Free Math Worksheets

- Notice how \(A\) and \(B\) are positioned relative to the center. - Recall the measure of an inscribed angle that subtends a diameter. - The legs are the sides adjacent to the right angle.

1. Angle \(\angle ASB\) is an inscribed angle that subtends diameter \(\overline{AB}\), so \(\gamma = 90^\circ\). 2. The legs are the two sides that form the right angle at \(S\): \(\overline{AS}\) and \(\overline{BS}\).

\(\gamma = 90^\circ\). The legs are \(\overline{AS}\) and \(\overline{BS}\).

- Recall the measure of an inscribed angle that subtends a diameter. - Use the triangle angle sum. - Compare an angle subtending the same segment from a point on the circle and from a point inside it.

1. An inscribed angle that subtends a diameter is a right angle, so \(\gamma = 90^\circ\). 2. By the triangle angle-sum theorem, \(\alpha = 180^\circ - 90^\circ - 38^\circ = 52^\circ\). 3. When \(C\) is moved inside the circle, the angle subtending fixed segment \(\overline{AB}\) becomes greater than \(90^\circ\). Therefore, \(\gamma\) becomes obtuse.

a) \(\gamma = 90^\circ\), by the inscribed-angle theorem for a diameter. b) \(\alpha = 52^\circ\). c) \(\gamma > 90^\circ\), so the angle becomes obtuse.

- Both angles intercept arc \(BC\). - What is the relationship between a central angle and an inscribed angle that intercept the same arc?

1. An inscribed angle has half the measure of the central angle that intercepts the same arc. 2. Therefore, \(\alpha = \frac{1}{2} \cdot 112^\circ = 56^\circ\).

\(\alpha = 56^\circ\)

- Which is larger: the central angle or the inscribed angle intercepting the same arc? - Recall the fixed relationship between those two angle measures.

1. A central angle has twice the measure of an inscribed angle that intercepts the same arc. 2. Therefore, \(\mu = 2 \cdot 34^\circ = 68^\circ\).

\(\mu = 68^\circ\)

- Find the angle that subtends the diameter. - Use the sum of the interior angles of a triangle.

1. Since \(\angle ACB\) subtends diameter \(\overline{AB}\), \(\angle ACB = 90^\circ\). 2. Use the triangle angle sum: \(\beta = 180^\circ - 90^\circ - 28^\circ = 62^\circ\).

\(\beta = 62^\circ\)

- What is the special relationship between opposite angles of a quadrilateral inscribed in a circle? - What do the measures of a pair of opposite angles add to?

1. Opposite angles of a cyclic quadrilateral are supplementary. 2. Therefore, \(\delta=180^\circ-104^\circ=76^\circ\).

\(\delta=76^\circ\)

- What is true of inscribed angles that intercept the same arc? - How are a central angle and an inscribed angle intercepting the same arc related? - Identify which angles in the diagram intercept arc \(AB\).

1. Inscribed angles that intercept the same arc are congruent, so \(\beta=\alpha=40^\circ\). 2. A central angle has twice the measure of an inscribed angle intercepting the same arc. 3. Therefore, \(\mu=2\cdot40^\circ=80^\circ\).

\(\beta=40^\circ\) and \(\mu=80^\circ\)

- Identify the arc intercepted by each marked angle. - What is true of inscribed angles that intercept the same arc? - Look for pairs of angles with the same arc endpoints. - Name the theorem involving angles whose vertices lie on a circle.

1. Angles \(\angle CAD\) and \(\alpha=\angle CBD\) are inscribed angles that intercept the same arc \(CD\). 2. Inscribed angles intercepting the same arc are congruent, so \(\alpha=75^\circ\). 3. Angles \(\angle ABD\) and \(\beta=\angle ACD\) intercept the same arc \(AD\). 4. Therefore, \(\beta=35^\circ\). 5. The theorem used is the Inscribed Angle Theorem, specifically the result that inscribed angles intercepting the same arc are congruent.

\(\alpha=75^\circ\) and \(\beta=35^\circ\). The theorem is the Inscribed Angle Theorem.

- Identify the intercepted arc for each unknown inscribed angle. - Compare each unknown angle with a given inscribed angle that intercepts the same arc.

1. Angles \(\angle LNM\) and \(\angle LKM\) are inscribed angles that intercept the same arc \(LM\). Therefore, \(\gamma = 40^\circ\). 2. Angles \(\angle MLN\) and \(\angle MKN\) are inscribed angles that intercept the same arc \(MN\). Therefore, \(\delta = 70^\circ\).

\(\gamma = 40^\circ\) and \(\delta = 70^\circ\)

- Find the midpoint and radius of the circle with diameter \(\overline{AB}\). - An angle subtending a diameter is a right angle. - Intersect the circle with lines parallel to \(\overline{AB}\) at the required distance.

1. The midpoint \(M\) of \(\overline{AB}\) is \(3\,\text{cm}\) from each endpoint. Draw the circle centered at \(M\) with radius \(3\,\text{cm}\). 2. By the inscribed-angle theorem, every point on the circle other than \(A\) and \(B\) forms a right angle subtending diameter \(\overline{AB}\). 3. For the triangle to be isosceles, \(C\) must also lie on the perpendicular bisector of \(\overline{AB}\). The two intersections of that line with the circle are the possible points \(C\). 4. Each such point is \(3\,\text{cm}\) from \(\overline{AB}\), so \(h_c = 3\,\text{cm}\). 5. Points with altitude \(2\,\text{cm}\) lie on the two lines parallel to \(\overline{AB}\) at distance \(2\,\text{cm}\). Each line intersects the circle twice, so there are four possible points \(D\).

a) The circle has center at the midpoint of \(\overline{AB}\) and radius \(3\,\text{cm}\). b) There are two possible points \(C\), and \(h_c = 3\,\text{cm}\). c) There are four possible points \(D\).

- Every possible right-angle vertex lies on the circle with the hypotenuse as diameter. - The farthest point on the circle from the diameter is one radius away. - Use \(A = \frac{1}{2}bh\) with the hypotenuse as the base.

1. Draw \(\overline{AB}\) with \(AB = 8\,\text{cm}\), construct its midpoint \(M\), and draw the circle centered at \(M\) with radius \(4\,\text{cm}\). 2. At \(A\), construct a \(40^\circ\) ray. Its second intersection with the circle is \(C\). Because \(\overline{AB}\) is a diameter, \(\angle C = 90^\circ\). 3. Any possible vertex \(C\) lies on this circle. The altitude \(h_c\) is the perpendicular distance from \(C\) to \(\overline{AB}\), and the greatest such distance is the circle's radius, \(4\,\text{cm}\). 4. The maximum area is \(\frac{1}{2}\cdot 8\cdot 4 = 16\,\text{cm}^2\). 5. Since \(20\,\text{cm}^2 > 16\,\text{cm}^2\), such a triangle is impossible.

a) Point \(C\) is the second intersection of the \(40^\circ\) ray with the circle whose diameter is \(\overline{AB}\). b) The maximum altitude is the radius, \(4\,\text{cm}\). c) No. The maximum possible area is \(16\,\text{cm}^2\).

- Use the locus of vertices of right angles subtending a fixed segment. - An isosceles triangle's vertex lies on the perpendicular bisector of its base. - Compare the radius and diameter of the same circle.

1. The locus is the circle with diameter \(\overline{AB}\), excluding endpoints \(A\) and \(B\). Its center is midpoint \(M\), and its radius is \(4\,\text{cm}\). 2. For the triangle to be isosceles with base \(\overline{AB}\), point \(C\) must lie on the perpendicular bisector of \(\overline{AB}\). Its intersections with the circle are \(4\,\text{cm}\) from \(\overline{AB}\), so \(h_c = 4\,\text{cm}\). 3. Since \(C\) lies on the circle with center \(M\), \(MC\) is a radius. Segment \(\overline{AB}\) is the diameter, so \(AB = 2MC\). Therefore, \(MC = \frac{1}{2}AB\).

a) All such points lie on the circle with diameter \(\overline{AB}\), excluding \(A\) and \(B\). The radius is \(4\,\text{cm}\). b) \(h_c = 4\,\text{cm}\). c) \(MC\) is a radius and \(AB\) is the diameter, so \(MC = \frac{1}{2}AB\).

- An inscribed angle that subtends a diameter is a right angle. - A point reflection across the center sends a point on the circle to the opposite endpoint of a diameter. - Examine the diagonals and interior angles of the resulting quadrilateral.

1. Since \(\overline{AB}\) is a diameter and \(C\) lies on the circle, \(\angle ACB = 90^\circ\). 2. The third angle is \(\angle ABC = 180^\circ - 90^\circ - 60^\circ = 30^\circ\). 3. Reflecting \(C\) across \(M\) makes \(\overline{CD}\) another diameter. Thus, diagonals \(\overline{AB}\) and \(\overline{CD}\) of quadrilateral \(ACBD\) bisect each other at \(M\) and are congruent. 4. Each interior angle of \(ACBD\) subtends one of the diameters and therefore measures \(90^\circ\). Hence, \(ACBD\) is a rectangle.

a) \(\angle ACB = 90^\circ\) and \(\angle ABC = 30^\circ\). b) \(ACBD\) is a rectangle because its diagonals are congruent diameters that bisect each other, and each interior angle is a right angle.

- The circle with diameter \(\overline{AB}\) has center \((3,1)\) and radius \(2\). - Points inside the circle form obtuse angles, points on it form right angles, and points outside it form acute angles. - Compare each point’s distance from the center with the radius.

1. The circle with diameter \(\overline{AB}\) has center \((3,1)\) and radius \(2\). 2. Point \((3,1.5)\) is \(0.5\) unit from the center, so it is inside the circle and \(\angle APB\) is obtuse. 3. Point \((3,3)\) is \(2\) units from the center, so it lies on the circle and \(\angle APB\) is right. 4. Point \((3,5)\) is \(4\) units from the center, so it is outside the circle and \(\angle APB\) is acute. 5. Therefore, the angle decreases from obtuse to right to acute as \(P\) moves upward. The right angle occurs at \(P(3,3)\).

a) At \(P(3,1.5)\), the angle is obtuse; at \(P(3,3)\), it is right; and at \(P(3,5)\), it is acute. The angle decreases as \(P\) moves upward. b) \(P(3, 3)\)

- Use the symmetry of the two post locations. - Consider the circle with diameter \(\overline{AB}\). - Find where the walking path touches that circle.

1. The circle with diameter \(\overline{AB}\) has center \((6,2)\) and radius \(4\). 2. The walking path is the horizontal line segment on \(y=6\). This line is tangent to the circle at \(P(6,6)\). 3. At the tangent point, \(P\) lies on the circle, so the angle subtending diameter \(\overline{AB}\) is \(90^\circ\). 4. Every other point on the walking path lies outside the circle and forms an angle smaller than \(90^\circ\). Therefore, the maximum occurs at \(P(6,6)\).

a) \(P(6, 6)\) b) A right angle measuring \(90^\circ\)

- Which theorem gives the measure of an angle formed by two secants outside a circle? - Find the difference of the intercepted arc measures, then divide by \(2\).

1. The measure of an angle formed by two secants outside a circle equals half the positive difference of the intercepted arcs. 2. Therefore, \(m\angle P=\frac{1}{2}(116^\circ-34^\circ)=\frac{1}{2}(82^\circ)=41^\circ\).

\(41^\circ\)

- Use the relationship between an exterior angle and its intercepted arcs. - Solve the resulting equation for the unknown arc measure.

1. For an angle formed by a tangent and a secant outside a circle, the angle measure is half the positive difference of the intercepted arcs. 2. Let \(x\) be the measure of the larger arc. Then \(35^\circ=\frac{1}{2}(x-42^\circ)\). 3. Multiply by \(2\): \(70^\circ=x-42^\circ\). 4. Therefore, \(x=112^\circ\).

\(112^\circ\)

- What is true about an angle inscribed in a semicircle? - Determine the length of the diameter before using the Pythagorean theorem.

1. An angle inscribed in a semicircle is a right angle, so the diameter is the hypotenuse and has length \(2\cdot 5=10\,\text{cm}\). 2. Find the other leg: \(b^2=10^2-6^2=64\), so \(b=8\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\cdot 6\cdot 8=24\,\text{cm}^2\).

The area is \(24\,\text{cm}^2\).

- An angle inscribed in a semicircle is a right angle. - Use the two right angles at \(D\) to determine the positions of \(A\), \(D\), and \(C\). - Apply the Pythagorean theorem in both right triangles.

1. Because \(AB\) is a diameter, \(\angle ADB = 90^{\circ}\). Because \(BC\) is a diameter, \(\angle BDC = 90^{\circ}\). 2. These adjacent angles form \(180^{\circ}\), so \(A\), \(D\), and \(C\) are collinear. 3. In right triangle \(ABD\), \(AD = \sqrt{AB^2 - BD^2} = \sqrt{10^2 - 8^2} = 6\,\text{cm}\). 4. In right triangle \(BCD\), \(DC = \sqrt{BC^2 - BD^2} = \sqrt{17^2 - 8^2} = 15\,\text{cm}\). 5. Therefore, \(AC = AD + DC = 6 + 15 = 21\,\text{cm}\).

\(AC = 21\,\text{cm}\)

- A point on the circle with diameter \(\overline{AB}\) forms a right angle subtending \(\overline{AB}\). - Points inside that circle form obtuse angles, and points outside it form acute angles. - Substitute \(y=3\) into the circle equation to find the right-angle positions.

1. The circle with diameter \(\overline{AB}\) has center \((4,0)\), radius \(4\), and equation \((x-4)^2+y^2=16\). 2. For \(P_1(0,3)\), \((0-4)^2+3^2=25>16\), so \(P_1\) is outside the circle and \(\angle AP_1B\) is acute. 3. For \(P_2(4,3)\), \((4-4)^2+3^2=9<16\), so \(P_2\) is inside the circle and \(\angle AP_2B\) is obtuse. 4. For \(P_3(8,3)\), \((8-4)^2+3^2=25>16\), so \(P_3\) is outside the circle and \(\angle AP_3B\) is acute. 5. On the path \(y=3\), right-angle positions lie on the circle and satisfy \((x-4)^2+9=16\). Thus, \(x=4\pm\sqrt{7}\). 6. Both values lie between \(0\) and \(8\), so the two positions are \(\left(4-\sqrt{7},3\right)\) and \(\left(4+\sqrt{7},3\right)\).

a) \(P_1\): acute \(P_2\): obtuse \(P_3\): acute b) The right-angle positions are \(\left(4-\sqrt{7},3\right)\) and \(\left(4+\sqrt{7},3\right)\).

- Use the midpoint and the relationship \(AB=2CD\) to find \(AM\) and \(MB\). - Compare \(AM\), \(MB\), and \(DM\). What circle can you identify? - Use the angle formed by an inscribed triangle whose side is a diameter. - Then apply the Pythagorean theorem.

1. Since \(AB=2CD\) and \(M\) is the midpoint of \(AB\), \(AM=MB=CD=7.5\,\text{cm}\). The problem also gives \(DM=7.5\,\text{cm}\). 2. Thus \(A\), \(B\), and \(D\) lie on a circle centered at \(M\), and \(AB\) is a diameter. Therefore, \(\angle ADB\) is a right angle. 3. The diameter is \(AB=15\,\text{cm}\). In right triangle \(ADB\), \(9^2+BD^2=15^2\). 4. Therefore, \(BD^2=144\), so \(BD=12\,\text{cm}\).

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

- Let \(M\) be the midpoint of \(AB\) and connect \(M\) to \(C\). - Use \(AB=2CD\) to compare \(AM\), \(MB\), and \(CD\). - Identify a parallelogram and then a circle with diameter \(AB\). - Apply the Pythagorean theorem to the resulting right triangle.

1. Let \(M\) be the midpoint of \(AB\). Since \(AB=2CD\), \(AM=MB=CD\). 2. Because \(AM\parallel CD\) and \(AM=CD\), quadrilateral \(AMCD\) is a parallelogram. Therefore, \(MC=AD\). 3. The problem gives \(AD=CD\), so \(MA=MB=MC\). Thus \(A\), \(B\), and \(C\) lie on a circle centered at \(M\), with \(AB\) as a diameter. Therefore, \(\angle ACB\) is a right angle. 4. Apply the Pythagorean theorem: \(AB^2=15^2+20^2=625\), so \(AB=25\,\text{cm}\). 5. Since \(AB=2CD\), \(CD=12.5\,\text{cm}\).

\(AB=25\,\text{cm}\) and \(CD=12.5\,\text{cm}\)

- Use the right angles formed by angles inscribed in semicircles. - Determine where \(M\) lies relative to diagonal \(QS\). - Compute the area of triangle \(PQS\) using two different base-height pairs.

1. Since \(PQ\) is a diameter, \(\angle PMQ = 90^{\circ}\). Since \(PS\) is a diameter, \(\angle PMS = 90^{\circ}\). 2. Therefore, \(Q\), \(M\), and \(S\) are collinear, and \(PM\) is the altitude to hypotenuse \(QS\) in right triangle \(PQS\). 3. The diagonal is \(QS = \sqrt{12^2 + 5^2} = 13\,\text{cm}\). 4. Compute the triangle’s area in two ways: \(\frac{1}{2}(12)(5) = \frac{1}{2}(13)(PM)\). 5. Thus, \(60 = 13(PM)\), so \(PM = \frac{60}{13}\,\text{cm} \approx 4.62\,\text{cm}\).

\(PM = \frac{60}{13}\,\text{cm} \approx 4.62\,\text{cm}\)