Mathematics • Year 8 • Unit 3 • Lesson 2
Missing Legs — Mixed Challenge
Mix finding a leg with finding the hypotenuse — decide before you start whether to add or subtract. Six mixed problems, one "find the mistake", and one open-ended verification puzzle.
1. Mixed problems — add or subtract?
Watch out: some of these ask for the hypotenuse, others for a leg. Decide which BEFORE you write your formula. Show your working. 3 marks each
1.1 A right triangle has c = 20 cm and b = 12 cm. Find a.
1.2 A right triangle has legs 7 cm and 24 cm. Find the hypotenuse.
1.3 A right triangle has c = 50 cm and b = 14 cm. Find a. (Hint: 14-48-50 = 7-24-25 × 2.)
1.4 A right triangle has c = 6 cm and one leg b = 4 cm. Find the other leg to 2 decimal places.
1.5 A 13 m flagpole is supported by a wire that reaches the ground 5 m from the base. Find the length of the wire.
1.6 A 6.5 m ladder leans against a wall. Its base is 2.5 m from the wall. How high up the wall does it reach?
2. Find the mistake
A student has been asked to find the missing leg of a right triangle with c = 17 and b = 15. Their working is below. Exactly one line contains a mistake — spot it, explain why, and re-do the working. 3 marks
Student's working — find a when c = 17 and b = 15:
Line 1: a² = c² − b²
Line 2: a² = 17² − 15² = 289 − 225 = 64
Line 3: a = 64 ÷ 2 = 32
Line 4: So the missing leg is 32 cm.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? The opposite of squaring is taking the SQUARE ROOT, not dividing by 2. √64 = 8, not 32.3. Open-ended challenge — find the missing leg, three ways
This question has more than one valid answer. 4 marks
3.1 A right-angled triangle has a hypotenuse of 25 cm. Find three different valid combinations of integer legs (a, b) where the hypotenuse is 25.
For each combination:
(i) State the values of a and b.
(ii) Show the check: a² + b² = 25² = 625.
(iii) Identify whether the triangle is a "primitive" triple or a multiple of a smaller triple.
Bonus: at least one of your three combinations must include a primitive Pythagorean triple (not just multiples of 3-4-5 or 5-12-13).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — c = 20, b = 12
a² = 20² − 12² = 400 − 144 = 256, so a = √256 = 16 cm. (12-16-20 = 3-4-5 × 4.)
1.2 — Legs 7 and 24
c² = 7² + 24² = 49 + 576 = 625, so c = √625 = 25 cm. (7-24-25 triple.)
1.3 — c = 50, b = 14
a² = 50² − 14² = 2500 − 196 = 2304, so a = √2304 = 48 cm. (14-48-50 = 7-24-25 × 2.)
1.4 — c = 6, b = 4
a² = 6² − 4² = 36 − 16 = 20, so a = √20 ≈ 4.47 cm (to 2 d.p.).
1.5 — Flagpole wire
This asks for the HYPOTENUSE (the wire). c² = 13² + 5² = 169 + 25 = 194, so c = √194 ≈ 13.93 m. (Note: this is NOT a 5-12-13 triple — the wire is the hypotenuse, not the pole.)
1.6 — Ladder
a² = 6.5² − 2.5² = 42.25 − 6.25 = 36, so a = √36 = 6 m up the wall. (2.5-6-6.5 = 5-12-13 × 0.5.)
2 — Find the mistake
(a) The mistake is on Line 3 (carried into Line 4).
(b) The student divided 64 by 2 to "undo" the square, but the inverse of squaring is the SQUARE ROOT, not dividing by 2. √64 = 8, not 32.
(c) Corrected working:
a² = c² − b² = 17² − 15² = 289 − 225 = 64
a = √64 = 8 cm. ✓
Sanity check: 8-15-17 is a core Pythagorean triple. Verify: 8² + 15² = 64 + 225 = 289 = 17² ✓
3 — Find three (a, b) pairs giving c = 25 (sample solution)
We need integer a, b with a² + b² = 625. There are several:
Pair 1: a = 7, b = 24. Check: 49 + 576 = 625 ✓. This is a primitive triple (7-24-25).
Pair 2: a = 15, b = 20. Check: 225 + 400 = 625 ✓. This is 3-4-5 × 5 (a multiple).
Pair 3: a = 20, b = 15. Check: 400 + 225 = 625 ✓. Same triangle as Pair 2 but with the legs swapped — accept as a separate "labelling" or push students to find one of the cases below.
Other valid pairs (any of these counts as a third unique triangle):
- If allowing non-integer legs, e.g. a = 5, b = √600 ≈ 24.49 — but only integer legs were asked.
Note: with integer legs and c = 25, only two distinct (a, b) triangles exist (7-24-25 and 15-20-25). If a student lists both orderings of one triangle, that's acceptable for full marks since the question said "combinations".
Marking: 1 mark per valid (a, b) pair with shown check (up to 3 marks). 1 bonus mark for including 7-24-25 as the primitive. Accept (7,24) and (24,7) as the same pair; (15,20) and (20,15) as the same pair. The two distinct triangles ARE 7-24-25 and 15-20-25 — that's all the integer solutions, so award 3 marks for finding both, and 1 bonus mark for naming the primitive.