Lessons 23–24 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
24
Chapter review: prism and cylinder
Textbook, Part 1: pp. 184–187
GoalReview the chapter on the prism and the cylinder: inscribed and circumscribed figures and surface and volume problems.
New words
circumscribed cylinder · tashqi chizilgan silindrinscribed cylinder · ichki chizilgan silindrequivalent solid · tengdosh jismnet · yoyilma
Explanation
Main formulas of the chapter: prism V = S_base · h, S_lat = P · h (right), S_total = S_lat + 2S_base; cuboid d² = a² + b² + c², V = abc; cylinder S_lat = 2πrh, S_total = 2πr(r + h), V = πr²h. If a prism is circumscribed about a cylinder, the cylinder’s base is the circle inscribed in the prism’s base (its radius is the inradius); if a prism is inscribed in a cylinder, its base is a polygon inscribed in the base circle and the cylinder’s radius equals the circumradius. In a right triangle, for example, the circumradius is half the hypotenuse. In problems with composite solids, cut them into simple pieces: the volume is the sum of the parts, or subtract a removed part from the large solid. Keep units consistent (cm, dm, m and litres). Check the result: radii or edges must not be negative or inappropriate roots of a quadratic.
Worked examples
A right prism has a right-triangle base with legs 5 and 12 (hypotenuse 13) and height 4. The inscribed cylinder: r = (5 + 12 − 13)/2 = 2 (the inradius of a right triangle), V = π · 4 · 4 = 16π. A regular quadrangular prism circumscribed about a cylinder (r = 3, h = 7): the base side is 2r = 6, so S_lat = 4 · 6 · 7 = 168.
A square hole with side 2 is drilled through a cube of edge 10 along its axis between opposite faces. The remaining volume is 1000 − 4 · 10 = 960. The cylinder circumscribed about a regular quadrangular prism (base side 4, height 9): R = 2√2 (half the base diagonal), V = π · 8 · 9 = 72π.
Class activity
“Can and box”: students take a (cleaned) cylindrical can and a square-based box that fits it, find the relation between the box’s inner dimensions and the can’s radius, and compute what percentage of the box’s volume the can fills (πr²h : (2r)²h = π/4 ≈ 78.5 %).
Practice
1
A regular hexagonal prism has base side 4 and height 10. Find the volume of the circumscribed cylinder.
R = 4 (in a regular hexagon the circumradius equals the side), V = π · 16 · 10 = 160π
2
Find the volume of the cylinder circumscribed about a cube of edge 2 (its bases are circumscribed about two opposite faces).
R is half the face diagonal: √2; V = π · 2 · 2 = 4π
3
A square hole with side 3 is cut through a cube of edge 8 between opposite faces. Find the volume that remains.
440
4
What percentage of the volume of a circumscribed square-based prism does the cylinder fill? Explain.
The cylinder has volume πr²h and the prism (2r)²h = 4r²h; the ratio is π/4 ≈ 78.5 %. The heights are equal, so the ratio of base areas decides.