A=πr^2 + 2 πrh+πrl and r^2 +h 2 =l^2
42 + 92 =l^2
l≈ 9. 8A≈π 42 + 2 π∗ 4 ∗ 12 +π∗ 4 ∗ 9. 8
A≈475 cm^2- A cone of radius 7 cm is carved out of a square prism of the same height. The square base of the prism has area
225 cm^2 and height 30 cm. What is the volume of this composite solid?
Answer:
V=Volume of Prism[U+0080][U+0093]Volume of ConeV= 225 ∗ 30 [U+0080][U+0093]1
3
∗π 72 ∗ 30 = 6750 [U+0080][U+0093] 490 πcm^3 ≈5210 cm^3Spheres
Expand on Circles –Students learned about circles earlier in the course. Review and expand on this knowledge as
they learn about spheres. Ask the students what they know about circles. Being able to demonstrate their knowledge
will build their confidence and activate their minds. Now modify the definitions that the students have provided to
fit the three-dimensional sphere. Students will learn the new material quickly and will remember it because it is now
neatly filed away with their knowledge of circles.
Explore Cross-Sections –One of the goals of this chapter is to develop the students’ ability to think about three-
dimensional objects. Most students will need a significant amount of practice before becoming competent at this
skill. Take some time and ask the students to think about what the cross-sections of a sphere and a plane will look
like. Explore trends. What happens to the cross-section as the plane moves farther away from the center of the
circle? A cross-section that passes through the center of the sphere makes the largest possible circle, or the great
circle of the sphere.
Cylinder to Sphere –It would be a good exercise for students to take the formula for the surface area of a cylinder
and derive the formula for the surface area of a sphere. It is just a matter of switching a few variables, but it would
be a good exercise for them. During the lesson, ask them to do it in their notes, wait a few minutes and then do it on
the board or ask one of them to put their work on the board. It should look something like this:
Acylinder=bases+lateral area
Acylinder= 2 ∗πr^2 + 2 πrh
Asphere= 2 ∗πr^2 + 2 πr∗r substitute inh=r
Asphere= 2 ∗πr^2 + 2 πr^2
Asphere= 4 πr^2Note to students that in the last line the like terms where combine. This can only be done because both terms had
the factorπr^2. The coefficients could have been different, but to combine terms using the distributive property they
2.11. Surface Area and Volume