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Get Ready: Module 3 — Investigating Proportionality

Every architect lays a floor before raising the building. This page is your floor. If Module 3 feels intimidating, that's okay — it just means a few earlier skills need a little more practice first. We'll build them up, one small step at a time.

Everyone starts somewhere. There's no rush and no grade here. Work through the skills below at your own pace, click "Show answer" whenever you want to check yourself, and use the free practice links to drill anything that feels shaky. When the skills feel comfortable, you'll be ready for the module.

Foundations · Prerequisite Skills for Module 03


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Skills to Build First

Five building blocks. Read the "what & why," follow the worked example, then try one yourself.

01 Ratios & Proportions

What & why A ratio compares two amounts, like \(3\) tall to \(2\) wide. A proportion says two ratios are equal, like \(\tfrac{3}{2}=\tfrac{6}{4}\). Module 3 is all about proportionality — when you shrink or stretch a shape, the side ratios stay equal. To find a missing side, you'll set up a proportion and solve it. Solving proportions is the single most-used skill in this module.

Worked example — solve a proportion

Solve for \(x\): \(\;\dfrac{3}{4}=\dfrac{x}{20}\)

  1. Cross-multiply. Multiply each top by the opposite bottom: \(3\times 20 = 4\times x\).
  2. Simplify. \(60 = 4x\).
  3. Undo the multiplication. Divide both sides by \(4\): \(x=\dfrac{60}{4}\).

Answer: \(x = 15\). Check: \(\tfrac{15}{20}=\tfrac{3}{4}\) ✓

Try it

Solve for \(n\): \(\;\dfrac{n}{6}=\dfrac{10}{15}\)

Show answer
Cross-multiply: \(15\,n = 6\times 10 = 60\). Divide both sides by \(15\): \(n=\dfrac{60}{15}=4\). \(n=4\).

A photo is \(4\) in wide and \(6\) in tall. You enlarge it so the width is \(10\) in. How tall is it now? Set up \(\dfrac{4}{6}=\dfrac{10}{h}\).

Show answer
Cross-multiply: \(4h = 6\times 10 = 60\). Divide by \(4\): \(h=15\). The enlarged photo is \(15\) in tall.
Practice free ▶ Khan Academy ▶ IXL

02 Fractions

What & why A fraction is just a division: \(\tfrac{3}{4}\) means "\(3\) out of \(4\)" or \(3\div 4\). Ratios are fractions, and every trig ratio later in the module (like \(\sin\theta=\tfrac{\text{opp}}{\text{hyp}}\)) is a fraction. Being able to simplify a fraction and turn it into a decimal lets you recognize when two ratios are really the same value.

Worked example — simplify, then make a decimal

Simplify \(\dfrac{12}{16}\) and write it as a decimal.

  1. Find a common factor. Both \(12\) and \(16\) divide by \(4\).
  2. Divide top and bottom by \(4\). \(\dfrac{12\div 4}{16\div 4}=\dfrac{3}{4}\).
  3. Divide to get a decimal. \(3\div 4 = 0.75\).

Answer: \(\dfrac{12}{16}=\dfrac{3}{4}=0.75\).

Try it

Simplify \(\dfrac{18}{24}\).

Show answer
Both divide by \(6\): \(\dfrac{18\div 6}{24\div 6}=\dfrac{3}{4}\). \(\dfrac{3}{4}\) (which is \(0.75\)).

Write \(\dfrac{5}{8}\) as a decimal.

Show answer
\(5\div 8 = 0.625\). \(\dfrac{5}{8}=0.625\).
Practice free ▶ Khan Academy ▶ IXL

03 Square Roots

What & why A square root undoes squaring. Since \(5^2 = 25\), we say \(\sqrt{25}=5\) — "what number times itself gives \(25\)?" When you use the Pythagorean theorem to find a triangle's side, the last step is almost always a square root. You'll also meet answers that don't come out even, like \(\sqrt{20}\), so it helps to estimate them.

Worked example — exact and estimated roots

Find \(\sqrt{49}\), then estimate \(\sqrt{20}\).

  1. For a perfect square, ask "what squared?" \(7\times 7 = 49\), so \(\sqrt{49}=7\).
  2. For \(\sqrt{20}\), find the perfect squares around it. \(16 < 20 < 25\), so \(\sqrt{16} < \sqrt{20} < \sqrt{25}\), i.e. \(4 < \sqrt{20} < 5\).
  3. Since \(20\) is closer to \(16\), a good estimate is \(\sqrt{20}\approx 4.5\) (calculator: \(4.47\)).

Answer: \(\sqrt{49}=7\) exactly; \(\sqrt{20}\approx 4.5\).

Try it

Find \(\sqrt{81}\).

Show answer
\(9\times 9 = 81\), so \(\sqrt{81}=9\).

Between which two whole numbers does \(\sqrt{50}\) fall?

Show answer
\(49 < 50 < 64\), so \(\sqrt{49} < \sqrt{50} < \sqrt{64}\), i.e. between \(7\) and \(8\) (about \(7.07\)).
Practice free ▶ Khan Academy ▶ IXL

04 The Pythagorean Theorem

What & why In any right triangle (one \(90^\circ\) corner), the two short sides (legs \(a\) and \(b\)) and the longest side (hypotenuse \(c\)) are linked by \(a^2 + b^2 = c^2\). Module 3 leans on right triangles constantly, and this rule lets you find a missing side from the other two. It combines squaring and square roots — the two skills above.

Worked example — find the hypotenuse

A right triangle has legs \(a=3\) and \(b=4\). Find the hypotenuse \(c\).

  1. Write the rule. \(a^2 + b^2 = c^2\).
  2. Put in the numbers. \(3^2 + 4^2 = c^2\), so \(9 + 16 = c^2\).
  3. Add. \(25 = c^2\).
  4. Square-root both sides. \(c=\sqrt{25}=5\).

Answer: \(c = 5\).

Try it

A right triangle has legs \(6\) and \(8\). Find the hypotenuse.

Show answer
\(6^2 + 8^2 = 36 + 64 = 100\), and \(\sqrt{100}=10\). The hypotenuse is \(10\).

A right triangle has one leg \(a=5\) and hypotenuse \(c=13\). Find the other leg \(b\).

Show answer
\(5^2 + b^2 = 13^2 \Rightarrow 25 + b^2 = 169 \Rightarrow b^2 = 144\). Then \(b=\sqrt{144}=12\). The other leg is \(12\).
Practice free ▶ Khan Academy ▶ IXL

05 Basic Triangle Properties

What & why A triangle's three angles always add up to \(180^\circ\). A right angle is \(90^\circ\), the longest side sits across from the biggest angle, and the smallest side sits across from the smallest angle. In Module 3 you'll decide when two triangles are the same shape (similar). The big shortcut is AA: if two angles match, the triangles are similar — and knowing angles sum to \(180^\circ\) lets you fill in a missing angle to check.

Worked example — find a missing angle

A triangle has angles \(90^\circ\) and \(35^\circ\). Find the third angle.

  1. All three angles add to \(180^\circ\). So \(90^\circ + 35^\circ + x = 180^\circ\).
  2. Add what you know. \(125^\circ + x = 180^\circ\).
  3. Subtract from both sides. \(x = 180^\circ - 125^\circ\).

Answer: \(x = 55^\circ\).

Try it

A triangle has angles \(60^\circ\) and \(80^\circ\). Find the third angle.

Show answer
\(60^\circ + 80^\circ = 140^\circ\), and \(180^\circ - 140^\circ = 40^\circ\). The third angle is \(40^\circ\).

Triangle A has angles \(50^\circ\) and \(60^\circ\). Triangle B has angles \(60^\circ\) and \(70^\circ\). Are they similar?

Show answer
Triangle A's third angle is \(180^\circ-50^\circ-60^\circ=70^\circ\), so A's angles are \(50^\circ, 60^\circ, 70^\circ\). Triangle B's third angle is \(180^\circ-60^\circ-70^\circ=50^\circ\), so B's angles are \(50^\circ, 60^\circ, 70^\circ\) too. The angles match, so by AA similarity, yes — they are similar.
Practice free ▶ Khan Academy ▶ IXL

Quick Readiness Check

Six short questions across all five skills. No pressure — just a self-check.

Solve for \(x\): \(\;\dfrac{x}{8}=\dfrac{9}{12}\).

Show answer
Cross-multiply: \(12x = 8\times 9 = 72\), so \(x=\dfrac{72}{12}=6\). \(x=6\).

Simplify \(\dfrac{14}{21}\).

Show answer
Both divide by \(7\): \(\dfrac{2}{3}\). \(\dfrac{2}{3}\).

What is \(\sqrt{64}\)?

Show answer
\(8\times 8 = 64\), so \(\sqrt{64}=8\).

A right triangle has legs \(9\) and \(12\). Find the hypotenuse.

Show answer
\(9^2+12^2 = 81+144 = 225\), and \(\sqrt{225}=15\). \(15\).

A triangle has angles \(90^\circ\) and \(48^\circ\). Find the third angle.

Show answer
\(180^\circ - 90^\circ - 48^\circ = 42^\circ\). \(42^\circ\).

Write \(\dfrac{3}{5}\) as a decimal, then say whether it equals the ratio \(\dfrac{6}{10}\).

Show answer
\(3\div 5 = 0.6\), and \(6\div 10 = 0.6\) as well, so yes — \(\dfrac{3}{5}=\dfrac{6}{10}\). They are a proportion.

If these feel comfortable, you're ready for the module. If a few felt tricky, that's completely normal — scroll back up, re-read that skill's worked example, and use the free practice links to drill it.


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When You're Ready →

Take what you've built into the module. You've got this.

Module 3 Overview

Head to the main Module 3 page — Investigating Proportionality, with the Visual Lab and the student edition.

Go to Module 3

Visual Lab

Jump straight to the interactive Similar Triangles & Trigonometry lab and drag the triangle yourself.

Open the Visual Lab

Course Syllabus

See how Module 3 fits into the full Geometry journey, policies, and the studio learning environment.

Read the Syllabus