Practice Maths

Solutions: Area of a Triangle

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  1. Fluency
    1. \[\text{Area} = \tfrac{1}{2}(8)(12)\sin 40° = 48 \times 0.6428 \approx \mathbf{30.85 \text{ cm}^2}\]
    2. \[\text{Area} = \tfrac{1}{2}(15)(10)\sin 75° = 75 \times 0.9659 \approx \mathbf{72.44 \text{ cm}^2}\]
    3. \[\text{Area} = \tfrac{1}{2}(6)(6)\sin 120° = 18 \times 0.8660 \approx \mathbf{15.59 \text{ cm}^2}\]
  2. Fluency
    1. \[\text{Area} = \tfrac{1}{2}(14)(20)\sin 62° = 140 \times 0.8829 \approx \mathbf{123.6 \text{ cm}^2}\]

    2. Third side c (opposite the 62° angle):

      \[c^2 = 14^2 + 20^2 - 2(14)(20)\cos 62° = 196 + 400 - 560 \times 0.4695 = 596 - 262.9 = 333.1\] \[c = \sqrt{333.1} \approx 18.25 \text{ cm}\]

      Perimeter = 14 + 20 + 18.25 = 52.25 cm

  3. Fluency

    Area = ½ab sin C  ⇒  42 = ½(9)(14) sin C = 63 sin C

    \[\sin C = \frac{42}{63} = \frac{2}{3} \approx 0.6667\] \[C = \sin^{-1}(0.6667) \approx 41.8°\]

    Also valid: C = 180° − 41.8° = 138.2° (obtuse triangle also gives Area = 42 m²).

    The included angle is approximately 41.8° (or 138.2° for an obtuse triangle).

  4. Understanding

    1. Sides a = 18, b = 22, c = 30. Find angle C (largest, opposite c = 30).

      \[\cos C = \frac{18^2 + 22^2 - 30^2}{2(18)(22)} = \frac{324 + 484 - 900}{792} = \frac{-92}{792} = -0.1162\] \[C = \cos^{-1}(-0.1162) \approx 96.7°\] \[\text{Area} = \tfrac{1}{2}(18)(22)\sin 96.7° = 198 \times 0.9933 \approx \mathbf{196.7 \text{ m}^2}\]

    2. Heron’s formula: s = (18 + 22 + 30)/2 = 35

      \[\text{Area} = \sqrt{35 \times (35-18) \times (35-22) \times (35-30)}\] \[= \sqrt{35 \times 17 \times 13 \times 5} = \sqrt{38675} \approx \mathbf{196.7 \text{ m}^2} \checkmark\]
  5. Understanding

    1. A rhombus = 2 congruent triangles. Each triangle has two sides of 12 cm and included angle 65°.

      \[\text{Area of rhombus} = 2 \times \tfrac{1}{2}(12)(12)\sin 65° = 144 \times 0.9063 \approx \mathbf{130.5 \text{ cm}^2}\]

    2. Diagonal d₁ is the diagonal that spans the 65° angle (the shorter diagonal):

      \[d_1^2 = 12^2 + 12^2 - 2(12)(12)\cos 65° = 288 - 288 \times 0.4226 = 288 - 121.7 = 166.3\] \[d_1 = \sqrt{166.3} \approx 12.90 \text{ cm}\]

      Diagonal d₂ spans the other angle = 180° − 65° = 115°:

      \[d_2^2 = 288 - 288\cos 115° = 288 - 288 \times (-0.4226) = 288 + 121.7 = 409.7\] \[d_2 = \sqrt{409.7} \approx 20.24 \text{ cm}\]
    3. \[\text{Area} = \tfrac{1}{2}d_1 d_2 = \tfrac{1}{2} \times 12.90 \times 20.24 \approx \mathbf{130.5 \text{ cm}^2} \checkmark\]
  6. Understanding
    1. \[\text{Area} = \tfrac{1}{2}(42)(58)\sin 70° = 1218 \times 0.9397 \approx \mathbf{1144.5 \text{ m}^2}\]
    2. \[BC^2 = 42^2 + 58^2 - 2(42)(58)\cos 70° = 1764 + 3364 - 4872 \times 0.3420 = 5128 - 1666.2 = 3461.8\] \[BC = \sqrt{3461.8} \approx \mathbf{58.83 \text{ m}}\]

    3. Perimeter = 42 + 58 + 58.83 = 158.83 m

      \[\text{Cost} = 158.83 \times \$85 \approx \mathbf{\$13{,}500.55}\]
  7. Understanding
    1. \[\text{Area of parallelogram} = ab\sin\theta = (15)(22)\sin 48° = 330 \times 0.7431 \approx \mathbf{245.2 \text{ cm}^2}\]

    2. Each triangle = half the parallelogram.

      \[\text{Area of each triangle} = \tfrac{1}{2} \times 245.2 \approx \mathbf{122.6 \text{ cm}^2}\]

    3. Shorter diagonal (spans 48° angle):

      \[d_1^2 = 15^2 + 22^2 - 2(15)(22)\cos 48° = 225 + 484 - 660 \times 0.6691 = 709 - 441.6 = 267.4\] \[d_1 \approx 16.35 \text{ cm}\]

      Longer diagonal (spans 132° angle):

      \[d_2^2 = 225 + 484 - 660\cos 132° = 709 - 660 \times (-0.6691) = 709 + 441.6 = 1150.6\] \[d_2 \approx 33.92 \text{ cm}\]
  8. Understanding

    1. Bearing A to B = 050°. Bearing B to C = 140°.

      At point B, the hiker was travelling on bearing 050° (from A). The bearing from B to C is 140°. The change in direction = 140° − 050° = 90°. So the exterior angle at B is 90°, meaning the interior angle ABC = 180° − 90° = 90°.

    2. Since angle B = 90°:

      \[AC = \sqrt{AB^2 + BC^2} = \sqrt{4.5^2 + 6^2} = \sqrt{20.25 + 36} = \sqrt{56.25} = \mathbf{7.5 \text{ km}}\]
    3. \[\text{Area} = \tfrac{1}{2}(4.5)(6)\sin 90° = \tfrac{1}{2}(4.5)(6)(1) = \mathbf{13.5 \text{ km}^2}\]
  9. Problem Solving
    1. \[\text{Area}_{ABD} = \tfrac{1}{2}(85)(110)\sin 68° = 4675 \times 0.9272 \approx \mathbf{4334.7 \text{ m}^2}\]
    2. \[\text{Area}_{BCD} = \tfrac{1}{2}(95)(75)\sin 54° = 3562.5 \times 0.8090 \approx \mathbf{2882.1 \text{ m}^2}\]

    3. Total area = 4334.7 + 2882.1 = 7216.8 m²

    4. BD from triangle ABD (A = 68° is included angle between AB = 85 and AD = 110):

      \[BD^2 = 85^2 + 110^2 - 2(85)(110)\cos 68° = 7225 + 12100 - 18700 \times 0.3746\] \[= 19325 - 7003 = 12322\] \[BD = \sqrt{12322} \approx \mathbf{111.0 \text{ m}}\]

    5. BD from triangle BCD (C = 54° is included angle between BC = 95 and CD = 75):

      \[BD^2 = 95^2 + 75^2 - 2(95)(75)\cos 54° = 9025 + 5625 - 14250 \times 0.5878\] \[= 14650 - 8377 = 6273\] \[BD = \sqrt{6273} \approx 79.2 \text{ m}\]

      Note: the two BD values differ because ABCD is not constrained — each triangle specifies BD independently. The problem has two different triangles sharing diagonal BD, but unless BD is consistent, this is a two-triangle problem, not a single quadrilateral. The values given produce BD ≈ 111.0 m (from ABD) and BD ≈ 79.2 m (from BCD). This shows the parcel data specifies two separate triangular parcels, each with its own area calculation. Total area = 4334.7 + 2882.1 = 7216.8 m² as calculated above.

  10. Problem Solving

    1. A is 120 m above sea level. Angle of depression to B = 12°.

      Vertical drop = 120 m (from A to sea level). tan 12° = vertical/horizontal.

      \[\text{Horizontal AB} = \frac{120}{\tan 12°} = \frac{120}{0.2126} \approx \mathbf{564.4 \text{ m}}\]

    2. Angle of depression to C = 8°.

      \[\text{Horizontal AC} = \frac{120}{\tan 8°} = \frac{120}{0.1405} \approx \mathbf{854.1 \text{ m}}\]

    3. Bearing of B from A = 080°, bearing of C from A = 160°.

      Angle BAC = 160° − 080° = 80°

    4. \[\text{Area} = \tfrac{1}{2}(564.4)(854.1)\sin 80° = \tfrac{1}{2} \times 482\,149 \times 0.9848 \approx \mathbf{237{,}400 \text{ m}^2}\]

      (approximately 23.7 hectares)