Geometry II — Area, Perimeter & Scaling Laws

By the end of this chapter you will:

1. Compute areas of squares, rectangles, triangles, parallelograms, circles from memory
2. Break composite & shaded regions into add/subtract pieces
3. Use the scaling law: length $\times k$ → area $\times k^2$ → volume $\times k^3$
4. Solve perimeter problems that hide a one-variable equation

6.0 The BC gap

BC baseline: G5 teaches area of squares and rectangles, plus volume informally. G6 adds triangle, parallelogram, trapezoid areas and volume in cm³ / m³ / mL / L. What BC does not teach: the scaling law ($k \to k^2 \to k^3$), composite shaded regions, surface area, nets, painted-cube casework, and grid-counting formulas (toothpicks). All four are Gauss staples.

6.0.1 The two formulas to never forget

$$\text{Area of a square with side } s = s^2 \qquad \text{Area of a circle with radius } r = \pi r^2$$

Everything else builds from these.

2024 Q5 — square area

A square with side length $5$ has an area of …

$5^2 = \mathbf{25}$. Answer (E). ✅

6.1 Area formulas

Perimeter (the boundary length):

• Rectangle: $2\ell + 2w$
• Circle (circumference): $2 \pi r$

6.2 Composite & shaded regions

The two moves: add simple regions, or subtract simpler-from-bigger.

Example

A square of side $10$ has a circle of radius $5$ inscribed in it (touching all four sides). Find the shaded area outside the circle but inside the square.

Shaded = Square − Circle = $100 - 25\pi$.

🔑 The subtraction approach is almost always cleaner than direct. When you see an annulus, an L-shape, a square minus a triangle, try whole minus inner.

6.3 The scaling law — the Part B move

If every length scales by a factor $k$, then every area scales by $k^2$, and every volume scales by $k^3$.

2024 Q15 — radius tripled

A circle has radius $2$. If the radius is tripled, the area of the original divided by the area of the new is …

Original radius $r$, new radius $3r$. Lengths scaled by $k = 3$, so areas scale by $k^2 = 9$.

$$\frac{\text{original area}}{\text{new area}} = \frac{1}{9}$$

Answer (C). ✅

🔑 Spotting scaling: any sentence with "doubled / tripled / halved / scaled" applied to a length. Don't compute both areas separately — use $k^2$.

6.4 Perimeter as a one-variable equation

Perimeter problems often hide a linear equation. Set up, solve.

2024 Q12 (revisited from Ch. 4)

Length is twice the width, perimeter is $120$ cm.

$P = 2\ell + 2w = 2(2w) + 2w = 6w = 120 \Rightarrow w = 20$ cm.

6.5 3D — nets, surface area, volume, painted cubes

BC G6 introduces volume in cm³ and m³ but does NOT teach surface area or 3D casework. Gauss expects both.

Surface area of a rectangular box

A box with edges $\ell, w, h$ has surface area

$$\text{SA} = 2\ell w + 2\ell h + 2 w h$$

For a cube of edge $s$: $\text{SA} = 6s^2$, $V = s^3$.

Nets

A net is an unfolded view of a 3D solid. Practice: a cube net has $6$ squares in a cross or T arrangement. Other prisms unfold into rectangles + the two cap shapes.

Recognize from sight: a cube has $11$ distinct nets (up to rotation/reflection). You don't need to memorize them — just know "if I fold these squares back up, do they form a cube?"

Painted-cube casework — 2024 Q23 (worked in full)

A rectangular prism with integer edges $a \times b \times c$ is painted on all 6 faces, then cut into unit cubes. Fifty unit cubes have no paint. What is the mean of all possible values of $V = abc$?

The "no paint" cubes are exactly the interior cubes — those not touching any face. The interior dimensions are $(a - 2) \times (b - 2) \times (c - 2)$ (each face peels off one layer on each side). So we need

$$(a - 2)(b - 2)(c - 2) = 50$$

with $a, b, c \ge 3$ (so the interior dimensions are $\ge 1$).

Factor 50: $50 = 2 \cdot 5^2$. Unordered triples of positive integers with product $50$:

Distinct volumes: $\{468, 324, 252, 196\}$. Mean $= (468 + 324 + 252 + 196) / 4 = 1240 / 4 = 310$.

Answer (B) 310. ✅

🔑 The interior-dimension trick is the entire problem. The moment you write "$(a-2)(b-2)(c-2) = 50$", it's a factor-counting exercise.

6.6 Grid-counting formulas — 2024 Q22

In a $2 \times 3$ grid of squares, $17$ toothpicks are used: $10$ outer and $7$ inner. If toothpicks make a $20 \times 24$ grid, what % are inner?

Toothpick count for an $m \times n$ grid (rows × columns of squares):

Verify on $2 \times 3$: total $= 2(6) + 2 + 3 = 17$ ✓ outer $= 4 + 6 = 10$ ✓ inner $= 12 - 5 = 7$ ✓.

Apply to $20 \times 24$:

• Total: $2(20)(24) + 20 + 24 = 960 + 44 = 1004$
• Outer: $40 + 48 = 88$
• Inner: $1004 - 88 = 916$

Percentage inner: $\dfrac{916}{1004} \approx 91.2\%$. Answer (E) 91%. ✅

🔑 Grid-counting generalizes to lattice points, gridlines, paths — always count the horizontals and verticals separately, then add.

6.7 Trap Alerts ⚠️

1. Area scales by $k^2$, NOT $k$. Doubling all sides quadruples area.
2. Volume scales by $k^3$. Doubling all dimensions of a box gives $8\times$ volume.
3. Triangle's $b$ and $h$ must be perpendicular. The base and the height to that base form a right angle — not any two sides.
4. The "area divided by area" of similar figures equals $k^2$, not $k$.
5. Circumference vs area — $2\pi r$ is a length, $\pi r^2$ is an area. Don't swap.

6.8 Mnemonic

"$s^2$, $\pi r^2$, $\tfrac{1}{2}bh$ — and $k$ becomes $k^2$ becomes $k^3$."

Practice Set

1. (Part A) Find the area of a rectangle with length $8$ and width $5$.
2. (Part B) A triangle has base $12$ and height $7$. Find its area.
3. (Part B) The side of a square is doubled. The new area is how many times the old?
4. (Part C) The radius of a sphere is multiplied by $\tfrac{2}{3}$. The new volume divided by the original is …

Answers: 1) 40; 2) 42; 3) 4; 4) $(\tfrac{2}{3})^3 = \tfrac{8}{27}$.

End of chapter. Next: Patterns, Sequences & Modular Periodicity.