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 \(7\,\text{cm}\).

- 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.

a) A plane parallel to an end face produces a cross section congruent to that face, so the cross section is a triangle. b) A plane perpendicular to the end faces and parallel to the prism's long edges produces 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 \(4\,\text{cm}\).

- 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, the leg on the axis becomes the cone's height. - 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.

a) Rotating about the \(8\,\text{cm}\) leg produces a cone with height \(8\,\text{cm}\) and radius \(3\,\text{cm}\). b) Rotating about the \(3\,\text{cm}\) leg produces a cone with height \(3\,\text{cm}\) and radius \(8\,\text{cm}\). c) 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 rotation about the \(3\,\text{cm}\) leg 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.

- 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.

- Draw or use the perpendicular from the triangle's apex to the rotation axis. - Think about what each half of the triangular region generates when it rotates. - Identify which segment becomes the radius of the shared circular cross section before combining the two solid volumes.

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 \(8\,\text{cm}\) 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 \(2\left(\frac{1}{3}\pi(3^2)(4)\right)=24\pi\,\text{cm}^3\).

The student 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 \(24\pi\,\text{cm}^3\).