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 \(\overline{BS}\).

- Identify the arc intercepted by \(\angle ACB\). - How is an inscribed angle related to the measure of its intercepted arc? - Point \(C\) is on the major arc, so the angle intercepts the minor arc \(AB\).

1. Angle \(ACB\) is an inscribed angle that intercepts minor arc \(AB\). 2. An inscribed angle measures half its intercepted arc, so \(m\angle ACB=\frac{1}{2}\cdot86^\circ=43^\circ\).

\(43^\circ\)

- Identify the arc intercepted by both marked angles. - What is the relationship between a central angle and an inscribed angle that intercept the same arc?

1. The inscribed angle and the marked central angle intercept the same arc \(BC\). 2. An inscribed angle has half the measure of the central angle that intercepts the same arc, so \(\alpha=\frac{1}{2}\cdot112^\circ=56^\circ\).

\(\alpha=56^\circ\)

- Identify the arc intercepted by the marked inscribed angle. - Which is larger: the central angle or an inscribed angle intercepting the same arc?

1. The marked inscribed angle and central angle intercept the same arc \(PQ\). 2. A central angle has twice the measure of an inscribed angle that intercepts the same arc, so \(\mu=2\cdot34^\circ=68^\circ\).

\(\mu=68^\circ\)

- Find the angle that subtends the diameter. - Use the angle marked in the diagram with the triangle angle sum.

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 special relationship holds between opposite angles of a quadrilateral inscribed in a circle? - Combine that relationship with the angle measure shown in the diagram.

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? - Use the numerical angle marked in the diagram.

1. The diagram shows \(\alpha=40^\circ\). 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. The diagram shows \(m\angle CAD=75^\circ\). 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. The diagram also shows \(m\angle ABD=35^\circ\). 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 arc intercepted by each angle of the quadrilateral. - Recall the angle measure subtended by a diameter. - Distinguish the conditions for a rectangle from the stronger conditions for a square.

1. Angle \(\angle ABC\) intercepts diameter \(\overline{AC}\), so it is a right angle. 2. Angle \(\angle BCD\) intercepts diameter \(\overline{BD}\), so it is a right angle. 3. Similarly, \(\angle CDA\) intercepts diameter \(\overline{AC}\), and \(\angle DAB\) intercepts diameter \(\overline{BD}\). Both are right angles. 4. Therefore, \(ABCD\) has four right angles and is a rectangle. 5. Perpendicular diameters are not required; they would only make the rectangle a square.

Each vertex angle intercepts one of the two diameters and is therefore \(90^\circ\). Thus \(ABCD\) is always a rectangle. It is a square only when the diameters are perpendicular.

- What locus contains all points equidistant from the endpoints of one side? - The desired center must be the same distance from all three vertices. - Two suitable loci are enough to determine their common intersection.

1. Construct the perpendicular bisector of \(\overline{AB}\) and the perpendicular bisector of a second side, such as \(\overline{BC}\). 2. Their intersection point \(O\) is equidistant from \(A\) and \(B\) because it lies on the perpendicular bisector of \(AB\). 3. It is also equidistant from \(B\) and \(C\) because it lies on the perpendicular bisector of \(BC\). 4. Therefore, \(OA=OB=OC\). A circle centered at \(O\) with radius \(OA\) passes through all three vertices.

Construct the perpendicular bisectors of any two sides. Their intersection is the circumcenter because it is equidistant from \(A\), \(B\), and \(C\). Center a circle there and use the distance to any vertex as the radius.

- The center of an inscribed circle must be equally distant from all three sides, not from the vertices. - What locus consists of points equidistant from the two sides of an angle? - Once the center is known, how can you obtain the shortest distance from it to a side?

1. Construct the angle bisectors of two angles of the triangle, such as \(\angle P\) and \(\angle Q\). 2. Their intersection \(I\) is the incenter. A point on an angle bisector is equidistant from the two sides of that angle, so \(I\) has the same perpendicular distance to all three sides. 3. Drop a perpendicular from \(I\) to any side, say \(\overline{PQ}\), and let the foot be \(T\). Use \(IT\) as the radius. 4. The circle centered at \(I\) with radius \(IT\) reaches each side at a perpendicular distance equal to \(IT\), so it is tangent to all three sides.

Construct two angle bisectors; their intersection \(I\) is the incenter. Drop a perpendicular from \(I\) to any side and use that perpendicular distance as the radius. The resulting circle is tangent to all three sides.

- What point must be the center of a circle whose diameter is \(\overline{AB}\)? - An angle subtending a diameter is a right angle; what additional condition makes the triangle isosceles? - Points with a fixed altitude to \(\overline{AB}\) lie on lines at a fixed perpendicular distance from it.

1. The midpoint \(M\) of \(\overline{AB}\) is \(3\,\text{cm}\) from each endpoint. The required circle is 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) Construct the midpoint of \(\overline{AB}\), then use that midpoint as the center and \(3\,\text{cm}\) as the radius. b) The two possible points \(C\) are the intersections of the circle with the perpendicular bisector of \(\overline{AB}\), 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. - How can the given angle at \(A\) select the correct point on that circle? - 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. Construct the midpoint \(M\) of \(\overline{AB}\) and the circle centered at \(M\) with radius \(4\,\text{cm}\). 2. From \(A\), construct the two rays that make a \(40^\circ\) angle with \(\overline{AB}\), one on each side of the line. Each ray has a second intersection with the circle, giving the two mirror-image possible locations of \(C\). Because \(\overline{AB}\) is a diameter, \(\angle C=90^\circ\) in each triangle. 3. Any possible right-angle 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 \(\frac{1}{2}\cdot8\,\text{cm}\cdot4\,\text{cm}=16\,\text{cm}^2\). 5. Since \(20\,\text{cm}^2>16\,\text{cm}^2\), such a triangle is impossible.

a) Construct the circle with diameter \(\overline{AB}\). The two possible points \(C\) are the second intersections of that circle with the two rays from \(A\) that make \(40^\circ\) with \(\overline{AB}\), one on each side. b) The maximum altitude is the radius, \(4\,\text{cm}\). c) No. The maximum possible area is \(16\,\text{cm}^2\).

- Recall the measure of an inscribed angle that subtends a diameter. - Use the triangle angle sum after identifying the right angle. - 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.

- 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.

- First determine the circle whose diameter is \(\overline{AB}\). - How does the position of \(P\) relative to that circle determine whether \(\angle APB\) is acute, right, or obtuse? - Compare each listed position with the circle without calculating the angle directly.

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) \(P(3, 3)\)

- Start with the midpoint and length of \(\overline{AB}\). - Compare the distance from the circle's center to the horizontal walking path with the radius. - What angle is subtended by a diameter from a point on its circle? - How do points outside that diameter circle classify the angle subtending \(\overline{AB}\)?

1. The midpoint of \(A(2,2)\) and \(B(10,2)\) is \((6,2)\), and \(AB=8\), so the circle with diameter \(\overline{AB}\) has center \((6,2)\) and radius \(4\). 2. The walking path lies on \(y=6\). Its distance from the center \((6,2)\) is \(4\), equal to the radius, so the line is tangent to the circle at \(P(6,6)\). 3. At \(P(6,6)\), point \(P\) lies on the circle with diameter \(\overline{AB}\), so \(\angle APB=90^\circ\). 4. Every other point of the walking segment lies outside that circle, so the angle subtending \(\overline{AB}\) is acute there. Therefore, the greatest viewing angle occurs at \(P(6,6)\) and measures \(90^\circ\).

a) Center: \((6,2)\); radius: \(4\) b) The path \(y=6\) is tangent to the circle at \(P(6,6)\). c) At the tangency point, \(P\) lies on the circle with diameter \(\overline{AB}\), so \(\angle APB\) is a right angle. All other positions on the path are outside the circle and give acute angles. The maximum is \(90^\circ\) at \(P(6,6)\).

- Use the diagram to identify which intercepted arc joins the farther points and which joins the nearer points. - Recall how an exterior angle formed by two secants is related to those two arc measures. - Check that your angle is smaller than either intercepted arc.

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 diagram to distinguish the tangent point from the two secant intersection points. - Recall how an exterior tangent-secant angle is related to the difference of the intercepted arcs. - Represent the unknown larger arc with a variable before solving.

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, \(x=112^\circ\).

\(112^\circ\)

- Which arc does \(\angle ABC\) intercept? - What theorem connects an inscribed angle with a diameter? - After establishing the right angle, determine the diameter and use the labeled side in the diagram. - How can the two perpendicular legs be used to find the area?

1. Because \(AC\) is a diameter and \(B\) lies on the circle, \(\angle ABC\) intercepts a semicircle. Therefore, \(\angle ABC=90^\circ\). 2. The diameter has length \(AC=2\cdot5=10\,\text{cm}\). 3. The diagram gives \(BC=6\,\text{cm}\). In right triangle \(ABC\), \(AB^2=10^2-6^2=64\), so \(AB=8\,\text{cm}\). 4. The area is \(\frac{1}{2}\cdot6\cdot8=24\,\text{cm}^2\).

a) \(\angle ABC=90^\circ\) because an inscribed angle that intercepts a diameter is a right angle. b) \(24\,\text{cm}^2\)

- For part a), focus on what arc each angle at \(D\) intercepts. - How does an inscribed angle behave when its intercepted chord is a diameter? - After establishing both right triangles, use the three labeled lengths in the diagram. - Combine the two collinear pieces to obtain \(AC\).

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

a) Each angle is an inscribed angle that intercepts a diameter, so \(\angle ADB=\angle BDC=90^\circ\). b) \(AC=21\,\text{cm}\)

- Relate the copied chord equality to intercepted arcs. - Identify two inscribed angles that use those arcs. - View \(\overline{AC}\) as a transversal of the two chord lines.

1. Because the app marks \(AD=BC\), the congruent chords intercept congruent arcs. Thus arc \(AD\) is congruent to arc \(BC\). 2. Inscribed angle \(\angle ACD\) intercepts arc \(AD\), and inscribed angle \(\angle CAB\) intercepts arc \(CB\). 3. Congruent arcs have congruent inscribed angles, so \(\angle ACD\cong\angle CAB\). 4. These angles are alternate interior angles formed by transversal \(\overline{AC}\) with lines \(CD\) and \(AB\). 5. By the converse of the alternate interior angles theorem, \(CD\parallel AB\).

Since \(AD=BC\), arcs \(AD\) and \(BC\) are congruent, so \(\angle ACD=\angle CAB\). These are alternate interior angles, hence \(CD\parallel AB\).

- Use the altitude relationships to identify the angles at \(E\) and \(F\). - Ask what single segment is subtended by both of those right angles. - Once you identify a diameter, determine the circle's center from it.

1. Since \(A\), \(E\), and \(C\) are collinear while \(E\) and \(H\) lie on the altitude from \(B\), \(AE\perp EH\). Thus \(\angle AEH=90^\circ\). 2. Since \(A\), \(F\), and \(B\) are collinear while \(F\) and \(H\) lie on the altitude from \(C\), \(AF\perp FH\). Thus \(\angle AFH=90^\circ\). 3. Points that form a right angle with endpoints \(A\) and \(H\) lie on the circle with diameter \(\overline{AH}\). Therefore, both \(E\) and \(F\) lie on that circle. 4. Construct the midpoint of \(\overline{AH}\) and draw the circle centered there through \(A\). It also passes through \(E\), \(H\), and \(F\).

Angles \(AEH\) and \(AFH\) are right angles, so \(E\) and \(F\) lie on the circle with diameter \(\overline{AH}\). Construct that circle from the midpoint of \(\overline{AH}\).

- Identify the arc intercepted by each of the two opposite inscribed angles. - How are the two arcs joining the same pair of endpoints related to a full circle? - Apply the Inscribed Angle Theorem to each angle before adding them.

1. Inscribed angle \(\angle DAB\) intercepts arc \(DB\) that passes through \(C\), so \(m\angle DAB=\frac{1}{2}m\widehat{DCB}\). 2. Inscribed angle \(\angle BCD\) intercepts the other arc \(BD\), the one that passes through \(A\), so \(m\angle BCD=\frac{1}{2}m\widehat{BAD}\). 3. Those two intercepted arcs make the full circle, so \(m\widehat{DCB}+m\widehat{BAD}=360^\circ\). 4. Therefore, \(m\angle DAB+m\angle BCD=\frac{1}{2}(360^\circ)=180^\circ\).

Each opposite inscribed angle is half of one of the two arcs joining \(B\) and \(D\). Those arcs total \(360^\circ\), so the two angle measures total \(180^\circ\). Thus opposite angles of a cyclic quadrilateral are supplementary.

- 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 \(AB=2CD\) to compare \(AM\), \(MB\), and \(CD\). - Combine those relationships with \(DM=CD\). What does that tell you about the distances from \(M\) to \(A\), \(B\), and \(D\)? - Once you have identified the circle, what special role does \(AB\) have? - Only after proving the right angle should you use the dimensions in the diagram to find \(BD\).

1. The diagram shows \(CD=7.5\,\text{cm}\). Since \(AB=2CD\) and \(M\) is the midpoint of \(AB\), \(AM=MB=CD=7.5\,\text{cm}\). Because \(DM=CD\), \(DM=7.5\,\text{cm}\) as well. 2. Thus \(MA=MB=MD\), so \(A\), \(B\), and \(D\) lie on a circle centered at \(M\). Because \(M\) is the midpoint of \(AB\), segment \(AB\) is a diameter of that circle. Therefore, \(\angle ADB=90^\circ\). 3. The diameter is \(AB=15\,\text{cm}\). In right triangle \(ADB\), \(9^2+BD^2=15^2\). 4. Therefore, \(BD^2=144\), so \(BD=12\,\text{cm}\).

a) \(MA=MB=MD\), so \(A\), \(B\), and \(D\) lie on a circle centered at \(M\). Since \(AB\) is a diameter, \(\angle ADB=90^\circ\). b) \(BD=12\,\text{cm}\)

- Use \(AB=2CD\) and the midpoint definition to compare \(AM\), \(MB\), and \(CD\). - Since \(AM\parallel CD\) and \(AM=CD\), what can you conclude about quadrilateral \(AMCD\)? - Use \(AD=CD\) to compare \(MA\), \(MB\), and \(MC\). - Only after proving the right angle should you use the two dimensions shown in the diagram.

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. From the diagram, \(AC=15\,\text{cm}\) and \(BC=20\,\text{cm}\). Apply the Pythagorean theorem: \(AB^2=15^2+20^2=625\), so \(AB=25\,\text{cm}\). 5. Since \(AB=2CD\), \(CD=12.5\,\text{cm}\).

a) \(MA=MB=MC\), so \(A\), \(B\), and \(C\) lie on a circle centered at \(M\). Since \(AB\) is a diameter, \(\angle ACB=90^\circ\). b) \(AB=25\,\text{cm}\) and \(CD=12.5\,\text{cm}\)

- What angle does diameter \(PQ\) subtend at \(M\)? What angle does diameter \(PS\) subtend there? - If both \(MQ\) and \(MS\) are perpendicular to the same line through \(M\), what does that tell you about \(Q\), \(M\), and \(S\)? - After proving that \(PM\) is an altitude of triangle \(PQS\), find \(QS\) from the side lengths shown. - Compare two base-height formulas for the area of triangle \(PQS\).

1. Since \(PQ\) is a diameter of the first circle, \(\angle PMQ=90^\circ\). Since \(PS\) is a diameter of the second circle, \(\angle PMS=90^\circ\). 2. Rays \(MQ\) and \(MS\) are therefore both perpendicular to \(MP\) at \(M\). They lie on the same line through \(M\), so \(Q\), \(M\), and \(S\) are collinear. Hence \(PM\perp QS\). 3. From the diagram, \(PQ=12\,\text{cm}\) and \(PS=5\,\text{cm}\), so \(QS=\sqrt{12^2+5^2}=13\,\text{cm}\). 4. Compute the area of right triangle \(PQS\) in two ways: \(\frac{1}{2}(12)(5)=\frac{1}{2}(13)(PM)\). 5. Thus \(60=13(PM)\), so \(PM=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\).

a) \(\angle PMQ=\angle PMS=90^\circ\) because they subtend diameters. Therefore \(Q\), \(M\), and \(S\) are collinear and \(PM\perp QS\). b) \(PM=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\)