This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Step 1: Solve Exercise 18, Question 1a). The triangle is right-angled at B. We are given the hypotenuse AC = 40 cm and a leg BC = 32 cm. We need to find the other leg AB. Using the Pythagorean theorem AB^2 + BC^2 = AC^2: AB^2 + 32^2 = 40^2 AB^2 + 1024 = 1600 AB^2 = 1600 - 1024 AB^2 = 576 AB = sqrt(576) AB = 24 cm The unknown side is 24 cm. Step 2: Solve Exercise 18, Question 1b). The triangle is right-angled at D. We are given the hypotenuse EF = 24 cm and a leg DE = 10 cm. We need to find the other leg DF. Using the Pythagorean theorem DE^2 + DF^2 = EF^2: 10^2 + DF^2 = 24^2 100 + DF^2 = 576 DF^2 = 576 - 100 DF^2 = 476 DF = sqrt(476) To simplify sqrt(476), we find its prime factors: 476 = 4 × 119. DF = sqrt(4 × 119) = sqrt(4) × sqrt(119) DF = 2sqrt(119) cm The unknown side is 2sqrt(119) cm. Step 3: Solve Exercise 18, Question 1c). The triangle is right-angled at P. We are given the hypotenuse QR = 25 cm and a leg PQ = 15 cm. We need to find the other leg PR. Using the Pythagorean theorem PQ^2 + PR^2 = QR^2: 15^2 + PR^2 = 25^2 225 + PR^2 = 625 PR^2 = 625 - 225 PR^2 = 400 PR = sqrt(400) PR = 20 cm The unknown side is 20 cm. Step 4: Solve Exercise 18, Question 2a). The triangle is right-angled at A. We are given the hypotenuse BC = 18 cm and a leg AC = 12 cm. We need to find the other leg AB. Using the Pythagorean theorem AB^2 + AC^2 = BC^2: AB^2 + 12^2 = 18^2 AB^2 + 144 = 324 AB^2 = 324 - 144 AB^2 = 180 AB = sqrt(180) To simplify sqrt(180), we find its prime factors: 180 = 36 × 5. AB = sqrt(36 × 5) = sqrt(36) × sqrt(5) AB = 6sqrt(5) cm The unknown side is 6sqrt(5) cm. Step 5: Solve Exercise 18, Question 2b). The triangle is right-angled at N. We are given the hypotenuse MO = 21 cm and a leg NO = 15 cm. We need to find the other leg MN. Using the Pythagorean theorem MN^2 + NO^2 = MO^2: MN^2 + 15^2 = 21^2 MN^2 + 225 = 441 MN^2 = 441 - 225 MN^2 = 216 MN = sqrt(216) To simplify sqrt(216), we find its prime factors: 216 = 36 × 6. MN = sqrt(36 × 6) = sqrt(36) × sqrt(6) MN = 6sqrt(6) cm The unknown side is 6sqrt(6) cm. Step 6: Solve Exercise 18, Question 2c). The triangle is right-angled at F. We are given the hypotenuse GH = 12 cm and a leg FG = 7 cm. We need to find the other leg FH. Using the Pythagorean theorem FG^2 + FH^2 = GH^2: 7^2 + FH^2 = 12^2 49 + FH^2 = 144 FH^2 = 144 - 49 FH^2 = 95 FH = sqrt(95) cm The number 95 has no perfect square factors (95 = 5 × 19), so it cannot be simplified further. The unknown side is sqrt(95) cm. Step 7: Solve Exercise 18, Question 3. The ladder, wall, and ground form a right-angled triangle. The length of the ladder is the hypotenuse (c = 4.25 m). The distance from the wall to the foot of the ladder is one leg (a = 2 m). The height above the ground is the other leg (b). Using the Pythagorean theorem a^2 + b^2 = c^2: 2^2 + b^2 = 4.25^2 4 + b^2 = 18.0625 b^2 = 18.0625 - 4 b^2 = 14.0625 b = sqrt(14.0625) b = 3.75 m The height above the ground is 3.75 m. Step 8: Solve Exercise 18, Question 4. To determine if a triangle is right-angled, we check if the square of the longest side is equal to the sum of the squares of the other two sides (c^2 = a^2 + b^2). The right angle is opposite the longest side. a)* ABC with a=9 cm, b=7 cm and c=4 cm. The longest side is a=9 cm. a^2 = 9^2 = 81. b^2 + c^2 = 7^2 + 4^2 = 49 + 16 = 65. Since 81 ≠ 65, the triangle is not right-angled. b)* DEF with d=5 cm, e=13 cm and f=12 cm. The longest side is e=13 cm. e^2 = 13^2 = 169. d^2 + f^2 = 5^2 + 12^2 = 25 + 144 = 169. Since 169 = 169, the triangle is right-angled. The right angle is opposite side e (which is DF), so the right angle is at E. c)* STR with s=20 cm, t=16 cm and r=10 cm. The longest side is s=20 cm. s^2 = 20^2 = 400. t^2 + r^2 = 16^2 + 10^2 = 256 + 100 = 356. Since 400 ≠ 356, the triangle is not right-angled. d)* PQR with p=24 cm, q=25 cm and r=7 cm. The longest side is q=25 cm. q^2 = 25^2 = 625. p^2 + r^2 = 24^2 + 7^2 = 576 + 49 = 625. Since 625 = 625, the triangle is right-angled. The right angle is opposite side q (which is PR), so the right angle is at Q. Step 9: Solve Exercise 18, Question 5. The ladder, wall, and ground form a right-angled triangle. The height against the wall is one leg (a = 6 m). The distance from the wall to the foot of the ladder is the other leg (b = 2.5 m). The length of the ladder is the hypotenuse (c). Using the Pythagorean theorem c^2 = a^2 + b^2: c^2 = 6^2 + 2.5^2 c^2 = 36 + 6.25 c^2 = 42.25 c = sqrt(42.25) c = 6.5 m The ladder must be 6.5 m long. What's next?