14.2 3D Solid Geometry (Katı Cisimler): Prisms, Cylinders, Pyramids, Cones & Spheres

14.2 3D Solid Geometry (Katı Cisimler): Prisms, Cylinders, Pyramids, Cones & Spheres

Quick Overview: Instead of memorizing dozens of isolated 3D formulas, classify every solid into one of three families: Prisms & Cylinders (uniform cross-section from base to top, $V = A_{\text{base}} \cdot h$), Pyramids & Cones (tapering to a single apex, $V = \frac{1}{3} A_{\text{base}} \cdot h$), and Spheres ($V = \frac{4}{3}\pi R^3$). When a question asks for the shortest path of an ant crawling along the surface of a solid, unroll (açınım) the faces into a flat 2D net and draw a straight line segment.

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1. Universal Volume & Surface Area Classification

Uniform Cross-Section Solids: Prisms (Prizmalar) & Cylinders (Silindirler)

Any right prism or cylinder has two congruent parallel bases connected by vertical lateral faces:

• Volume (Hacim):

$$V = A_{\text{base}} \cdot h$$

• Lateral Area (Yanal Alan):

$$A_{\text{lateral}} = P_{\text{base}} \cdot h \quad (\text{Base Perimeter} \times \text{Height})$$

• Total Surface Area (Tüm Alan):

$$A_{\text{total}} = 2 A_{\text{base}} + A_{\text{lateral}}$$

Pointed Apex Solids: Pyramids (Piramitler) & Cones (Koniler)

Any pyramid or cone tapers linearly from a base of area $A_{\text{base}}$ to a single apex at perpendicular height $h$:

• Volume (Hacim):

$$V = \frac{1}{3} A_{\text{base}} \cdot h$$

• Total Surface Area (Tüm Alan):

$$A_{\text{total}} = A_{\text{base}} + A_{\text{lateral}}$$

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2. Rectangular Cuboids (Dikdörtgenler Prizması) & Cubes (Küp)

Rectangular Cuboid (Dikdörtgenler Prizması)

For a rectangular box with mutually perpendicular edge lengths $a, b, c$:

• Sum of all 12 edges: $L_{\text{edges}} = 4(a + b + c)$.
• Volume: $V = a \cdot b \cdot c$.
• Total Surface Area: $A_{\text{surface}} = 2(ab + bc + ac)$.
• Space Diagonal (Cisim Köşegeni): By applying the Pythagorean theorem twice (first across a base face $\sqrt{a^2 + b^2}$, then vertically with $c$):

$$e = \sqrt{a^2 + b^2 + c^2}$$

• The Polynomial-Cuboid Bridge: Squaring the sum of the three dimensions links the edge sum, space diagonal, and surface area directly:

$$(a + b + c)^2 = \underbrace{a^2 + b^2 + c^2}_{e^2} + \underbrace{2(ab + bc + ac)}_{A_{\text{surface}}} \implies (a + b + c)^2 = e^2 + A_{\text{surface}}$$

Cube (Küp)

Setting $a = b = c$ gives a cube with $6$ congruent square faces and $12$ equal edges:

• Volume: $V = a^3$
• Total Surface Area: $A_{\text{total}} = 6a^2$
• Face Diagonal (Yüzey Köşegeni): $f = a\sqrt{2}$
• Space Diagonal (Cisim Köşegeni): $e = a\sqrt{3}$

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3. Right Circular Cylinders (Dik Dairesel Silindir)

A cylinder of base radius $r$ and height $h$ has circular bases of area $\pi r^2$ and circumference $2\pi r$. Unrolling its lateral surface produces a rectangle of width $2\pi r$ and height $h$:

• Volume: $V = \pi r^2 h$
• Lateral Area: $A_{\text{lateral}} = 2\pi r h$
• Total Surface Area: $A_{\text{total}} = 2\pi r^2 + 2\pi r h = 2\pi r(r + h)$

Shortest Ant Path (En Kısa Karınca Yolu) & Tilted Cylinders

1. Shortest Wrapping Path: If an ant starts at the bottom rim and spirals $k$ full turns around a cylinder of radius $r$ to reach the top rim at height $h$, unroll the lateral surface $k$ times side by side into a flat rectangle of horizontal width $W = k \cdot (2\pi r)$ and vertical height $h$. The shortest path is the hypotenuse:

$$d = \sqrt{(2\pi r \cdot k)^2 + h^2}$$

1. Tilted Cylinder Rule (Eğik Silindirde Su Hacmi): When a partially filled cylinder is tilted by an angle $\alpha$, the water surface remains parallel to the horizontal ground (forming alternate interior angles $\alpha$ with the cylinder's base). In the upper cylindrical zone bounded by the lowest and highest water contact points (whose difference along the cylinder axis is $\Delta h = 2r\tan\alpha$ when its axis tilts by $\alpha$ from vertical), exactly half of that upper zone is water and half is empty air:

$$V_{\text{empty top zone}} = V_{\text{water in top zone}} = \frac{1}{2}\pi r^2 \Delta h$$

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4. Cones (Dik Koni), Frustum Scaling, Spheres (Küre) & Regular Pyramids

Right Circular Cone (Dik Dairesel Koni)

In a right circular cone with base radius $r$ and vertical height $h$, the slant height (ana doğru) $l$ satisfies:

$$l = \sqrt{r^2 + h^2}$$

• Volume: $V = \frac{1}{3}\pi r^2 h$
• Lateral Area (Yanal Alan): $A_{\text{lateral}} = \pi r l$
• Total Surface Area: $A_{\text{total}} = \pi r^2 + \pi r l = \pi r(r + l)$
• Unrolled Sector Angle (Açınım Merkez Açısı $\alpha$) Golden Formula: When the lateral surface of a cone is unrolled into a flat circular sector of radius $l$, its arc length equals the base circumference $2\pi r$. Equating $2\pi l \cdot \frac{\alpha}{360^\circ} = 2\pi r$ yields:

$$\frac{r}{l} = \frac{\alpha}{360^\circ}$$

Volume Similarity Scaling (Hacimler Oranı $= k^3$)

When any cone or pyramid is sliced by a plane parallel to its base at distance $h_1$ from the apex (out of total height $h_2$), the top small solid is similar to the whole solid with linear similarity ratio $k = \frac{h_1}{h_2} = \frac{r_1}{r_2}$:

• Cross-Sectional & Lateral Area Ratio: $\frac{A_1}{A_2} = k^2$
• Volume Ratio: $\frac{V_1}{V_2} = k^3$
• Equal-Height Slicing (kesik koni / kesik piramit): If the altitude is divided into equal segments ($1 : 2 : 3 : \dots$), cumulative volumes from the apex scale as $1^3 : 2^3 : 3^3 = 1 : 8 : 27$, so the individual slices from top to bottom have volumes:

$$V, \quad 7V, \quad 19V, \quad 37V, \dots$$

Sphere (Küre), Regular Square Pyramid & Regular Tetrahedron (Düzgün DörtYüzlü)

• Sphere (Küre) of radius $R$:

$$V = \frac{4}{3}\pi R^3, \qquad A_{\text{surface}} = 4\pi R^2$$

Slicing a sphere at distance $d$ from the center creates a circular cross-section (ara kesit daire) of radius $r = \sqrt{R^2 - d^2}$.

• Regular Square Pyramid (Düzgün Kare Piramit): Base is a square of side $a$, apex projects to the center of the square. The slant height of a triangular face $h_{\text{face}}$ satisfies $h^2 + \left(\frac{a}{2}\right)^2 = h_{\text{face}}^2$, and a lateral edge $l$ satisfies $h^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 = l^2$.
• Regular Tetrahedron (Düzgün DörtYüzlü): A pyramid bounded by 4 congruent equilateral triangles of edge $a$:

$$h = \frac{a\sqrt{6}}{3}, \qquad A_{\text{total}} = a^2\sqrt{3}, \qquad V = \frac{a^3\sqrt{2}}{12}$$

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Summary Reference Table: 3D Solids

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Worked Examples

Worked Example 1: Rectangular Cuboid Edge Sum & Space Diagonal

Problem: The sum of the lengths of all $12$ edges of a rectangular cuboid (dikdörtgenler prizması) is $96\text{ cm}$, and its space diagonal (cisim köşegeni) measures $e = 14\text{ cm}$. What is the total surface area of the cuboid?

Step-by-Step Solution:

1. Find $a + b + c$ from the total edge sum: A rectangular cuboid has $4$ edges of each dimension $a, b, c$:

$$4(a + b + c) = 96 \implies a + b + c = 24\text{ cm}$$

1. Use the algebraic cuboid identity: Recall that $(a + b + c)^2 = e^2 + A_{\text{surface}}$, where $e^2 = a^2 + b^2 + c^2 = 14^2 = 196$ and $A_{\text{surface}} = 2(ab + bc + ac)$:

$$24^2 = 14^2 + A_{\text{surface}} \implies 576 = 196 + A_{\text{surface}} \implies A_{\text{surface}} = 380\text{ cm}^2$$

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Worked Example 2: Right Circular Cone Unrolled Sector & Frustum Volume

Problem: A right circular cone has base radius $r = 6\text{ cm}$ and vertical height $h = 8\text{ cm}$.

1. Find its slant height $l$, total surface area, and the central angle $\alpha$ of its unrolled lateral sector.
2. If the cone is cut parallel to its base halfway down from the apex, find the volume of the bottom frustum (kesik koni).

Step-by-Step Solution:

1. Compute $l$, surface area, and sector angle $\alpha$:

$$l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = 10\text{ cm}$$ $$A_{\text{total}} = \pi r^2 + \pi r l = \pi(36) + \pi(6)(10) = 96\pi\text{ cm}^2$$

Using the unrolled sector formula $\frac{r}{l} = \frac{\alpha}{360^\circ}$:

$$\frac{6}{10} = \frac{\alpha}{360^\circ} \implies \alpha = \frac{3}{5} \times 360^\circ = 216^\circ$$

1. Find the frustum volume via $k^3$ similarity scaling: The full cone volume is:

$$V_{\text{total}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(36)(8) = 96\pi\text{ cm}^3$$

Slicing at the midpoint ($k = \frac{1}{2}$) splits the volume in the ratio $1 : 7$ (top mini-cone $V = \frac{1}{8}V_{\text{total}} = 12\pi\text{ cm}^3$ and bottom frustum $7V$). Thus:

$$V_{\text{frustum}} = \frac{7}{8}(96\pi) = 84\pi\text{ cm}^3$$

The half-filled tilted-cylinder zone formula assumes that the water plane intersects every side generator before meeting either end cap; it cannot be applied unchanged when a base clips that zone. The path formula assumes the same angular position at the two endpoints and a specified number of turns along the lateral surface. Without the turn constraint, a vertical path can be shorter.

Review the Method

Slicing a pyramid or cone parallel to its base with linear similarity ratio k = h_1 / h_2 scales cross-sectional areas by k^2 and volumes by k^3 (dividing a cone at mid-height into volumes V and 7V)