Mathematics • Year 9 • Unit 3 • Lesson 2
Finding the Hypotenuse
Build fluency with the four-step method for finding the hypotenuse: label the sides → substitute into $c^2 = a^2 + b^2$ → compute the sum → take the square root and round to 2 d.p. From a fully worked example through guided practice to independent problems.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. A right-angled triangle has legs $a = 5$ cm and $b = 12$ cm. Find the hypotenuse $c$.
Step 1 — Label.
$a = 5$, $b = 12$, $c = $ ? (the unknown hypotenuse).
Reason: identify which side is which BEFORE substituting. $c$ is the side opposite the 90° — the longest.
Step 2 — Substitute into $c^2 = a^2 + b^2$.
$c^2 = 5^2 + 12^2$
Reason: Pythagoras' theorem rearranged to make $c^2$ the subject.
Step 3 — Compute.
$c^2 = 25 + 144 = 169$
Reason: square each leg first, THEN add. Never add before squaring.
Step 4 — Square root.
$c = \sqrt{169} = 13$ cm.
Reason: $\sqrt{}$ undoes squaring. 169 is a perfect square, so no rounding needed here.
Answer: $c = \mathbf{13}$ cm (this is the 5-12-13 triple).
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. A right-angled triangle has legs $a = 1.8$ m and $b = 3.2$ m. Find the hypotenuse $c$ to 2 d.p.
Step 1 — Label: $a = $ ____, $b = $ ____, $c = $ __________ (unknown longest side).
Step 2 — Substitute into $c^2 = a^2 + b^2$:
$c^2 = \_\_\_^2 + \_\_\_^2$
Step 3 — Compute:
$c^2 = \_\_\_\_ + \_\_\_\_ = \_\_\_\_\_$
Step 4 — Square root, round to 2 d.p.:
$c = \sqrt{\_\_\_\_} \approx \_\_\_\_$ m
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (use known triples — no calculator). The middle two are standard (calculator + round to 2 d.p.). The last two are extension (decimal legs, sanity check).
Foundation — known triples (no calculator)
3.1 Legs 3 cm and 4 cm. Find the hypotenuse. 1 mark
3.2 Legs 6 cm and 8 cm. Find the hypotenuse. 1 mark
3.3 Legs 8 cm and 15 cm. Find the hypotenuse. 1 mark
3.4 Legs 9 cm and 12 cm. Find the hypotenuse. 1 mark
Standard — calculator, round to 2 d.p.
3.5 Legs 4 m and 7 m. Find the hypotenuse to 2 d.p. 2 marks
3.6 Legs 1 cm and 1 cm. Find the hypotenuse to 2 d.p. 2 marks
Extension — decimal legs and sanity checks
3.7 Legs 2.5 m and 6 m. Find the hypotenuse to 2 d.p. and state explicitly whether your answer passes the sanity check “hypotenuse must be longer than each leg”. 3 marks
3.8 Legs 56.7 inches and 31.9 inches. Find the hypotenuse to 2 d.p. (This is the diagonal of a typical 65-inch TV.) 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded 1.8 m and 3.2 m)
Step 1: $a = \mathbf{1.8}$, $b = \mathbf{3.2}$, $c = $ hypotenuse.
Step 2: $c^2 = \mathbf{1.8}^2 + \mathbf{3.2}^2$.
Step 3: $c^2 = \mathbf{3.24} + \mathbf{10.24} = \mathbf{13.48}$.
Step 4: $c = \sqrt{\mathbf{13.48}} \approx \mathbf{3.67}$ m.
3.1 — Legs 3 and 4
$c^2 = 3^2 + 4^2 = 9 + 16 = 25$, so $c = \sqrt{25} = \mathbf{5}$ cm. (3-4-5 triple.)
3.2 — Legs 6 and 8
$c^2 = 36 + 64 = 100$, so $c = \mathbf{10}$ cm. (This is $2\times$(3-4-5).)
3.3 — Legs 8 and 15
$c^2 = 64 + 225 = 289$, so $c = \sqrt{289} = \mathbf{17}$ cm. (8-15-17 triple.)
3.4 — Legs 9 and 12
$c^2 = 81 + 144 = 225$, so $c = \sqrt{225} = \mathbf{15}$ cm. (This is $3\times$(3-4-5).)
3.5 — Legs 4 and 7
$c^2 = 4^2 + 7^2 = 16 + 49 = 65$, so $c = \sqrt{65} \approx \mathbf{8.06}$ m. (Sanity: 8.06 > 7 and > 4 ✓.)
3.6 — Legs 1 and 1
$c^2 = 1 + 1 = 2$, so $c = \sqrt{2} \approx \mathbf{1.41}$ cm. (This $\sqrt{2}$ is a famous irrational number — the diagonal of a unit square.)
3.7 — Legs 2.5 m and 6 m
$c^2 = 2.5^2 + 6^2 = 6.25 + 36 = 42.25$, so $c = \sqrt{42.25} = \mathbf{6.50}$ m (exact — 42.25 is $6.5^2$).
Sanity check: $6.50 > 6$ ✓ and $6.50 > 2.5$ ✓ — pass.
3.8 — Legs 56.7 and 31.9
$c^2 = 56.7^2 + 31.9^2 = 3214.89 + 1017.61 = 4232.5$, so $c = \sqrt{4232.5} \approx \mathbf{65.06}$ inches. (This matches the 65-inch TV diagonal from the lesson's worked example.)