Lessons 39–40 · 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
40
Chapter review: pyramid and cone
Textbook, Part 2: pp. 140–145
GoalReview the chapter on the pyramid and the cone: inscribed and circumscribed solids, similar solids and the surface and volume of composite shapes.
New words
inscribed pyramid · ichki chizilgan piramidacircumscribed cone · tashqi chizilgan konusaxial section · o‘q kesimsimilar solids · o‘xshash jismlar
Explanation
Main formulas of the chapter: pyramid V = ⅓ S_base H, regular pyramid S_lat = ½ P l; cone S_lat = πrl, V = ⅓ πr²h; for frustums V = (H/3)(Q₁ + √(Q₁Q₂) + Q₂), where for a cone Q = πr². If a pyramid is inscribed in a cone, they share the same height and the pyramid’s base is a polygon inscribed in the cone’s base (the cone’s radius is the circumradius); if a pyramid is circumscribed about a cone, the cone’s base is the circle inscribed in the polygon (the radius is the inradius). For a cone with an equilateral axial section, l = 2r and h = r√3. For similar solids with linear ratio k, the areas are in ratio k² and the volumes in ratio k³. Split a composite solid into pieces or subtract a removed part from a larger solid; keep units consistent and check the result with a sketch or an extreme case.
Worked examples
A regular quadrangular pyramid is circumscribed about a cone with r = 3, h = 4. The pyramid’s base side is 2r = 6, V_pyramid = ⅓ · 36 · 4 = 48, V_cone = ⅓ π · 9 · 4 = 12π. The ratio V_cone : V_pyramid = 12π/48 = π/4 ≈ 0.785.
A cone’s axial section is an equilateral triangle with side 4. r = 2, l = 4, h = 2√3. S_lat = π · 2 · 4 = 8π, S_total = 8π + 4π = 12π, V = ⅓ π · 4 · 2√3 = 8√3π/3.
Class activity
“Formula card”: on one A4 sheet write the surface and volume formulas for the pyramid, frustum of a pyramid, cone and frustum of a cone and add one numerical example next to each. Swap with a friend and check each other’s examples.
Practice
1
What is the ratio (pyramid : cone) of the volume of a square-based pyramid inscribed in a cone to the volume of the cone?
2/π (the square inscribed in a circle of radius r has area 2r², the disc has area πr², and the heights are equal)
2
Find the volume of the cone whose axial section is an equilateral triangle with side 6.
r = 3, h = 3√3, V = ⅓ π · 9 · 3√3 = 9√3π
3
A cone has volume 54. A plane parallel to the base divides the height in the ratio 1 : 2 counting from the apex. What is the volume of the small cone cut off?
2
4
Why can a circle be circumscribed about the base polygon if the pyramid is inscribed in a cone?
The vertices of the pyramid’s base lie on the circle of the cone’s base, so a single circle (the cone’s base) passes through all vertices of the polygon — which is exactly what a circumscribed circle is.