Free math worksheets

- Identify the solid and the shape of its bases from the diagram. - Imagine moving a plane parallel to a base through the solid. - Ask whether the cylinder's radius changes from one height to another.

1. The solid shown is a right circular cylinder. 2. Every slice parallel to a cylinder's bases is a circle congruent to either base, so the cross section has the same radius and area as a base.

A circle congruent to either base.

- Track what happens to a segment perpendicular to the axis as the region turns. - The farthest distance from the axis determines one dimension of the solid. - The length along the axis determines the other main dimension.

1. Rotating the rectangle about one side makes circular disks perpendicular to that side, so the solid is a right circular cylinder. 2. The distance from the axis to the opposite side is \(4\,\text{cm}\), so the cylinder's radius is \(4\,\text{cm}\). 3. The side on the axis has length \(7\,\text{cm}\), so the cylinder's height is \(7\,\text{cm}\).

A right circular cylinder with radius \(4\,\text{cm}\) and height \(7\,\text{cm}\).

- Notice where the cutting plane passes relative to the sphere's center. - Compare this cut with the sphere's widest circular slice.

1. The cutting plane passes through the sphere's center \(O\). 2. A plane through the center of a sphere makes the largest possible circular cross-section, whose radius equals the sphere's radius.

The cross-section is a circle of radius \(r\).

- For part a, compare the cutting plane directly with the prism's end faces. - For part b, picture the cut extending along the prism's length rather than across it. - Focus on the boundary segments where each plane meets the faces of the prism.

1. A plane parallel to an end face produces a cross section congruent to that face, so the cross section in part a is a triangle. 2. A plane perpendicular to the end faces and parallel to the prism's long edges extends along the prism's length, so the cross section in part b is a rectangle.

a) Triangle. b) Rectangle.

- Imagine the curved edge rotating around the diameter. - Track points that are the same distance from the midpoint of the diameter. - The greatest distance from the axis becomes the radius of the three-dimensional solid.

1. Every point of the semicircle rotates around the diameter. 2. The semicircular region sweeps out a sphere. 3. The maximum distance from the axis is the semicircle's radius, \(4\,\text{cm}\), so the sphere also has radius \(4\,\text{cm}\).

A sphere with radius \(4\,\text{cm}\).

- An axial plane passes through both the cone's vertex and the center of its circular base. - Determine what full base length appears in that plane. - Use the cone's perpendicular height as the triangle's height.

1. An axial cut through a right circular cone produces an isosceles triangle. 2. The triangle's base is the cone's diameter, \(2r=10\,\text{cm}\), and its perpendicular height is \(12\,\text{cm}\). 3. Its area is \(A=\frac{1}{2}(10)(12)=60\,\text{cm}^2\).

The cross-section is an isosceles triangle with area \(60\,\text{cm}^2\).

- Track the paths of the endpoints farthest from the axis. - The two distances from the axis are different, so the two circular ends will have different radii. - Compare the result with a cone whose tip has been cut off parallel to its base.

1. The two horizontal sides are perpendicular to the axis and have different lengths. 2. When the trapezoid rotates, those sides sweep out two circular bases with different radii. 3. The slanted side sweeps out the tapered lateral surface joining those circles. 4. Therefore, the solid of revolution is a conical frustum.

The rotation produces a conical frustum. Each horizontal side sweeps out a circular base, and the side's length becomes that base's radius.

- Think about how far the shaded region extends on each side of the rotation axis. - Compare rotating a quarter disk with rotating a semicircle about a diameter. - Track the constant distance from the axis in the rectangular panel.

1. In panel a), the shaded region is a quarter disk. Rotating it about one radius sweeps out exactly one half of a sphere, so it generates a hemisphere. 2. In panel b), the shaded region is a semicircle. Rotating it about its diameter sweeps out a full sphere. 3. In panel c), the shaded region is a rectangle. Rotating it about one side sweeps out a cylinder.

Panel a) generates the hemisphere. Panel b) generates a sphere, and panel c) generates a cylinder.

- Think about all points in the cutting plane that are also on the sphere. - Use the perpendicular distance from the sphere's center to the plane. - Distinguish an actual geometric section from how a circle can look in perspective.

1. Let the sphere have radius \(r\), and let the cutting plane be at perpendicular distance \(d\) from the center. 2. When \(0\le d<r\), every boundary point of the section is the same in-plane distance \(\sqrt{r^2-d^2}\) from the foot of that perpendicular, so the section is a circle. 3. When \(d=r\), the plane is tangent and the cross-section is a single point. When \(d>r\), the plane does not meet the sphere. 4. Therefore, a noncircular ellipse is not an exact plane cross-section of a sphere; an ellipse may only be how a circular section appears in a perspective drawing.

No. A plane that cuts through a sphere makes a circular cross-section; a tangent plane meets it at one point, and a plane farther away does not intersect it. A noncircular ellipse is not an exact plane cross-section of a sphere.

- Think about how slices of a right circular cone change as the cutting plane moves from the base toward the apex. - Compare the symmetry of the desired circular cross section with the symmetry of the cone. - For the second part, consider what happens when the cutting plane contains the point where all of the cone's generators meet.

1. The solid shown is a right circular cone. A plane parallel to its circular base and located between the apex and the base produces a smaller circular cross section. 2. Such a plane does not pass through the apex. A nondegenerate plane through the apex meets the cone along straight generator segments rather than forming a circular slice.

Use a plane parallel to the base between the apex and the base. A nondegenerate plane through the apex cannot produce the circle because its intersection follows straight generator segments of the cone rather than a circular layer.

- A plane parallel to a pyramid's base produces a cross section similar to the base. - Compare the apex-to-cut distance with the full perpendicular height of the pyramid. - Use the resulting linear scale factor before finding the area of the cross section.

1. Because the cutting plane is parallel to the square base, the cross section is a square. 2. The small pyramid above the cross section has height \(3\,\text{cm}\), compared with \(9\,\text{cm}\) for the full pyramid, so its linear scale factor is \(\frac{3}{9}=\frac{1}{3}\). 3. The cross-section side length is \(12\cdot\frac{1}{3}=4\,\text{cm}\). 4. Its area is \(4^2=16\,\text{cm}^2\).

The cross section is a square with area \(16\,\text{cm}^2\).

- In each panel, decide which leg lies on the rotation axis. - The other perpendicular leg sweeps out the circular base and therefore determines the radius. - Use the same cone-volume formula for both rotations before comparing the exact results.

1. In panel a, the leg on the axis becomes the cone's height, \(8\,\text{cm}\), and the perpendicular leg becomes its radius, \(3\,\text{cm}\). 2. In panel b, the leg on the axis becomes the cone's height, \(3\,\text{cm}\), and the perpendicular leg becomes its radius, \(8\,\text{cm}\). 3. The first volume is \(\frac{1}{3}\pi(3^2)(8)=24\pi\,\text{cm}^3\). The second is \(\frac{1}{3}\pi(8^2)(3)=64\pi\,\text{cm}^3\), so panel b produces the greater volume.

a) Cone: radius \(3\,\text{cm}\), height \(8\,\text{cm}\). b) Cone: radius \(8\,\text{cm}\), height \(3\,\text{cm}\). c) The volumes are \(24\pi\,\text{cm}^3\) and \(64\pi\,\text{cm}^3\), respectively; panel b produces the greater volume.

- Identify which two base vertices lie in the cutting plane. - The triangle's base is a diagonal of the square base. - The pyramid's perpendicular height lies in this plane because the diagonal passes through the base center.

1. The plane contains the apex and two opposite base vertices, so the cross-section is an isosceles triangle. 2. Its base is a diagonal of the square: \(d=6\sqrt{2}\,\text{cm}\). 3. Because the base diagonal passes through the square's center, the pyramid's \(4\,\text{cm}\) perpendicular height lies in the cutting plane and is perpendicular to that diagonal. 4. The cross-sectional area is \(A=\frac{1}{2}(6\sqrt{2})(4)=12\sqrt{2}\,\text{cm}^2\).

The cross-section is an isosceles triangle with exact area \(12\sqrt{2}\,\text{cm}^2\).

- Distances from the rotation axis become radii after rotation. - Because the rectangle does not touch the axis, determine both an inner and an outer radius. - Model the volume as an outer cylinder with an inner cylinder removed.

1. The near side of the rectangle is \(2\,\text{cm}\) from the axis, so the rotation leaves a cylindrical hole of radius \(2\,\text{cm}\). 2. The rectangle is \(3\,\text{cm}\) wide, so the outer radius is \(2+3=5\,\text{cm}\). Its height is \(7\,\text{cm}\). 3. The solid is a hollow right circular cylinder. Its volume is the outer cylinder minus the inner cylinder: \(V=\pi(5^2)(7)-\pi(2^2)(7)=147\pi\,\text{cm}^3\).

The rotation produces a hollow right circular cylinder with exact volume \(147\pi\,\text{cm}^3\).

- Use a vertical cross-section through the apex and the midpoints of opposite base sides. - In that cross-section, the sphere becomes a circle tangent to all three sides of a triangle. - Find the equal side lengths of the triangle with the Pythagorean theorem. - Use the relationship \(A=rs\) for a triangle's area, inradius, and semiperimeter.

1. Take a vertical cross-section through the apex and the midpoints of two opposite sides of the square base. The cross-section is an isosceles triangle with base \(160\,\text{m}\) and height \(60\,\text{m}\). The sphere appears as the triangle's incircle. 2. Each equal side of the triangle has length \(\sqrt{80^2+60^2}=100\,\text{m}\). 3. The triangle's area is \(\frac12\cdot160\cdot60=4800\,\text{m}^2\), and its semiperimeter is \(\frac{160+100+100}{2}=180\,\text{m}\). 4. For a triangle with inradius \(r\), \(A=rs\). Therefore, \(r=\frac{4800}{180}=\frac{80}{3}\,\text{m}\approx26.67\,\text{m}\). 5. Because the chamber is tangent to the floor, its center is one radius above the floor. Its center is therefore \(\frac{80}{3}\,\text{m}\approx26.67\,\text{m}\) high.

The radius is \(\frac{80}{3}\,\text{m}\approx26.67\,\text{m}\), and the center is the same distance above the floor.

- Use the altitude to think about the triangular region in two parts. - Track what happens to each part during a full rotation about the base. - Identify the radius and axial length needed to compute the resulting volume.

1. The \(3\,\text{cm}\) altitude meets the midpoint of the \(8\,\text{cm}\) base, splitting the region into two right triangles with base length \(4\,\text{cm}\). 2. When the region rotates about the axis, each right triangle generates a cone with radius \(3\,\text{cm}\) and height \(4\,\text{cm}\). 3. The two cones share the same circular base and point in opposite directions, so the solid is a double cone rather than one cone. 4. Its volume is \(2\left(\frac{1}{3}\pi(3^2)(4)\right)=24\pi\,\text{cm}^3\).

Youssef overlooks that the altitude divides the generating region into two right triangles. The rotation produces two congruent cones joined at a common circular base, with total volume \(24\pi\,\text{cm}^3\).