). 3. Therefore, \(MP=13\,\text{cm}\).

a) Radius \(MT\) is perpendicular to the tangent at \(T\), so \(\angle MTP=90^\circ\). b) \(MP=13\,\text{cm}\)

- Which chord is shared by the marked inscribed angle and the tangent-chord angle? - Recall how an angle formed by a tangent and a chord is related to an inscribed angle intercepting the same chord. - Use the numerical angle shown in the diagram after identifying that relationship.

1. By the Tangent-Chord Theorem, the angle formed by a tangent and a chord equals an inscribed angle that intercepts the same chord. 2. The diagram gives \(m\angle BAC=42^\circ\). Therefore, \(m\angle CBE=m\angle BAC=42^\circ\).

\(42^\circ\)

- Which segment is a radius to the point where the tangent touches the circle? - What relationship must that radius have to the tangent? - After justifying the angle at \(B\), use the numerical angle marked in the diagram and the triangle angle sum.

1. A radius to a point of tangency is perpendicular to the tangent, so \(m\angle OBA=90^\circ\). 2. The diagram gives \(m\angle OAB=32^\circ\). The angle measures of triangle \(OBA\) sum to \(180^\circ\). 3. Therefore, \(m\angle AOB=180^\circ-90^\circ-32^\circ=58^\circ\).

a) Radius \(OB\) is perpendicular to the tangent at \(B\), so \(\angle OBA=90^\circ\). b) \(m\angle AOB=58^\circ\)

- A radius is perpendicular to a tangent at the point of tangency. - Consider the angle sum of quadrilateral \(AMBQ\). - Two of the quadrilateral's angles are right angles; use the remaining marked angle from the diagram.

1. Radius \(\overline{MA}\) is perpendicular to tangent segment \(\overline{QA}\), and radius \(\overline{MB}\) is perpendicular to tangent segment \(\overline{QB}\). Thus, \(\angle MAQ=90^\circ\) and \(\angle MBQ=90^\circ\). 2. The diagram gives \(m\angle AMB=120^\circ\). The interior angles of quadrilateral \(AMBQ\) sum to \(360^\circ\). 3. Therefore, \(120^\circ+90^\circ+\alpha+90^\circ=360^\circ\). 4. Solving gives \(\alpha=60^\circ\).

\(\alpha=60^\circ\)

- Compare the two triangles formed by the center, the midpoint, and the chord endpoints. - Identify equal lengths coming from the circle and from the midpoint. - Use the relationship between the two adjacent angles at \(M\).

1. Radii \(\overline{OA}\) and \(\overline{OB}\) are congruent, so \(OA=OB\). 2. Since \(M\) is the midpoint of \(\overline{AB}\), \(AM=MB\). 3. Segment \(\overline{OM}\) is common to \(\triangle OMA\) and \(\triangle OMB\). 4. Therefore, \(\triangle OMA\cong\triangle OMB\) by SSS. 5. Corresponding angles \(\angle OMA\) and \(\angle OMB\) are congruent and form a linear pair. 6. Congruent supplementary angles each measure \(90^\circ\), so \(OM\perp AB\).

The triangles on either side of \(\overline{OM}\) are congruent by SSS. Their angles at \(M\) are congruent and supplementary, so each is \(90^\circ\). Therefore, \(OM\perp AB\).

- Use the radius relationships in each circle separately. - Identify the locus of points equidistant from the chord endpoints. - Use the fact that the two centers are distinct.

1. Points \(A\) and \(B\) lie on the circle centered at \(O_1\), so \(O_1A=O_1B\). 2. Therefore, \(O_1\) lies on the perpendicular bisector of \(\overline{AB}\). 3. Points \(A\) and \(B\) also lie on the circle centered at \(O_2\), so \(O_2A=O_2B\). 4. Therefore, \(O_2\) lies on the same perpendicular bisector of \(\overline{AB}\). 5. The line through the two distinct points \(O_1\) and \(O_2\) is that perpendicular bisector.

Each center is equidistant from \(A\) and \(B\), so both centers lie on the perpendicular bisector of \(\overline{AB}\). Hence line \(O_1O_2\) is the perpendicular bisector of the common chord.

- The center must be equidistant from the two sides of the angle. - Use the perpendicular through \(A\) to identify one center-to-side distance. - Recognize the special right triangle formed with the angle bisector.

1. The angle bisector makes \(45^\circ\) with each ray. Triangle \(OAC\) is a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle because \(AC\perp OX\). 2. Since \(OA=4\,\text{cm}\), the equal legs give \(AC=4\,\text{cm}\). 3. Point \(C\) lies on the angle bisector, so its perpendicular distances to the two rays are equal. 4. One of those distances is \(AC=4\,\text{cm}\), so both distances are \(4\,\text{cm}\). 5. Therefore, the circle centered at \(C\) with radius \(4\,\text{cm}\) is tangent to both rays and lies inside the angle.

Triangle \(OAC\) is a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle, so \(AC=OA=4\,\text{cm}\). Since \(C\) lies on the angle bisector, it is \(4\,\text{cm}\) from both rays. Thus the circle is tangent to both rays inside the angle.

- Draw the radii to the two tangent points and identify the resulting right triangles. - Compare the two right triangles that share \(\overline{PO}\). - A congruence statement about the angles at \(P\) will locate the center relative to \(\angle TPU\).

1. Radii to points of tangency are perpendicular to the tangents, so \(OT\perp PT\) and \(OU\perp PU\). 2. Right triangles \(PTO\) and \(PUO\) share hypotenuse \(\overline{PO}\), and \(OT=OU\) because both are radii. 3. The triangles are congruent by HL. 4. Therefore, \(\angle TPO\cong\angle OPU\). 5. Hence ray \(\overrightarrow{PO}\) bisects \(\angle TPU\), so the center lies on its angle bisector.

The right triangles \(PTO\) and \(PUO\) are congruent by HL, so \(\angle TPO=\angle OPU\). Therefore, \(O\) lies on the angle bisector of \(\angle TPU\).

- Compare the two triangles formed by the center and the congruent chords. - Identify the equal radii and the shared segment. - Use the resulting triangle congruence to compare the two angles at \(A\).

1. In triangles \(AOB\) and \(AOC\), \(AB=AC\) by the given chord congruence. 2. Also, \(OB=OC\) because they are radii, and \(AO\) is shared. 3. Therefore, the triangles are congruent by SSS. 4. Corresponding angles \(\angle BAO\) and \(\angle OAC\) are congruent. 5. Hence ray \(\overrightarrow{AO}\) bisects \(\angle BAC\).

Triangles \(AOB\) and \(AOC\) are congruent by SSS, so \(\angle BAO=\angle OAC\). Therefore, \(\overrightarrow{AO}\) is the angle bisector.

- Pair the two pieces of each complete chord. - What product relationship holds when two chords intersect inside a circle? - Substitute the three labeled lengths only after writing that relationship.

1. For two chords intersecting inside a circle, the products of the two segment lengths are equal: \(AP\cdot PB=CP\cdot PD\). 2. From the diagram, \(AP=3\), \(PB=8\), and \(CP=4\). Let \(PD=x\). 3. Then \(3\cdot8=4x\), so \(24=4x\). 4. Therefore, \(PD=6\).

\(PD=6\)

- Draw the radii to the two points of tangency and connect \(M\) to \(P\). - What equal lengths do you know from radii and from tangent segments drawn from one external point? - Compare the two triangles on either side of \(\overline{MP}\); what does triangle congruence tell you about the angle at \(P\)? - After that, identify a right triangle that contains the radius and \(MP\).

1. Let the tangent points be \(S\) and \(T\). Then \(MS=MT\) because both are radii, \(PS=PT\) because tangent segments from the same external point are congruent, and \(MP\) is common to triangles \(MPS\) and \(MPT\). 2. By SSS, \(\triangle MPS\cong\triangle MPT\). Therefore, \(MP\) bisects the \(60^\circ\) angle between the tangents, so each resulting angle at \(P\) is \(30^\circ\). 3. A radius to a point of tangency is perpendicular to the tangent. In right triangle \(MPT\), \(\sin(30^\circ)=\frac{MT}{MP}=\frac{2}{MP}\). 4. Since \(\sin(30^\circ)=\frac{1}{2}\), \(MP=4\,\text{cm}\).

\(MP=4\,\text{cm}\)

- A radius and tangent are perpendicular at the point of tangency. - The two acute angles in a right triangle are complementary. - Compare a fixed opposite leg with an increasing adjacent leg.

1. A tangent is perpendicular to the radius at the point of tangency, so \(MB \perp BA\). Therefore, \(\angle MBA = 90^\circ\), and the right-angle vertex is \(B\). 2. The acute angles of a right triangle are complementary, so \(\angle MAB = 90^\circ - 58^\circ = 32^\circ\). 3. As \(A\) moves farther from \(B\), leg \(AB\) increases while leg \(MB\) remains fixed. 4. From vertex \(A\), the ratio of the opposite leg to the adjacent leg decreases. Therefore, \(\angle MAB\) decreases and approaches \(0^\circ\).

a) The triangle is right at \(B\) because the tangent is perpendicular to radius \(\overline{MB}\). b) \(\angle MAB = 32^\circ\). c) \(\angle MAB\) decreases toward \(0^\circ\) as \(A\) moves farther from \(B\).

- Identify the locus determined by each condition separately. - Compare distance from a point with perpendicular distance from a line. - Determine how each parallel line meets the circle.

1. The points exactly \(4\,\text{cm}\) from \(P\) form a circle centered at \(P\) with radius \(4\,\text{cm}\). 2. The points exactly \(4\,\text{cm}\) from \(g\) lie on two lines parallel to \(g\), one on each side. 3. Because \(P\) lies on \(g\), each parallel line is exactly one radius from the center. Each line is tangent to the circle at one point. Therefore, there are \(2\) points that satisfy both conditions.

1. A circle centered at \(P\) with radius \(4\,\text{cm}\) 2. Two lines parallel to \(g\), each \(4\,\text{cm}\) from \(g\) 3. \(2\) points

- Find the distance between two points on the same horizontal line. - Recall the relationship between a tangent and a radius at the point of tangency. - A line perpendicular to a horizontal line is vertical. - Locate the point opposite \(P\) across the center.

1. Points \(M\) and \(P\) have the same y-coordinate, so \(MP = 10 - 6 = 4\,\text{cm}\). Thus, \(r = 4\,\text{cm}\). 2. Radius \(\overline{MP}\) is horizontal. A tangent is perpendicular to the radius at the point of tangency, so \(t_1\) is vertical. 3. The point of tangency for the parallel tangent is the point opposite \(P\) on the horizontal diameter. Move \(4\) units left from \(M(6, 4)\) to get \(Q(2, 4)\).

a) \(r = 4\,\text{cm}\) b) \(\overline{MP}\) is horizontal, and a tangent is perpendicular to the radius at the point of tangency, so \(t_1\) is vertical. c) \(Q(2, 4)\)

- Interpret each “at most” condition as a closed disk. - Determine when two closed disks have exactly one common point. - Find the current distance between the centers. - Compare the current distance with the sum of the radii.

1. The current distance between \(A\) and \(B\) is \(11 - 2 = 9\,\text{cm}\). 2. The two conditions describe closed disks with radii \(4\,\text{cm}\) and \(2\,\text{cm}\). 3. The disks have exactly one common point when they are externally tangent, so the distance between their centers must be \(4\,\text{cm} + 2\,\text{cm} = 6\,\text{cm}\). 4. Therefore, the minimum leftward shift is \(9\,\text{cm} - 6\,\text{cm} = 3\,\text{cm}\).

Point \(B\) must be moved \(3\,\text{cm}\) to the left.

- For external tangency, relate the center distance to the sum of the radii. - For containment, consider the point of the smaller disk farthest from the larger center. - Compare the center distance plus the smaller radius with the larger radius.

1. For external tangency, the center distance equals the sum of the radii: \(d = r_1 + r_2 = 2\,\text{cm} + 5\,\text{cm} = 7\,\text{cm}\). 2. For the smaller closed disk to remain inside the larger one, the center distance plus the smaller radius cannot exceed the larger radius: \(d + r_1 \le r_2\). 3. Substitute the radii: \(d + 2\,\text{cm} \le 5\,\text{cm}\), so \(d \le 3\,\text{cm}\). Therefore, the greatest possible value is \(3\,\text{cm}\).

a) \(d = 7\,\text{cm}\) b) \(d = 3\,\text{cm}\) is the greatest possible value because \(d + 2\,\text{cm} \le 5\,\text{cm}\).

- Interpret each distance condition as a closed disk. - Compare the center distance with the sum of the radii. - Decide when two disks overlap instead of only touching.

1. The conditions describe closed disks centered at \(M\) and \(N\) with radii \(4\,\text{cm}\) and \(2\,\text{cm}\). 2. The center distance is \(6\,\text{cm}\), and the sum of the radii is \(4\,\text{cm} + 2\,\text{cm} = 6\,\text{cm}\). 3. Therefore, the disks are externally tangent and have exactly one common point. It lies on \(\overline{MN}\), \(4\,\text{cm}\) from \(M\) and \(2\,\text{cm}\) from \(N\). 4. For the intersection to have positive area, the disks must overlap rather than merely touch. Therefore, the center distance must be less than \(6\,\text{cm}\).

a) Exactly one point. It lies on \(\overline{MN}\), \(4\,\text{cm}\) from \(M\) and \(2\,\text{cm}\) from \(N\). b) The distance between \(M\) and \(N\) must be less than \(6\,\text{cm}\).

- What theorem relates a radius to a tangent at the point where they meet? - After justifying the two angles at \(B\) and \(C\), consider quadrilateral \(ABMC\). - Use the marked angle in the diagram with the interior-angle sum of a quadrilateral.

1. A radius drawn to a point of tangency is perpendicular to the tangent, so \(m\angle MBA=m\angle MCA=90^\circ\). 2. The diagram gives \(m\angle BAC=44^\circ\). The angle measures in quadrilateral \(ABMC\) sum to \(360^\circ\). 3. Therefore, \(44^\circ+90^\circ+m\angle BMC+90^\circ=360^\circ\). 4. Thus, \(m\angle BMC=360^\circ-224^\circ=136^\circ\).

a) Each angle is formed by a radius and a tangent at the point of tangency, so \(\angle MBA=\angle MCA=90^\circ\). b) \(m\angle BMC=136^\circ\)

- Focus on the two tangent segments that meet at each vertex. - Give the equal tangent lengths temporary variables and express each side as a sum of two of them. - Compare the sums of opposite sides only after deriving the relationship. - Then substitute the three lengths shown in the diagram.

1. Tangent segments from the same external point to a circle have equal lengths. 2. Let the tangent lengths from vertices \(A\), \(B\), \(C\), and \(D\) be \(p\), \(q\), \(r\), and \(s\), respectively. Then \(AB=p+q\), \(BC=q+r\), \(CD=r+s\), and \(DA=s+p\). 3. Therefore, \(AB+CD=(p+q)+(r+s)=(q+r)+(s+p)=BC+DA\). 4. Read the side lengths from the diagram: \(8+10=11+d\). 5. Thus \(18=11+d\), so \(d=7\).

a) Tangent segments from the same external point are equal. b) \(AB+CD=BC+DA\) c) \(d=7\)

- What are the angle measures in an equilateral triangle? - What angle is formed by a radius and a tangent at the point of tangency? - Which quadrilateral in the diagram has a known angle sum?

1. Because triangle \(ABP\) is equilateral, \(m\angle APB=60^\circ\). 2. A radius is perpendicular to a tangent at the point of tangency, so \(m\angle MAP=m\angle MBP=90^\circ\). 3. The angle measures of quadrilateral \(AMBP\) sum to \(360^\circ\). 4. Therefore, \(m\angle AMB=360^\circ-90^\circ-90^\circ-60^\circ=120^\circ\).

\(m\angle AMB=120^\circ\)

- Translate the perpendicular-bisector condition into a relationship between two chords. - Connect equal chords in one circle to their intercepted arcs. - Use the stated location of \(C\) on the minor arc.

1. Since \(C\) lies on the perpendicular bisector of \(\overline{AB}\), \(CA=CB\). 2. Congruent chords in the same circle intercept congruent minor arcs. 3. Therefore, minor arc \(AC\) is congruent to minor arc \(CB\). 4. Because \(C\) lies on the minor arc \(AB\), these two congruent arcs partition that arc. 5. Thus \(C\) is the midpoint of the minor arc \(AB\).

The perpendicular-bisector condition gives \(CA=CB\). Equal chords intercept equal arcs, so minor arc \(AC\) equals minor arc \(CB\). Therefore, \(C\) is the midpoint of minor arc \(AB\).

- Identify the complete locus of points equidistant from \(A\) and \(B\). - Intersect that locus with the given circle. - Use the perpendicular from the circle's center to the resulting chord.

1. Every point equidistant from \(A\) and \(B\) lies on the perpendicular bisector \(p\) of \(\overline{AB}\). 2. Therefore, the required points are the intersections \(P\) and \(Q\) of \(p\) with the circle. 3. Let \(M\) be the foot of the perpendicular from \(O\) to \(p\). Since \(OM=3\,\text{cm}\), the perpendicular from the center to chord \(\overline{PQ}\) bisects the chord. 4. In right triangle \(OMP\), \(MP=\sqrt{5^2-3^2}=4\,\text{cm}\). 5. Thus \(PQ=2MP=8\,\text{cm}\).

The points are the two intersections \(P\) and \(Q\) of the circle with the perpendicular bisector of \(\overline{AB}\). Their distance is \(8\,\text{cm}\).

- Express tangency to each line as a perpendicular-distance condition. - Recall the locus of points equidistant from two intersecting lines. - Compare the halves of two adjacent supplementary angles.

1. A circle tangent to both \(\ell\) and \(m\) has a center whose perpendicular distances to the two lines are equal. 2. The locus of points equidistant from two intersecting lines consists of the bisectors of the four angles formed by the lines. 3. The app displays one full bisector line through the internal bisectors of a pair of vertical angles and a second full bisector line through the other pair. 4. Every point on either bisector line, except \(O\), can serve as a center; its perpendicular distance to either original line is the circle's positive radius. 5. Adjacent angles formed by \(\ell\) and \(m\) are supplementary. Their half-measures sum to \(90^\circ\), so the two bisector lines are perpendicular.

The complete locus is the union of the internal and external angle-bisector lines through \(O\), excluding \(O\) for circles with positive radius. The two locus lines are perpendicular because they bisect adjacent supplementary angles.

- Compare the two right triangles formed by \(\overline{OM}\) and the two halves of the chord. - Which sides in those triangles are already known to be congruent? - After finding half the chord, relate it to the center-to-chord distance and a radius. - For the second chord, compare its length with \(\overline{AB}\) before doing any new calculation.

1. Triangles \(OMA\) and \(OMB\) are right triangles. Also, \(OA=OB\) because they are radii, and \(OM\) is a common leg. 2. By hypotenuse-leg congruence, \(\triangle OMA\cong\triangle OMB\). Therefore, \(AM=MB\), so \(M\) bisects \(\overline{AB}\). 3. Since \(AB=16\,\text{cm}\), each half is \(8\,\text{cm}\). With \(OM=6\,\text{cm}\), the radius is \(OA=\sqrt{6^2+8^2}=10\,\text{cm}\). 4. Chord \(CD\) has the same length as chord \(AB\). Congruent chords in the same circle are equidistant from the center, so the perpendicular distance from \(O\) to \(CD\) is also \(6\,\text{cm}\).

a) \(M\) bisects \(\overline{AB}\), so \(AM=MB=8\,\text{cm}\). b) \(10\,\text{cm}\) c) \(6\,\text{cm}\)

- Distinguish the external part \(PA\) from the whole secant \(PB\). - What length-product relationship connects a tangent and a secant from one external point? - The tangent length is used twice in that relationship.

1. For a tangent and a secant from the same external point, \(PT^2=PA\cdot PB\). 2. The diagram gives \(PT=12\) and \(PA=8\), so \(12^2=8\cdot PB\). 3. Thus \(144=8\cdot PB\), so \(PB=18\).

\(PB=18\)

- On each secant, distinguish the segment outside the circle from the entire segment from \(P\) to the farther intersection point. - The upper whole secant includes both labeled pieces on that ray. - Express the lower whole secant using the unknown \(CD\) before writing the product relationship. - Check that the two products use the same kind of pair: outside length and whole-secant length.

1. From the diagram, \(PA=4\,\text{cm}\) and \(AB=5\,\text{cm}\), so the whole upper secant is \(PB=4+5=9\,\text{cm}\). 2. Let \(CD=x\,\text{cm}\). Since \(PC=3\,\text{cm}\), the whole lower secant is \(PD=3+x\,\text{cm}\). 3. For two secants from the same external point, external segment times whole secant is equal: \(PA\cdot PB=PC\cdot PD\). 4. Thus \(4\cdot9=3(3+x)\), so \(36=9+3x\) and \(x=9\). 5. Multiplying the outside segment by only the inside segment is incorrect because the theorem uses the distance from \(P\) all the way to the farther intersection point on each secant.

\(CD=9\,\text{cm}\). The theorem uses external segment \(\times\) whole secant, not external segment \(\times\) inside segment.

- Use the radius-tangent perpendicularity theorem. - Apply the quadrilateral angle sum. - Identify the two pairs of adjacent congruent sides. - Recall the diagonal property of a kite.

1. Radii \(\overline{MS}\) and \(\overline{MT}\) are perpendicular to the tangent segments at \(S\) and \(T\). Thus, \(\angle MSP = 90^\circ\) and \(\angle MTP = 90^\circ\). 2. The interior angles of quadrilateral \(MSPT\) sum to \(360^\circ\). Therefore, \(124^\circ + 90^\circ + \angle SPT + 90^\circ = 360^\circ\). 3. Solving gives \(\angle SPT = 56^\circ\). 4. Since \(MS = MT\) as radii and \(PS = PT\) as tangent segments from the same external point, \(MSPT\) is a kite. 5. The diagonal joining the vertices where the congruent sides meet, \(\overline{MP}\), is the perpendicular bisector of the other diagonal. Therefore, \(ST \perp MP\).

a) The angles at \(S\) and \(T\) are \(90^\circ\). b) \(\angle SPT = 56^\circ\). c) \(ST \perp MP\). Quadrilateral \(MSPT\) is a kite, and its symmetry diagonal \(\overline{MP}\) is perpendicular to \(\overline{ST}\).

- In each panel, draw or imagine the perpendicular from the center to the chord. - What does that perpendicular do to the chord? - Use the radius and half the chord to find the center-to-chord distance. - Compare how the triangle's altitude is built in the two panels.

1. The perpendicular from the center to the chord bisects the \(10\,\text{cm}\) chord, so each half is \(5\,\text{cm}\). 2. Let \(d\) be the distance from the center to the chord. Then \(d=\sqrt{6^2-5^2}=\sqrt{11}\,\text{cm}\). 3. In panel a), the triangle contains the center, so its altitude is \(6+\sqrt{11}\). Thus \(s=\sqrt{(6+\sqrt{11})^2+5^2}=\sqrt{72+12\sqrt{11}}\,\text{cm}\approx10.57\,\text{cm}\). 4. In panel b), the triangle does not contain the center, so its altitude is \(6-\sqrt{11}\). Thus \(s=\sqrt{(6-\sqrt{11})^2+5^2}=\sqrt{72-12\sqrt{11}}\,\text{cm}\approx5.67\,\text{cm}\).

a) \(s=\sqrt{72+12\sqrt{11}}\,\text{cm}\approx10.57\,\text{cm}\) b) \(s=\sqrt{72-12\sqrt{11}}\,\text{cm}\approx5.67\,\text{cm}\)

- Look for a quadrilateral that has two right angles. - Compare the angle made by \(EF\) and \(AB\) with the angle made by the tangent and \(AB\). - Equal angles formed with the same transversal can establish that two lines are parallel.

1. Since \(BE\perp AC\) and \(CF\perp AB\), angles \(\angle BEC\) and \(\angle BFC\) are right angles. 2. Therefore, points \(B\), \(C\), \(E\), and \(F\) lie on the circle with diameter \(\overline{BC}\). 3. Opposite angles of cyclic quadrilateral \(BCEF\) are supplementary, so \(m\angle EFB+m\angle ECB=180^\circ\). 4. Points \(A\), \(F\), and \(B\) are collinear, so \(m\angle EFA+m\angle EFB=180^\circ\). Therefore, \(\angle EFA=\angle ECB\). 5. Because \(A\), \(E\), and \(C\) are collinear, \(\angle ECB=\angle ACB\). 6. By the Tangent-Chord Theorem, the acute angle between \(t\) and \(AB\) equals \(\angle ACB\). Thus \(EF\) and \(t\) make equal acute angles with \(AB\), so \(EF\parallel t\).

The altitude feet make \(BCEF\) cyclic. Supplementary-angle relationships give \(\angle EFA=\angle ACB\), and the tangent at \(A\) makes the same acute angle with \(AB\). Therefore, \(EF\parallel t\).