Ninth Grade Mathematics: Composite Solids & Material Displacement
In real-world engineering and architecture, objects rarely exist as pure isolated geometric shapes; instead, they are composite solids constructed by combining or hollowing out cylinders, cones, hemispheres, and prisms. In this chapter, you will master the calculation of surface areas and volumes of composite solids, understand Archimedes' principle of fluid displacement, and solve recasting problems.
The Logic of Composite Volumes versus Surface Areas
When analyzing a composite solid formed by joining two or more standard solids, volume and surface area behave in completely different ways:
Volume of Composite Solids: ADDITIVE
Volume is the measure of internal space:
Total Volume = Volume of Solid 1 + Volume of Solid 2 + ...
Surface Area of Composite Solids: NOT SIMPLY ADDITIVE!
Surface area measures ONLY EXTERNALLY EXPOSED SURFACES.
Any surfaces that are glued, fused, or touching inside the joint are NO LONGER EXPOSED
and must be EXCLUDED from the total surface area.
Example: A Toy Formed by Mounting a Cone atop a Hemisphere:
Cone
* Total Volume:
/|\ Volume = Vol(Cone) + Vol(Hemisphere)
/ | \ = (1/3)*pi*r^2*h + (2/3)*pi*r^3
*--+--*
. ' | ' . Total Surface Area:
/ * \ The flat circular base is INSIDE the joint!
; ; TSA = CSA(Cone) + CSA(Hemisphere)
\ / = pi * r * l + 2 * pi * r^2
. _ _ _ . '
Hemisphere
Notice that neither the circular base of the cone nor the circular base of the hemisphere is included in the surface area, because they are welded together inside the toy.
Archetypes of Composite Solids
Common composite figures in high school mathematics include:
Composite Figure Structural Components Total Surface Area Formula
-----------------------------------------------------------------------------------------
Circus Tent Cylinder base + Conical roof CSA(cylinder) + CSA(cone)
Storage Silo Cylinder body + Hemisphere cap Base + CSA(cylinder) + CSA(hemi)
Medicine Capsule Cylinder + Two Hemisphere ends CSA(cylinder) + 2 * CSA(hemi)
Wooden Test Tube Stand Prism block with cylindrical holes TSA(prism) + Cyl_CSA - Hole_areas
Material Conservation in Recasting Problems
When a metal object is melted down and remolded into a different shape (such as melting a copper cylinder to draw a long thin cylindrical wire), the physical mass and volume remain perfectly conserved:
The Recasting Conservation Principle:
Volume of Original Solid(s) = Volume of Recast Solid(s)
Example: Melting a metallic sphere of radius R to cast small cones of radius r and height h:
Volume of Sphere = N * (Volume of one cone)
(4/3) * pi * R^3 = N * [ (1/3) * pi * r^2 * h ]
Solve for N (number of cones).
Archimedes' Principle of Fluid Displacement
When an insoluble solid is completely submerged in a liquid container, it pushes aside a volume of liquid exactly equal to its own physical volume:
Volume of Submerged Solid = Volume of Liquid Displaced
For a cylindrical container of radius R:
Volume of Displaced Liquid = pi * R^2 * (Rise in water level)
Measuring the rise in liquid level inside a graduated cylinder is the standard laboratory method for determining the volume of irregular solid objects.
Chapter Exercises
Exercise 1: A solid toy is in the shape of a right circular cone mounted on a hemisphere with a common base radius of 3.5 cm. If the total height of the toy is 15.5 cm, calculate the volume of the toy and its total external surface area (use 22/7 for pi).
Exercise 2: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The total length of the entire capsule is 14 mm, and the diameter of the capsule is 5 mm. Find the surface area of the capsule.
Exercise 3: A cylindrical bucket of base radius 18 cm and height 32 cm is filled to the brim with sand. This sand is poured onto the ground to form a conical heap. If the height of the conical heap is 24 cm, find the base radius and slant height of the sand heap.
Solutions and Explanations
Solution 1: The common radius is r = 3.5 cm = 7/2 cm. The total height of the toy is 15.5 cm. The height of the hemispherical portion is simply its radius: h_hemi = 3.5 cm. Therefore, the vertical height of the conical portion is h_cone = 15.5 - 3.5 = 12 cm. First compute the slant height l of the cone: l = sqrt(r^2 + h^2) = sqrt((3.5)^2 + 12^2) = sqrt(12.25 + 144) = sqrt(156.25) = 12.5 cm. Volume of the toy = Volume of cone + Volume of hemisphere = (1/3)pir^2h + (2/3)pir^3 = (1/3)pir^2(h + 2r) = (1/3) * (22/7) * (7/2) * (7/2) * (12 + 2(3.5)) = (1/3) * (77 / 2) * (12 + 7) = (77 / 6) * 19 = 1,463 / 6 = 243.83 cubic cm. Total external surface area = CSA of cone + CSA of hemisphere = pirl + 2pir^2 = pir(l + 2r) = (22/7) * (7/2) * (12.5 + 2(3.5)) = 11 * (12.5 + 7) = 11 * 19.5 = 214.5 square cm.
Solution 2: The diameter is 5 mm, so radius r = 5 / 2 = 2.5 mm. The two hemispherical ends each have length equal to the radius (2.5 mm each). The length of the middle cylindrical part is h = Total length - 2 * r = 14 - 2(2.5) = 14 - 5 = 9 mm. The external surface area consists of the curved surface of the cylinder plus the curved surfaces of the two hemispheres (which combine into the surface area of one full sphere): Surface Area = CSA(cylinder) + 2 * CSA(hemisphere) = 2pirh + 4pir^2 = 2pir(h + 2r). Substitute values: Surface Area = 2 * (22/7) * (2.5) * (9 + 2*(2.5)) = (110 / 7) * (9 + 5) = (110 / 7) * 14 = 110 * 2 = 220 square mm.
Solution 3: Volume of sand in cylindrical bucket = pi * r_cyl^2 * h_cyl = pi * (18^2) * 32 = pi * 324 * 32 = 10,368 * pi cm^3. The sand is poured to form a cone of height h_cone = 24 cm and unknown radius R. Volume of conical heap = (1/3) * pi * R^2 * h_cone = (1/3) * pi * R^2 * 24 = 8 * pi * R^2. Equating the two volumes: 8 * pi * R^2 = 10,368 * pi. Divide both sides by 8*pi: R^2 = 10,368 / 8 = 1,296. Taking the square root gives base radius R = 36 cm. To find the slant height l of the conical heap, use the Pythagorean theorem: l = sqrt(R^2 + h^2) = sqrt(36^2 + 24^2) = sqrt(1296 + 576) = sqrt(1872) = sqrt(144 * 13) = 12 * sqrt(13) approx 12 * 3.605 = 43.27 cm.