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| 13 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
In the diagram, a cube is given. Try to identify how many different cross-sections can be formed. What geometric shapes do these cross-sections have?
These are the possible interactions with the presented applet.
Rotateor click on the circle and move it around.
Reset.
These are possible interactions with the presented applet.
These are the possible interactions with the presented applet.
Rotatebutton.
Resetbutton.
From the previous applet, it could be concluded that rotating a rectangle about one of its sides forms a right cylinder. What are the cross-sections of a right cylinder? There are several types depending on the position of an intersecting plane.
Case | Position of the Plane | Cross-Section |
---|---|---|
1 | Perpendicular to the base | Rectangle |
2 | Parallel to the base | Circle |
3 | Diagonal to the base | Ellipse |
The following applet illustrates each type of the mentioned cross-sections.
Analyze the shapes of the cross-sections closely. What conclusions can be drawn about the sides and angles of the cross-sections?
To pair each plane with the corresponding type of cross-section, each given case should be investigated.
The first intersecting plane is parallel to the cylinder's bases. Therefore, the cross-section it creates has the same shape as the bases, which are circles.
Hence, the cross-section is also a circle.
The intersecting plane is perpendicular to both bases of the cylinder. Since the cylinder is right, the cross-section has four right angles. Additionally, the longer sides of the cross-section are parallel to the height of the cylinder and have the same lengths.
Furthermore, the shorter sides are parallel chords of the cylinder's bases, which also have equal lengths. Therefore, the cross-section is a rectangle.
The last plane intersects only the curved surface of the cylinder, so it does not have straight sides. The cross-section has a circular shape, but it is not a circle.
Recall the possible cross-sections of a right cylinder depending on the position of the intersecting plane.
The Plane's Position | Cross-Section |
---|---|
Perpendicular to the base | Rectangle |
Parallel to the base | Circle |
Diagonal to the base | Ellipse |
Since the intersecting plane is diagonal to the bases of the cylinder, it can be concluded that the cross-section is an ellipse.
The given two-dimensional shape is a cross-section of a right cylinder. Identify whether the cross-section is parallel, perpendicular, or diagonal to the base of the cylinder.
These are the possible interactions with the presented applet.
Rotate.
Reset.
From the previous applet, it can be observed that when rotating a right triangle about its height, a right cone is formed. What are the cross-sections of a right cone? Here are some possible types depending on the position of an intersecting plane.
The Plane's Position | Cross-Section |
---|---|
Perpendicular to the base | Triangle |
Parallel to the base | Circle |
Diagonal to the base | Ellipse |
Analyze the shapes of the cross-sections closely. What conclusions can be drawn about the sides and angles of the cross-sections?
To pair each plane with the corresponding type of cross-section, each given case should be analyzed.
The first intersecting plane is parallel to the cone's base. Consequently, the cross-section it creates has the same shape as the base, which is a circle.
Therefore, the cross-section is also a circle.
The second plane goes through the curved surface of the cone, so it does not have any straight sides. The cross-section has a circular shape, but it is not a circle.
To identify the cross-section, the possible cross-sections of a right cone can be reviewed.
The Plane's Position | Cross-Section |
---|---|
Perpendicular to the base | Triangle |
Parallel to the base | Circle |
Diagonal to the base | Ellipse |
It can be noticed that the intersecting plane is diagonal to the bases of the cone. Therefore, the cross-section can be concluded to be an ellipse.
The last intersecting plane is perpendicular to the base of the cone and goes through the vertex of the cone. The cross-section has three vertices, which are connected by sides.
Therefore, the cross-section is a triangle.
The given 2D shape is a cross-section of a right cone. Identify whether the cross-section is parallel, perpendicular, or diagonal to the base of the cone.
3D Object | Intersecting Plane's Position | Cross-Section |
---|---|---|
Sphere | Any position | Circle |
Right Cylinder | Perpendicular to the base | Rectangle |
Parallel to the base | Circle | |
Diagonal to the base | Ellipse | |
Right Cone | Perpendicular to the base | Triangle |
Parallel to the base | Circle | |
Diagonal to the base | Ellipse |
What is the surface area of the solid created when the triangle is rotated around the y-axis? Give the answer in terms of π.
When we revolve the triangle about the y-axis, we get a solid of revolution that is a cone. However, the given triangle will produce a solid with a cone-shaped hole in the middle. In the following diagram, we see the resulting solid after completing the rotation about the y-axis.
The surface area of this composite solid is the surface area of the big cone, minus the circular area in the center, plus the lateral surface area of the hole. To calculate the surface area of a cone, we need its radius and slant height.
The big cone has a radius and height of 4 and 8 centimeters. The small cone has a radius and height of 1 and 8 centimeters. With this information, we can calculate the slant heights of each cone using the Pythagorean Theorem. Big Cone: (l_B)^2=4^2+8^2 ⇒ l_B = 4sqrt(5) [0.1em] Small Cone: (l_S)^2=1^2+8^2 ⇒ l_S =sqrt(65) Now we can calculate the areas we need.
Let's calculate the surface area of the big cone SA_B as if it did not have a hole.
As previously explained, to obtain the surface area of the composite solid, we have to add the lateral surface area of the hole and subtract the base area. Let's determine these parts separately. Base area:& π (1)^2 = π Lateral area:& π (1)(sqrt(65))= sqrt(65)π
Now we can find the surface area of the composite solid. SA_(CS)= 16π+16sqrt(5)π- π+ sqrt(65)π ⇓ SA_(CS)=15π+16sqrt(5)π+sqrt(65)π The surface area of the composite solid is 15π+16sqrt(5)π+sqrt(65)π square units.