Free math worksheets

- How are a circle’s radius and diameter related? - Which circle on a globe passes through the globe’s center? - Think about how to move from diameter to radius and from radius to diameter.

1. The radius is half the diameter, so \(24\,\text{cm} \div 2 = 12\,\text{cm}\). 2. The equator passes through the center of the globe, so its radius is the same as the globe’s radius: \(12\,\text{cm}\). 3. A diameter is twice the radius, so \(2 \cdot 3\,\text{cm} = 6\,\text{cm}\).

a) The radius of the globe is \(12\,\text{cm}\). b) The radius of the equator is \(12\,\text{cm}\). c) The diameter of the circle of latitude is \(6\,\text{cm}\).

- How many \(36^\circ\) angles fit into a full \(360^\circ\) turn? - Use the relationship between a regular polygon's number of sides and its central angle.

1. The central angles of a regular polygon sum to \(360^\circ\). 2. Divide the full angle by one central angle: \(n=\frac{360^\circ}{36^\circ}=10\).

The polygon has \(10\) sides.

- How many degrees are in a full circle? - Which operation divides a total into equal parts? - Divide the total angle measure by the number of slices.

1. A full circle measures \(360^\circ\). 2. The \(10\) slices have equal central angles, so each angle measures \(360^\circ\div10=36^\circ\).

\(36^\circ\)

- How many degrees are in a full circle? - What is true about all central angles of a regular polygon? - For part b, think about how many equal \(24^\circ\) central angles make one full turn.

1. The central angles around the center total \(360^\circ\). 2. For part a, \(360^\circ \div 12 = 30^\circ\). 3. For part b, the number of sides is \(360^\circ \div 24^\circ = 15\).

a) \(30^\circ\) b) \(15\) sides

- Interpret the phrase “at most.” - Decide whether interior points satisfy each condition. - Distinguish the boundary from the boundary together with the interior.

1. The points exactly \(4\,\text{cm}\) from \(M\) form a circle. This set contains only the boundary. 2. The points at most \(4\,\text{cm}\) from \(M\) form a closed disk. This set contains the circle and every point inside it. 3. Therefore, the circle is only the boundary, while the closed disk includes both the boundary and the interior.

The points exactly \(4\,\text{cm}\) from \(M\) form a circle. The points at most \(4\,\text{cm}\) from \(M\) form a closed disk, which includes the circle and its entire interior.

- Decide whether a point exactly \(3\,\text{cm}\) away satisfies Condition A. - Identify what Condition B adds. - Compare “less than” with “less than or equal to.”

1. Condition A describes the open disk centered at \(S\) with radius \(3\,\text{cm}\). It contains the interior but not the boundary circle. 2. Condition B describes the circle centered at \(S\) with radius \(3\,\text{cm}\). 3. Their union is the closed disk centered at \(S\) with radius \(3\,\text{cm}\). It contains both the interior and the boundary. 4. The union differs from Condition A alone because it includes the points exactly \(3\,\text{cm}\) from \(S\).

The union is the closed disk centered at \(S\) with radius \(3\,\text{cm}\). Unlike Condition A alone, it includes the boundary circle.

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

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

- Compare the circles using the same measurement: either two radii or two diameters. - How do you find a radius from a diameter? - Compare the two radii after converting.

1. Find the radius of the Oslo circle: \(r_{\text{Oslo}} = 10\,\text{cm} \div 2 = 5\,\text{cm}\). 2. The radius of the Cairo circle is \(8.5\,\text{cm}\). 3. Because \(8.5\,\text{cm} > 5\,\text{cm}\), the circle of latitude through Cairo has the greater radius.

The circle of latitude through Cairo has the greater radius. The Oslo circle has radius \(5\,\text{cm}\), while the Cairo circle has radius \(8.5\,\text{cm}\).

- What does a radius measure, and what does a diameter measure? - How are radius and diameter related? - Convert both measurements to radii or both to diameters before comparing.

1. The diameter of Annie’s circle is twice its radius: \(d = 2r = 2 \cdot 6\,\text{cm} = 12\,\text{cm}\). 2. Tom’s circle also has a diameter of \(12\,\text{cm}\). 3. The circles are the same size. Annie compared measurements that represent different parts of a circle.

Annie is not correct. Her circle has diameter \(2 \cdot 6\,\text{cm} = 12\,\text{cm}\), so the two circles are the same size.

- Does a compass opening represent a radius or a diameter? - First find the diameter of the original circle. - Then find the new diameter and convert it to a radius.

1. The first circle has diameter \(d_1 = 2r_1 = 2 \cdot 5\,\text{cm} = 10\,\text{cm}\). 2. The second circle must have diameter \(d_2 = 2d_1 = 2 \cdot 10\,\text{cm} = 20\,\text{cm}\). 3. A compass is set to the radius, so \(r_2 = 20\,\text{cm} \div 2 = 10\,\text{cm}\). 4. Felix is not correct; he should set the compass to \(10\,\text{cm}\).

Felix is not correct. The first diameter is \(10\,\text{cm}\), so the second diameter is \(20\,\text{cm}\). The required compass setting is the radius, \(10\,\text{cm}\).

- How many radii make one diameter? - Express the new diameter in terms of \(r + 5\). - You may also test the claim with a simple numerical example.

1. For every circle, \(d = 2r\). 2. After the change, the new diameter is \(2(r + 5) = 2r + 10\). 3. The original diameter was \(2r\), so the diameter increases by \(10\,\text{cm}\), not \(5\,\text{cm}\). 4. The claim is false.

The claim is false. Since \(d = 2r\), increasing the radius by \(5\,\text{cm}\) increases the diameter by \(2 \cdot 5\,\text{cm} = 10\,\text{cm}\).

- What do \(MA\) and \(MB\) have in common because \(M\) is the center? - After comparing the two radii, use the single pair of congruence marks in the diagram. - What can you conclude about a triangle whose three sides are congruent?

1. Points \(A\) and \(B\) lie on the circle centered at \(M\), so \(MA\) and \(MB\) are radii of the same circle. Therefore, \(MA=MB\). 2. The diagram marks \(AB\) congruent to \(MA\). Thus \(AB=MA=MB\), so triangle \(AMB\) is equilateral. 3. Every angle of an equilateral triangle measures \(60^\circ\), so \(m\angle AMB=60^\circ\).

a) \(MA=MB\) because both segments are radii of the same circle. b) Triangle \(AMB\) is equilateral, and \(m\angle AMB=60^\circ\).

- How are arc lengths related to their central angles in the same circle? - How many total parts are represented by the ratio \(3:4:5\)? - How many degrees correspond to one ratio part?

1. Arc lengths in the same circle are proportional to their central angle measures. 2. The ratio has \(3+4+5=12\) total parts. 3. One part represents \(360^\circ\div12=30^\circ\). 4. The three central angles are \(3\cdot30^\circ=90^\circ\), \(4\cdot30^\circ=120^\circ\), and \(5\cdot30^\circ=150^\circ\). 5. The largest central angle is \(150^\circ\).

\(150^\circ\)

- 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.

- Which arc lies inside the marked central angle? - How is the degree measure of a central angle related to the arc it intercepts? - How many degrees are in the entire circle?

1. The marked central angle \(\angle AOB\) measures \(128^\circ\). A central angle and its intercepted arc have the same degree measure, so \(m\widehat{AB}=128^\circ\). 2. A full circle measures \(360^\circ\), so the major arc has measure \(360^\circ-128^\circ=232^\circ\). 3. The central angle with sides through \(A\) and \(B\) is \(\angle AOB\).

a) \(128^\circ\) b) \(232^\circ\) c) \(\angle AOB\)

- 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 \(d + 2\,\text{cm} \le 5\,\text{cm}\).

- Add the two visibility ranges and compare the result with the distance between the lighthouses. - Model the connecting segment as a number line with \(L_1\) at \(0\). - Determine where the range from \(L_2\) begins on that number line.

1. The sum of the ranges is \(12\,\text{mi} + 15\,\text{mi} = 27\,\text{mi}\). Since \(27\,\text{mi} > 20\,\text{mi}\), the two illuminated closed disks overlap. 2. Measure positions along the segment from \(L_1\), with \(L_1\) at mile \(0\) and \(L_2\) at mile \(20\). 3. The light from \(L_1\) reaches through mile \(12\). 4. The light from \(L_2\) reaches \(15\) miles toward \(L_1\), beginning at mile \(20 - 15 = 5\). 5. Therefore, both lights are visible from mile \(5\) through mile \(12\), a segment \(12 - 5 = 7\,\text{mi}\) long.

a) Yes, because \(12\,\text{mi} + 15\,\text{mi} = 27\,\text{mi} > 20\,\text{mi}\). b) Measured from \(L_1\), both lights are visible from mile \(5\) through mile \(12\). This segment is \(7\,\text{mi}\) long.

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

- Count how many hexagon central angles separate each selected pair of vertices. - Relate equal central angles to their chords. - Split one central triangle into two right triangles to find the chord length.

1. Consecutive vertices of a regular hexagon are separated by \(60^\circ\) central angles. 2. Each side of \(\triangle ACE\) skips one hexagon vertex, so each subtends a \(120^\circ\) central angle. 3. Equal central angles intercept congruent chords, so \(AC=CE=EA\). Therefore, \(\triangle ACE\) is equilateral. 4. For chord \(\overline{AC}\), the two radii and the chord form an isosceles triangle with sides \(6\), \(6\), and included angle \(120^\circ\). 5. Bisecting that central triangle gives a right triangle with hypotenuse \(6\) and acute angle \(60^\circ\), so half the chord is \(3\sqrt3\,\text{cm}\). 6. Therefore, each side of \(\triangle ACE\) is \(6\sqrt3\,\text{cm}\).

a) The three sides subtend equal \(120^\circ\) central angles, so the three chords are congruent and \(\triangle ACE\) is equilateral. b) \(6\sqrt3\,\text{cm}\)

- What transformation can move one center to the other without changing size? - What transformation changes every distance from a center by the same factor? - Compare the two radii to determine that factor, then generalize it to radii \(r_1\) and \(r_2\).

1. Translate circle \(c_1\) so that its center \(A\) moves to \(B\). Translation preserves its radius, so the translated circle still has radius \(4\,\text{cm}\). 2. Dilate the translated circle about \(B\) by scale factor \(\frac{10}{4}=\frac{5}{2}\). Every distance from \(B\) is multiplied by \(\frac{5}{2}\), so the radius becomes \(10\,\text{cm}\). The image is \(c_2\). 3. In general, for circles with positive radii \(r_1\) and \(r_2\), first translate one center to the other and then dilate about the common center by scale factor \(\frac{r_2}{r_1}\). 4. A translation followed by a dilation is a similarity transformation, so any two circles are similar.

a) Translate \(A\) to \(B\), then dilate about \(B\) by scale factor \(\frac{5}{2}\). b) For radii \(r_1,r_2>0\), translate the centers together and dilate by \(\frac{r_2}{r_1}\); therefore, all circles are similar.