Free Math Worksheets

- Recall the theorem about tangent segments from one external point. - Look for symmetry in the figure.

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

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

- What theorem relates tangent segments drawn from the same external point? - Look for symmetry in the diagram.

1. Tangent segments drawn from the same external point to a circle are congruent, so \(PA=PB\). 2. Since \(PA=12\,\text{cm}\), \(PB=12\,\text{cm}\).

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

- 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, \(r_D = 6\,\text{cm} - 3\,\text{cm} = 3\,\text{cm}\).

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

- What angle is formed by a radius and a tangent at the point of tangency? - Use the resulting right triangle.

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

Point \(P\) is \(13\,\text{cm}\) from center \(M\).

- How is an angle formed by a tangent and a chord related to an inscribed angle intercepting the same chord? - Apply the Tangent–Chord Theorem.

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. Therefore, \(m\angle CBE=m\angle BAC=42^\circ\).

\(42^\circ\)

- What special angle is formed by a radius and a tangent at the point of tangency? - What is the sum of the angle measures in a triangle?

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

\(m\angle AOB=58^\circ\)

- 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.

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 interior angles of quadrilateral \(AMBQ\) sum to \(360^\circ\). 3. Therefore, \(120^\circ + 90^\circ + \alpha + 90^\circ = 360^\circ\). 4. Solving gives \(\alpha = 60^\circ\).

\(\alpha = 60^\circ\)

- Consider the radii to the two tangency points. - Use the symmetry of the two tangent segments. - Identify a right triangle in half of the figure. - The \(60^\circ\) angle is bisected by \(MP\).

1. The segment \(MP\) bisects the angle between the two tangents, so each resulting angle at \(P\) is \(30^\circ\). 2. A radius to a point of tangency is perpendicular to the tangent. In right triangle \(MPT\), \(\sin(30^\circ)=\frac{MT}{MP}=\frac{2}{MP}\). 3. 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 \(d + 2\,\text{cm} \le 5\,\text{cm}\).

- What angle is formed by a radius and a tangent at the point of tangency? - What is the sum of the interior angles of a quadrilateral? - Consider quadrilateral \(ABMC\).

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 angle measures in quadrilateral \(ABMC\) sum to \(360^\circ\). 3. Therefore, \(44^\circ+90^\circ+m\angle BMC+90^\circ=360^\circ\). 4. Thus, \(m\angle BMC=360^\circ-224^\circ=136^\circ\).

\(m\angle BMC=136^\circ\)

- 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, \(m\angle AMB=360^\circ-90^\circ-90^\circ-60^\circ=120^\circ\).

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

- 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 \(\overline{ST}\).

- Draw the perpendicular from the center to the chord. - Use the radius and half the chord to find the center-to-chord distance. - The two possible altitudes are the radius plus or minus that distance.

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. When the triangle contains the center, its altitude is \(6+\sqrt{11}\). Thus \(s=\sqrt{(6+\sqrt{11})^2+5^2}=\sqrt{72+12\sqrt{11}}\,\text{cm}\approx 10.57\,\text{cm}\). 4. When the triangle does not contain the center, its altitude is \(6-\sqrt{11}\). Thus \(s=\sqrt{(6-\sqrt{11})^2+5^2}=\sqrt{72-12\sqrt{11}}\,\text{cm}\approx 5.67\,\text{cm}\).

When the triangle contains the center, \(s=\sqrt{72+12\sqrt{11}}\,\text{cm}\approx 10.57\,\text{cm}\). When it does not, \(s=\sqrt{72-12\sqrt{11}}\,\text{cm}\approx 5.67\,\text{cm}\).

- 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 \(6\,\text{cm}\).