Mathematics • Year 8 • Unit 3 • Lesson 2
Finding a Shorter Side
Build fluency with the rearranged formula a = √(c² − b²). One worked example, one guided example with blanks, then eight independent problems with clean triples, decimals and verification checks.
1. I do — fully worked example
Notice the big change from Lesson 1: when the hypotenuse is known and we want a leg, we subtract instead of add.
Problem. A right-angled triangle has hypotenuse c = 10 cm and leg b = 6 cm. Find the other leg a.
Step 1 — Identify the hypotenuse.
c = 10 is the longest side (opposite the right angle).
Reason: the formula only works if c is correctly labelled as the hypotenuse first.
Step 2 — Rearrange the formula.
From c² = a² + b², subtract b²: a² = c² − b²
Reason: a leg's square equals the hypotenuse² MINUS the other leg². Subtract, not add.
Step 3 — Substitute and calculate.
a² = 10² − 6² = 100 − 36 = 64
Reason: square each known side, then subtract.
Step 4 — Take the square root, then verify.
a = √64 = 8 cm. Check: 8² + 6² = 64 + 36 = 100 = 10² ✓
Reason: always verify — substitute all three sides back into a² + b² = c².
Answer: a = 8 cm (6-8-10 = 3-4-5 × 2).
2. We do — fill in the missing steps
Same shape as Section 1, with the working faded. Fill in each blank. 4 marks
Problem. A right-angled triangle has hypotenuse c = 13 cm and leg b = 5 cm. Find the other leg a.
Step 1 — Hypotenuse check: c = ______ is the longest side.
Step 2 — Rearranged formula:
a² = c² ______ b²
Step 3 — Substitute and calculate:
a² = 13² − 5² = ______ − ______ = ______
Step 4 — Square root and verify:
a = √______ = ______ cm
Check: ______² + ______² = ______ = 13² ✓
3. You do — independent practice
Show all working. The first four are foundation (clean triples). The middle two are standard (verify your answer). The last two are extension (decimal answers and a small application).
Foundation — clean triples
3.1 c = 17 cm, b = 15 cm. Find a. 1 mark
3.2 c = 25 cm, b = 7 cm. Find a. 1 mark
3.3 c = 15 cm, b = 9 cm. Find a. 1 mark
3.4 c = 26 cm, b = 10 cm. Find a. (Hint: 10-24-26 = 5-12-13 × 2.) 1 mark
Standard — find and verify
3.5 c = 7.5 m, b = 4.5 m. Find a, then verify by substitution. 2 marks
3.6 c = 41 cm, b = 40 cm. Find a, then verify by substitution. 2 marks
Extension — decimals and application
3.7 c = 8 cm, b = 3 cm. Find a to 2 decimal places. 2 marks
3.8 A tent has a slant side of 3.5 m (this is the hypotenuse) and a half-width of 2.1 m (one leg). Find the tent's perpendicular height (the other leg). 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded 5-?-13)
Step 1: c = 13.
Step 2: a² = c² − b².
Step 3: a² = 169 − 25 = 144.
Step 4: a = √144 = 12 cm. Check: 12² + 5² = 169 = 13² ✓
3.1 — c = 17, b = 15
a² = 17² − 15² = 289 − 225 = 64, so a = √64 = 8 cm. (8-15-17 triple.)
3.2 — c = 25, b = 7
a² = 25² − 7² = 625 − 49 = 576, so a = √576 = 24 cm. (7-24-25 triple.)
3.3 — c = 15, b = 9
a² = 15² − 9² = 225 − 81 = 144, so a = √144 = 12 cm. (9-12-15 = 3-4-5 × 3.)
3.4 — c = 26, b = 10
a² = 26² − 10² = 676 − 100 = 576, so a = √576 = 24 cm. (10-24-26 = 5-12-13 × 2.)
3.5 — c = 7.5, b = 4.5
a² = 7.5² − 4.5² = 56.25 − 20.25 = 36, so a = √36 = 6 m. Verify: 6² + 4.5² = 36 + 20.25 = 56.25 = 7.5² ✓ (3-4-5 × 1.5.)
3.6 — c = 41, b = 40
a² = 41² − 40² = 1681 − 1600 = 81, so a = √81 = 9 cm. Verify: 9² + 40² = 81 + 1600 = 1681 = 41² ✓ (9-40-41 primitive triple.)
3.7 — c = 8, b = 3
a² = 8² − 3² = 64 − 9 = 55, so a = √55 ≈ 7.42 cm (to 2 d.p.).
3.8 — Tent height
a² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84, so height = √7.84 = 2.8 m. (2.1-2.8-3.5 = 3-4-5 × 0.7.)