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Mathematical Architects · Geometry · Foundations
Get Ready: Module 4 — Connecting Geometric & Algebraic Descriptions
Everyone starts somewhere. Before we connect pictures to equations, let's pour the floor —
the handful of skills the rest of the module quietly leans on. There's no rush and no quiz here:
read an example, try one yourself, peek at the answer when you want it, and move on when you feel ready.
Go at your own pace. You belong in this room.
Foundations · build the floor first
How to use this page: work through the five skill cards in order. Each one tells you
what it is, why Module 4 needs it, shows a fully worked example, then gives you a
couple to try. If a card already feels easy, skim it and keep going.
🧱
Skills to build first
Five small skills. Master these and the module feels like a downhill walk.
1
Area & perimeter of basic shapes
What it is: Perimeter is the distance all the way around a shape (add up the sides).
Area is the space inside it (square units). For a rectangle, area is length times width; for a
triangle it's half the base times the height.
Why Module 4 needs it: the module measures real figures — the space inside a circle,
the outside (surface area) of a box, the paint on a cylinder. All of it is just area and perimeter ideas
stretched onto curves and into 3-D.
Worked example
- Find the perimeter and area of a rectangle that is \(8\) cm long and \(3\) cm wide.
- Perimeter = add all four sides \(= 8 + 3 + 8 + 3 = 22\) cm. (Or \(2(\ell+w)=2(8+3)=22\) cm.)
- Area = length \(\times\) width \(= 8 \times 3 = 24\) square cm \(\;(24\text{ cm}^2)\).
Try it
a. A rectangle is \(10\) m by \(4\) m. Find its perimeter and area.
Show answer
Perimeter \(= 2(10+4) = 2(14) = 28\) m.
Area \(= 10 \times 4 = 40\) square meters \((40\text{ m}^2)\).
b. A triangle has base \(6\) in and height \(5\) in. Find its area.
Show answer
Area of a triangle \(= \tfrac12 \, b\,h = \tfrac12 \times 6 \times 5 = \tfrac12 \times 30 = 15\) square inches \((15\text{ in}^2)\).
2
Circle vocabulary: radius, diameter, & \(\pi\)
What it is: the radius \(r\) is the distance from the center to the edge. The
diameter \(d\) goes straight across through the center, so it's always twice the radius:
\(d = 2r\). The number pi \((\pi \approx 3.14)\) is how many diameters fit around the outside of
any circle — it never changes.
Why Module 4 needs it: circles are half the module. Circumference \(C = 2\pi r\), area
\(A = \pi r^2\), arc length, and the circle equation all start from knowing what \(r\), \(d\), and \(\pi\) mean.
Worked example
- A circle has a radius of \(5\) cm. Find its diameter, then its circumference (use \(\pi \approx 3.14\)).
- Diameter: \(d = 2r = 2 \times 5 = 10\) cm.
- Circumference: \(C = 2\pi r = 2 \times 3.14 \times 5 = 31.4\) cm. (Distance around the edge.)
Try it
a. A circle has a diameter of \(14\) in. What is its radius?
Show answer
Since \(d = 2r\), divide the diameter in half: \(r = \dfrac{d}{2} = \dfrac{14}{2} = 7\) in.
b. A circle has radius \(3\) m. Estimate its circumference using \(\pi \approx 3.14\).
Show answer
\(C = 2\pi r = 2 \times 3.14 \times 3 = 18.84\) m (about \(18.8\) m around).
3
Squaring numbers & exponents
What it is: an exponent is a shortcut for repeated multiplication. Squaring a number
means multiplying it by itself: \(5^2 = 5 \times 5 = 25\). The small \(2\) is the exponent; it does not
mean “times 2.”
Why Module 4 needs it: the area of a circle is \(A = \pi r^2\) and the circle equation is
\((x-h)^2 + (y-k)^2 = r^2\) — both square the radius. If \(r^2\) trips you up, those formulas will too,
so let's make squaring automatic.
Worked example
- Evaluate \(7^2\), and find the area of a circle with radius \(4\) (leave \(\pi\) in the answer).
- \(7^2\) means \(7 \times 7 = 49\). (Not \(7 \times 2 = 14\) — that's a common slip.)
- Area: \(A = \pi r^2 = \pi \times 4^2 = \pi \times 16 = 16\pi\). Square first, then multiply by \(\pi\).
Try it
a. Evaluate \(9^2\) and \(3^2\).
Show answer
\(9^2 = 9 \times 9 = 81\). \(3^2 = 3 \times 3 = 9\).
b. Find the area of a circle with radius \(6\), leaving \(\pi\) in your answer.
Show answer
\(A = \pi r^2 = \pi \times 6^2 = \pi \times 36 = 36\pi\) square units.
4
Substituting values into a formula
What it is: a formula is a recipe written with letters. To use it, you substitute
(swap each letter for its number), then simplify carefully using order of operations — exponents before
multiplication, multiplication before addition.
Why Module 4 needs it: almost every answer in this module comes from plugging numbers into
a formula: \(V = \ell w h\), \(SA = 2\pi r^2 + 2\pi r h\), \(C = 2\pi r\). The skill is purely “put the
right number in the right slot, then evaluate.”
Worked example
- Find the volume of a box using \(V = \ell w h\) with \(\ell = 5\), \(w = 3\), \(h = 2\).
- Substitute each letter: \(V = (5)(3)(2)\).
- Simplify left to right: \(5 \times 3 = 15\), then \(15 \times 2 = 30\). So \(V = 30\) cubic units.
Try it
a. Use \(V = \ell w h\) with \(\ell = 4\), \(w = 4\), \(h = 10\).
Show answer
\(V = (4)(4)(10)\). First \(4 \times 4 = 16\), then \(16 \times 10 = 160\). So \(V = 160\) cubic units.
b. Use \(C = 2\pi r\) with \(r = 7\) and \(\pi \approx 3.14\).
Show answer
\(C = 2 \times 3.14 \times 7\). First \(2 \times 3.14 = 6.28\), then \(6.28 \times 7 = 43.96\). So \(C \approx 43.96\) units.
5
The coordinate plane
What it is: the coordinate plane is a grid with a horizontal \(x\)-axis and a vertical
\(y\)-axis crossing at the origin \((0,0)\). A point is named by an ordered pair \((x, y)\):
go across by \(x\) first, then up or down by \(y\). Right and up are positive; left and down
are negative.
Why Module 4 needs it: this is where pictures become equations. A circle gets a center
\((h, k)\) plotted on this grid, and the equation \((x-h)^2 + (y-k)^2 = r^2\) only makes sense once you can
locate points confidently.
Worked example
- Plot the point \((3, -2)\).
- The \(x\) is \(3\): start at the origin and move \(3\) units to the right.
- The \(y\) is \(-2\): from there, move \(2\) units down (because it's negative). That spot is \((3, -2)\).
Try it
a. Describe how to plot \((-4, 5)\).
Show answer
Start at the origin. Move \(4\) units left (the \(x\) is negative), then \(5\) units
up (the \(y\) is positive). That point is in the upper-left region.
b. A circle's center is at \((2, -3)\). In which direction from the origin is the center?
Show answer
\(2\) units right and \(3\) units down from the origin (lower-right region).