Geometry Basics — 700-800

Geometry Basics: The 700-800 Patterns

Part 1 of 3 — The Archetypes Hard-Tier Items Are Built From

At the 700-800 level this topic is almost entirely solid geometry inside a word problem. The formulas are given to you on the reference sheet; the difficulty is in the layer wrapped around them — a density, a purchase, a rate, an already-elapsed time, a hidden face. Five archetypes cover the bank.

Archetype 1: The Scaling Trio (Linear, Area, Volume)

If every length is multiplied by $k$, then every area is multiplied by $k^{2}$ and every volume by $k^{3}$. Almost every "by what percent" item is this rule wearing a disguise.

• Radius $\times 1.5$ and height $\times 0.8$: volume factor $= (1.5)^{2}(0.8) = 1.8$, an 80 percent increase. Not $1.5 \times 0.8 = 1.2$, because the radius is squared.
• Surface area goes from $100\pi$ to $225\pi$: the AREA ratio is $2.25$, so the LINEAR ratio is $\sqrt{2.25} = 1.5$ and the volume ratio is $(1.5)^{3} = 3.375$ — a $237.5$ percent increase, not $337.5$.
• Model at scale $1:4$: bronze (a volume) scales by $64$; gold leaf (an area) scales by $16$. Two different exponents in one problem is the classic hard-item build.

Percent increase is the ratio minus one. A volume ratio of $3.375$ is an increase of $237.5$ percent. Reporting $337.5$ is the most common miss in the topic.

Archetype 2: Composite and Subtracted Solids

Silo = cylinder $+$ hemisphere. Toy = cube $+$ pyramid. Block $-$ drilled cylinder. Cube $-$ inscribed sphere. Planter = outer cylinder $-$ interior cylinder.

Two disciplines make these routine:

1. Hemisphere is $\frac{2}{3}\pi r^{3}$, not $\frac{4}{3}\pi r^{3}$ and not $\frac{1}{3}\pi r^{3}$. Both wrong versions are planted.
2. Surface-area composites lose faces. A cube under a pyramid contributes only its four vertical sides: its bottom rests on the table, its top is hidden at the joint. The choices that include $5$ faces or all $6$ are waiting for you.

Archetype 3: Density and Mass

Mass $=$ volume $\times$ density. Every hard item in this family plants the raw volume as an answer choice, because it is the last number you compute before the final multiplication.

The alloy variant is worth memorizing: melting two metals adds their volumes and adds their masses, so the new density is $\frac{\text{total mass}}{\text{total volume}}$. Averaging the two densities is valid only for equal volumes — and the average is always a choice.

Melting or melting-down never changes mass. Ice at $0.92$ g/cm³ becoming water at $1.00$ g/cm³: $900$ cm³ of ice keeps its $828$ g and shrinks to $828$ cm³ of water.

Archetype 4: Rates, Elapsed Time, and Unit Conversion

Volume $\div$ rate $=$ time. The hard version adds one wrinkle:

• "How much LONGER" — subtract the time already spent, or the volume already added.
• "How many HOURS" when the rate is per minute — the un-converted minute count is a choice.
• Liters vs cubic meters — $1$ m³ $= 1000$ L, so $250$ L/min $= 0.25$ m³/min.
• Cubic inches vs cubic feet — $1$ ft³ $= 1728$ in³, not $144$. Cubing the conversion factor is exactly what students forget.

Archetype 5: Similar-Solid Partial Fill

A cone held point-down filled to $\frac{3}{4}$ of its height contains a similar smaller cone, so the water is $\left(\frac{3}{4}\right)^{3} = \frac{27}{64}$ of the capacity — not $\frac{3}{4}$ of it. A pyramid cut halfway up gives a top piece of $\frac{1}{8}$ the volume and a bottom piece of $\frac{7}{8}$.

The same idea drives cone-filling rate problems: at depth $5$ in a cone that is $10$ deep with top radius $4$, the water's surface radius is $2$, not $4$. Scale the radius with the depth before you use $\frac{1}{3}\pi r^{2}h$.

The Two Structural Habits

Habit 1: circle the piece. Cube minus sphere: is the question about the sphere or the leftover? Cone filled to $\frac{3}{4}$: is it about the water or the empty space? The other piece is always a choice.

Habit 2: apply the last multiplier. Density, cost per unit, price per liter, tiles per square meter. The un-multiplied quantity is the single most-selected wrong answer in this topic.

Geometry Basics: Traps & Speed

Part 2 of 3 — Distractor Autopsy and Efficient Setups

The arithmetic in a hard solid-geometry item is never the hard part. The four answer choices are a menu of "places a reasonable person stops," and knowing the menu lets you eliminate before you compute.

The Six Distractor Species

1. The un-multiplied quantity. The volume when mass was asked. The area when cost was asked. The perimeter in meters when dollars were asked. If the problem ends in "per cubic meter," "per gram," or "per square meter," the number just before that multiplication is a choice.

2. The wrong piece. Cube minus sphere: the sphere's mass is offered. Cone filled to $\frac{3}{4}$ height: the water is offered when the empty space was asked. Silo: the cylinder alone is offered. Read the final noun.

3. Diameter used as radius. A tank "with base diameter $4$ feet" is $r = 2$. Because volume depends on $r^{2}$, this error multiplies the answer by $4$ — and both the wrong volume and the wrong time built from it appear as choices.

4. Slant height used as height. A cone with base radius $5$ and slant height $13$ has $h = \sqrt{13^{2} - 5^{2}} = 12$. Using $13$ directly is a choice; forgetting the $\frac{1}{3}$ is another.

5. The unit conversion left undone. Minutes offered when hours were asked. Cubic inches divided by $144$ instead of $1728$. A length of "$6$ feet" used as $6$ inches inside an inch-based cross-section. Liters not converted to cubic meters.

6. The wrong rounding direction. "Paint is sold only in whole liters." "Soil comes in bags." "How many whole panels fit." $17.3$ liters means you buy $18$; the choice built on $17$ leaves part of the tank bare. Purchases always round UP; things that must fit inside always round DOWN.

Speed Setups

Set up percent-change problems as pure factors. Do not pick $r = 100$ and grind. Write $(1.25)^{2}(0.64)$ and evaluate. Ten seconds, no arithmetic slips, and the structure is visible.

For scaling problems, ask "is this quantity a length, an area, or a volume?" for each cost line separately. Paint, leaf, fabric, coating $\rightarrow$ $k^{2}$. Metal, concrete, water, weight $\rightarrow$ $k^{3}$. Fencing, trim, edging $\rightarrow$ $k$.

Keep $\pi$ symbolic until the last line. $2\pi r^{2} + 2\pi rh = 18\pi + 48\pi = 66\pi$ is exact and error-free; decimalize once, at the end.

In composite-surface problems, sketch a two-column tally: faces counted, faces excluded. Table contact and hidden joints go in the second column, and the totals from the wrong tallies are the distractors.

Write the requested unit in the margin before you start — grams, dollars, hours, whole bags. Then the last line of your work has to carry that unit.

Geometry Basics: Timed Drill

Part 3 of 3 — Four Questions at Full Difficulty

Budget about 90 seconds per question. These items are long to read and short to solve, so spend your first fifteen seconds reading rather than computing.

Run this checklist on each one:

1. What solid is it? Name every piece and every piece that is removed.
2. What is the final unit? Grams, dollars, hours, whole bags. Write it in the margin.
3. Is there a last multiplier? Density, price, coverage rate. If so, the un-multiplied number is a trap.
4. Which direction does it round? Buying rounds up; fitting rounds down.

Keep $\pi$ symbolic until the final line and decimalize once.