Pythagoras’ Theorem: Finding a Shorter Side — Solutions

Click any answer to watch the solution video.

1. Q1 — Find the unknown shorter side (exact answers)

   1. (a) c=13, b=5: a²=169−25=144 → a=12 cm (5–12–13 triple)
   2. (b) c=17, b=8: a²=289−64=225 → a=15 cm (8–15–17 triple)
   3. (c) c=25, b=7: a²=625−49=576 → a=24 cm (7–24–25 triple)
   4. (d) c=10, b=6: a²=100−36=64 → a=8 cm (3–4–5 ×2)
   5. (e) c=15, b=9: a²=225−81=144 → a=12 cm (3–4–5 ×3)
   6. (f) c=26, b=10: a²=676−100=576 → a=24 cm (5–12–13 ×2)
   7. (g) c=√50, b=5: a²=50−25=25 → a=5 m
   8. (h) c=10, b=√19: a²=100−19=81 → a=9 m
   9. (i) c=29, b=20: a²=841−400=441 → a=21 cm (20–21–29 triple)
   10. (j) c=41, b=9: a²=1681−81=1600 → a=40 cm (9–40–41 triple)

2. Q2 — Find the unknown leg (decimal, 2 d.p.)

   1. (a) c=11, b=6: a=√(121−36)=√85≈9.22 cm
   2. (b) c=20, b=11: a=√(400−121)=√279≈16.70 cm
   3. (c) c=7.5, b=4.5: a=√(56.25−20.25)=√36=6.00 m
   4. (d) c=9, b=5: a=√(81−25)=√56≈7.48 cm
   5. (e) c=30, b=18: a=√(900−324)=√576=24.00 m
   6. (f) c=14.5, b=8.7: a=√(210.25−75.69)=√134.56≈11.60 cm
   7. (g) c=6.5, b=2.5: a=√(42.25−6.25)=√36=6.00 m
   8. (h) c=18, b=13: a=√(324−169)=√155≈12.45 cm

3. Q3 — Pythagorean triples

   1. (a) c=10, b=8: a=√(100−64)=6. 6–8–10 (3–4–5 ×2)
   2. (b) c=15, b=12: a=√(225−144)=9. 9–12–15 (3–4–5 ×3)
   3. (c) c=65, b=25: a=√(4225−625)=√3600=60. 25–60–65 (5–12–13 ×5)
   4. (d) c=34, b=16: a=√(1156−256)=√900=30. 16–30–34 (8–15–17 ×2)
   5. (e) c=50, b=30: a=√(2500−900)=√1600=40. 30–40–50 (3–4–5 ×10)

4. Q4 — True or False

   1. False. The correct formula is a² = c² − b² (subtract, not add).
   2. True. a²=100−36=64 → a=8. Check: 6²+8²=36+64=100=10² ✓
   3. True. The hypotenuse is always the longest side, so any leg is shorter than the hypotenuse.
   4. False. If hypotenuse=10 and legs are equal: a²+a²=100 → 2a²=100 → a=√50=5√2≈7.07, not 5.
   5. False. You cannot subtract lengths directly; you must square them first: a=√(c²−b²).
   6. True. Substituting back into a²+b²=c² should give an exact equality.

5. Q5 — Ladder problems

   1. h²=5²−2²=25−4=21. h=√21 (exact) ≈ 4.58 m
   2. Base = 2 m > 1.5 m minimum. Yes, the ladder is positioned safely.
   3. h²=6²−2²=36−4=32. h=√32=4√2≈5.66 m

6. Q6 — Gate diagonal brace

   1. h=√(4.5²−3.6²)=√(20.25−12.96)=√7.29=2.70 m
   2. Width = 3.6 m, height = 2.7 m. The width is longer than the height.
   3. h=√(3.0²−2.4²)=√(9−5.76)=√3.24=1.80 m

7. Q7 — Height of an isosceles triangle

   1. h=√(13²−5²)=√(169−25)=√144=12 cm (exact)
   2. Area = ½ × base × height = ½ × 10 × 12 = 60 cm²
   3. h=√(10²−6²)=√(100−36)=√64=8 cm. Area = ½ × 12 × 8 = 48 cm²

8. Q8 — Identify and correct errors

   1. Error: Student A added instead of subtracting. When the hypotenuse is known, use a²=c²−b²=169−25=144. Correct answer: a=12 cm.
   2. Error: Student B subtracted side lengths directly instead of squaring. Correct: a²=100−36=64 → a=8 cm.
   3. Error: Same mistake — subtracted lengths without squaring. Correct: a²=9²−4²=81−16=65 → a=√65≈8.06 cm.

9. Q9 — Rectangular field and tent

   1. w²=100²−80²=10000−6400=3600. w=60 m
   2. Area = 80 × 60 = 4800 m²
   3. Perimeter = 2(80+60) = 2 × 140 = 280 m
   4. Fencing cost = 280 × $45 = $12 600. Two diagonals = 2 × 100 = 200 m. Path cost = 200 × $12 = $2 400. Total = $15 000

10. Q10 — Coordinate geometry and tent

    1. Half-base = 2 m. h=√(2.5²−2²)=√(6.25−4)=√2.25=1.5 m
    2. Area = ½ × 4 × 1.5 = 3 m²
    3. AB²=k²+64=100 → k²=36 → k=6
    4. Second tent: h=√(3²−2²)=√(9−4)=√5≈2.24 m. This is taller than the first tent (1.5 m) because the slant sides are longer relative to the base.