Mathematics • Year 9 • Unit 3 • Lesson 3
Finding a Shorter Side
Build fluency with the rearranged Pythagoras formula $a^2 = c^2 - b^2$, used when the hypotenuse and one leg are known. Identify the hypotenuse, SUBTRACT (not add), then square-root. 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 hypotenuse 13 cm and one leg of 5 cm. Find the other leg.
Step 1, Identify the hypotenuse $c$.
$c = 13$ (the biggest), $b = 5$ (known leg), $a = $ ? (missing leg).
Reason: the hypotenuse is the LARGEST of the three sides, always check this first.
Step 2, Rearrange and substitute into $a^2 = c^2 - b^2$.
$a^2 = 13^2 - 5^2$
Reason: finding a LEG means SUBTRACTING. The hypotenuse squared goes FIRST (it's the bigger number).
Step 3, Compute.
$a^2 = 169 - 25 = 144$
Reason: positive result confirms we put the hypotenuse-squared first. If you ever get negative, you've swapped $c$ and $b$.
Step 4, Square root.
$a = \sqrt{144} = 12$ cm.
Reason: $\sqrt{}$ undoes squaring. Sanity check: $12 < 13$, so the leg is shorter than the hypotenuse ✓.
Answer: $a = \mathbf{12}$ cm (this completes 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 hypotenuse 10 cm and one leg of 7 cm. Find the other leg to 2 d.p.
Step 1, Identify the hypotenuse: $c = $ ____ (the __________ of the three), $b = $ ____ (known leg), $a = $ ?
Step 2, Rearrange to $a^2 = c^2 - b^2$ and substitute:
$a^2 = \_\_\_^2 - \_\_\_^2$
Step 3, Compute:
$a^2 = \_\_\_\_ - \_\_\_\_ = \_\_\_\_$
Step 4, Square root, round to 2 d.p.:
$a = \sqrt{\_\_\_\_} \approx \_\_\_\_$ cm
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). The last two are extension (sanity-check and reasoning).
Foundation, known triples (no calculator)
3.1 Hypotenuse 5 cm, one leg 3 cm. Find the other leg. 1 mark
3.2 Hypotenuse 10 cm, one leg 6 cm. Find the other leg. 1 mark
3.3 Hypotenuse 17 cm, one leg 8 cm. Find the other leg. 1 mark
3.4 Hypotenuse 25 cm, one leg 7 cm. Find the other leg. 1 mark
Standard, calculator, round to 2 d.p.
3.5 Hypotenuse 9 m, one leg 4 m. Find the other leg to 2 d.p. 2 marks
3.6 Hypotenuse 20 cm, one leg 11 cm. Find the other leg to 2 d.p. 2 marks
Extension, sanity check and reasoning
3.7 The hypotenuse of a right triangle is 65 inches and one leg is 56.7 inches. Find the other leg to 2 d.p. and state explicitly whether your answer passes the sanity check "leg must be shorter than the hypotenuse". 3 marks
3.8 A classmate writes $a^2 = 7^2 - 13^2 = 49 - 169 = -120$ when trying to find a leg with $c = 13$ and $b = 7$. (i) What sign mistake did they make? (ii) What is the correct value of the missing leg? 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (faded hyp 10, leg 7)
Step 1: $c = \mathbf{10}$ (the largest), $b = \mathbf{7}$, $a = $ ?
Step 2: $a^2 = \mathbf{10}^2 - \mathbf{7}^2$.
Step 3: $a^2 = \mathbf{100} - \mathbf{49} = \mathbf{51}$.
Step 4: $a = \sqrt{\mathbf{51}} \approx \mathbf{7.14}$ cm.
3.1, Hyp 5, leg 3
$a^2 = 5^2 - 3^2 = 25 - 9 = 16$, so $a = \sqrt{16} = \mathbf{4}$ cm. (3-4-5 triple.)
3.2, Hyp 10, leg 6
$a^2 = 10^2 - 6^2 = 100 - 36 = 64$, so $a = \sqrt{64} = \mathbf{8}$ cm. (This is $2\times$(3-4-5).)
3.3, Hyp 17, leg 8
$a^2 = 17^2 - 8^2 = 289 - 64 = 225$, so $a = \sqrt{225} = \mathbf{15}$ cm. (8-15-17 triple.)
3.4, Hyp 25, leg 7
$a^2 = 25^2 - 7^2 = 625 - 49 = 576$, so $a = \sqrt{576} = \mathbf{24}$ cm. (7-24-25 triple.)
3.5, Hyp 9, leg 4
$a^2 = 9^2 - 4^2 = 81 - 16 = 65$, so $a = \sqrt{65} \approx \mathbf{8.06}$ m. (Sanity: $8.06 < 9$ ✓.)
3.6, Hyp 20, leg 11
$a^2 = 20^2 - 11^2 = 400 - 121 = 279$, so $a = \sqrt{279} \approx \mathbf{16.70}$ cm. (Sanity: $16.70 < 20$ ✓.)
3.7, Hyp 65, leg 56.7
$a^2 = 65^2 - 56.7^2 = 4225 - 3214.89 = 1010.11$, so $a = \sqrt{1010.11} \approx \mathbf{31.78}$ inches.
Sanity check: $31.78 < 65$ ✓, pass. (This is the height of a typical 65-inch TV, matches the lesson's worked example.)
3.8, Classmate's sign mistake
(i) They put the smaller number FIRST: they wrote $7^2 - 13^2$ instead of $13^2 - 7^2$. The hypotenuse-squared (the biggest) must come first when subtracting; otherwise you get a negative number, and the square root of a negative is not a real number.
(ii) Correct working: $a^2 = 13^2 - 7^2 = 169 - 49 = 120$, so $a = \sqrt{120} \approx \mathbf{10.95}$ cm.