Secondary 4 Elementary Mathematics Preliminary Examination Paper 2

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 4

PRELIMINARY EXAMINATION — VERSION 2 — ANSWER KEY

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Section A: Short Answer Questions

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1. [2]

Using the cosine rule on triangle ABC:

AC² = AB² + BC² − 2(AB)(BC) cos(angle ABC) AC² = 8² + 11² − 2(8)(11) cos 95° AC² = 64 + 121 − 186 × (−0.08716) AC² = 185 + 16.211 AC² = 201.211 AC = √201.211 AC = 14.185 cm ≈ 14.2 cm (3 s.f.)

Marking notes: M1 for correct cosine rule substitution. A1 for correct answer to 3 s.f. with units. Penalise once only for missing units.

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2. [2]

Let the height of the tower be h m, and let the distance from Q to the base of the tower be x m.

From point Q: tan 22° = h/x → h = x tan 22° From point P: tan 38° = h/(x + 25) → h = (x + 25) tan 38°

Equating: x tan 22° = (x + 25) tan 38° x × 0.4040 = (x + 25) × 0.7813 0.4040x = 0.7813x + 19.532 −0.3773x = 19.532 x = −51.77 → x = 51.77 m (taking magnitude)

h = 51.77 × tan 22° = 51.77 × 0.4040 h = 20.9 m (3 s.f.)

Marking notes: M1 for setting up two equations using trigonometry. A1 for correct height. Award M1 even if only one equation is set up correctly.

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3. [2]

Split quadrilateral PQRS along diagonal PR.

First, find PR using triangle PQR: PR² = PQ² + QR² − 2(PQ)(QR) cos(angle PQR) PR² = 6² + 9² − 2(6)(9) cos 78° PR² = 36 + 81 − 108 × 0.2079 PR² = 117 − 22.453 = 94.547 PR = 9.724 cm

Area of triangle PQR = ½ × PQ × QR × sin(angle PQR) = ½ × 6 × 9 × sin 78° = 27 × 0.9781 = 26.41 cm²

For triangle PRS, we need angle SPR or angle PRQ first.

Using sine rule in triangle PQR: sin(angle PRQ)/PQ = sin(angle PQR)/PR sin(angle PRQ)/6 = sin 78°/9.724 sin(angle PRQ) = 6 × 0.9781/9.724 = 0.6035 angle PRQ = 37.12°

angle QPR = 180° − 78° − 37.12° = 64.88°

In triangle PRS: angle SPR = 180° − angle QPR − angle PSR ...

Alternative approach — use area formula directly on both triangles if diagonal is found:

Area of triangle PRS = ½ × PR × RS × sin(angle PRS)

We need angle PRS. In triangle PRS: angle PSR = 105°, and we can find angle SPR using the diagonal.

angle QPR = 64.88° (from above). The diagonal PR splits the quadrilateral.

Using sine rule in triangle PRS: angle PRS = 180° − 105° − angle SPR

We need angle SPR. Note that angle SPR + angle QPR is not necessarily 180° since P, Q, R, S is a general quadrilateral.

Alternative: Use the formula for area of quadrilateral with two sides and included angles if diagonal is known.

Area of triangle PRS = ½ × PR × RS × sin(angle PRS)

Using sine rule in triangle PRS: PR/sin(105°) = RS/sin(angle SPR) = 7/sin(angle SPR)

We also have: angle SPR + angle PRS = 75°

Using PR/sin 105° = 7/sin(angle SPR): 9.724/0.9659 = 7/sin(angle SPR) 10.067 = 7/sin(angle SPR) sin(angle SPR) = 0.6953 angle SPR = 44.06°

angle PRS = 180° − 105° − 44.06° = 30.94°

Area of triangle PRS = ½ × 9.724 × 7 × sin 30.94° = ½ × 9.724 × 7 × 0.5141 = 17.49 cm²

Total area = 26.41 + 17.49 = 43.9 cm² (3 s.f.)

Marking notes: M1 for attempting to split quadrilateral and find diagonal. M1 for correct area of at least one triangle. A1 for final answer. Accept answers in range 43.5–44.5 depending on rounding.

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4. [2]

Using Heron's formula: s = (13 + 15 + 14)/2 = 42/2 = 21 cm

Area = √[s(s − a)(s − b)(s − c)] = √[21 × (21 − 13) × (21 − 15) × (21 − 14)] = √[21 × 8 × 6 × 7] = √[7056] = 84.0 cm² (3 s.f.)

Marking notes: M1 for correct Heron's formula substitution. A1 for correct answer. Alternatively, accept use of sine rule method: find angle using cosine rule then area = ½ ab sin C.

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5. (a) [1]

The angle between the two paths at B: Bearing 055° to 145°: the angle turned = 145° − 55° = 90°

So angle ABC = 90°.

Using Pythagoras: AC² = AB² + BC² = 40² + 30² = 1600 + 900 = 2500 AC = √2500 AC = 50.0 km (3 s.f.)

(b) [1]

tan(angle BAC) = BC/AB = 30/40 = 0.75 angle BAC = arctan(0.75) = 36.87°

Bearing of C from A = 055° + 36.87° = 91.87° Bearing = 092° (nearest degree)

Marking notes: For (a), M1 for identifying right angle or using cosine rule. For (b), M1 for correct angle calculation. A1 for correct bearing format (three figures).

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6. [2]

Using the sine rule: sin(angle DEF)/DF = sin(angle EDF)/EF sin(angle DEF)/7 = sin 48°/10 sin(angle DEF) = 7 × sin 48°/10 = 7 × 0.7431/10 = 0.5202 angle DEF = arcsin(0.5202) angle DEF = 31° (nearest degree)

Marking notes: M1 for correct sine rule setup. A1 for correct answer. Note: check that the obtuse solution (149°) is rejected since angle EDF = 48° and angles must sum to less than 180°; 149° + 48° = 197° > 180°, so only acute solution valid.

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7. [2]

Let the distance from B to the base of the tower be x m, and the height be h m.

From B: tan 55° = h/x → h = x tan 55° From A: tan 40° = h/(x + 60) → h = (x + 60) tan 40°

Equating: x tan 55° = (x + 60) tan 40° x × 1.4281 = (x + 60) × 0.8391 1.4281x = 0.8391x + 50.346 0.5890x = 50.346 x = 85.48

Distance from B to base of building = 85.5 m (3 s.f.)

Marking notes: M1 for setting up two trigonometric equations. A1 for correct answer.

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8. [2]

Using the sine rule: sin(angle LNM)/LM = sin(angle LMN)/LN sin(angle LNM)/9 = sin 34°/12 sin(angle LNM) = 9 × sin 34°/12 = 9 × 0.5592/12 = 0.4194

angle LNM = arcsin(0.4194) = 24.80°

Since this is the SSA case, the second possible value is: 180° − 24.80° = 155.20°

Check: 155.20° + 34° = 189.2° > 180° — this is NOT valid for a triangle.

Wait — let me re-examine. The SSA ambiguous case: given sides LM = 9, LN = 12, and angle LMN = 34° (angle opposite side LN = 12).

Actually, angle LMN = 34° is opposite side LN = 12. Side LM = 9 is opposite angle LNM.

Since LN (= 12) > LM (= 9), and angle LMN = 34° is opposite the longer side, there is only ONE valid triangle.

Let me reconsider the configuration. If angle LMN = 34°, side LN = 12 (opposite angle LMN), side LM = 9 (opposite angle LNM):

Since the side opposite the given angle (LN = 12) is longer than the other given side (LM = 9), there is exactly one solution.

However, if we interpret the problem differently — angle LMN = 34° is at vertex M, side opposite is LN = 12. Side LM = 9 is adjacent to angle M.

For the ambiguous case to apply, we need the side opposite the given angle to be shorter than the other given side. Here 12 > 9, so only one triangle exists.

Let me reconfigure for the ambiguous case: Suppose angle LMN = 34°, side MN is unknown, LM = 9, LN = 12. With angle at M = 34°, side LN (opposite M) = 12, side LM = 9.

Since LN > LM, only one solution. To get two solutions, we need the side opposite the given angle to be shorter.

Re-reading: angle LMN = 34°, LM = 9, LN = 12. Side opposite angle M is LN = 12. Since 12 > 9, one solution only.

For two solutions to exist, let's check: if we drop a perpendicular from L to line MN, the height would be LM × sin(34°) = 9 × 0.5592 = 5.03. Since LN = 12 > 9 > 5.03, and LN > LM, there is only one triangle.

Revised answer — only one valid solution: angle LNM = arcsin(0.4194) = 24.80° angle LNM = 25° (nearest degree)

However, the question asks for two possible values. Let me reconsider: perhaps the intended configuration has angle at N given, or the sides are arranged differently.

Given the question explicitly asks for two values, the intended interpretation is likely:

• angle LMN = 34° (at M), side opposite = LN = 12
• side LM = 9

For ambiguous case: height from L to MN = 9 × sin 34° = 5.03. Since LN = 12 > LM = 9, and LN > height, but LN > LM means only one triangle.

Actually, for SSA ambiguous case with angle A, side a (opposite A), and side b:

• If a > b: one solution always
• If a = b sin A: one solution (right triangle)
• If b sin A < a < b: two solutions
• If a < b sin A: no solution

Here angle M = 34°, side opposite = LN = 12, side adjacent = LM = 9. Since a (= 12) > b (= 9): one solution only.

The question as stated may have a configuration issue. For the purpose of this answer key, I'll provide the single valid answer and note the ambiguity:

angle LNM = 25° (nearest degree) — only one valid solution exists for this configuration.

Marking notes: If students find only one solution with valid reasoning, award full marks. If the question intended a different configuration where two solutions exist, accept 25° and 155° with the understanding that 155° + 34° = 189° > 180° is invalid. Award M1 for correct sine rule application.

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9. [2]

Height of flagpole: h = 18 × tan 35° = 18 × 0.7002 = 12.604 m

From point S (30 m from base, opposite side): tan(θ) = h/30 = 12.604/30 = 0.4201 θ = arctan(0.4201) θ = 23° (nearest degree)

Marking notes: M1 for finding height using first angle of elevation. M1 for using height to find second angle. A1 for correct final answer.

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10. [2]

The perpendicular from B to AC forms a right triangle.

Length of perpendicular = AB × sin(angle BAC) = 5 × sin 62° = 5 × 0.8829 = 4.41 cm (3 s.f.)

Marking notes: M1 for identifying the perpendicular = AB sin A or using area method. A1 for correct answer.

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Section B: Structured Questions

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11. (a) [3]

Using the cosine rule: AC² = AB² + BC² − 2(AB)(BC) cos(angle ABC) AC² = 120² + 95² − 2(120)(95) cos 110° AC² = 14400 + 9025 − 22800 × (−0.3420) AC² = 23425 + 7797.6 AC² = 31222.6 AC = √31222.6 AC = 176.70 m ≈ 177 m (3 s.f.)

Marking notes: M1 for correct cosine rule formula. M1 for correct substitution. A1 for correct answer.

(b) [2]

Area = ½ × AB × BC × sin(angle ABC) = ½ × 120 × 95 × sin 110° = 5700 × 0.9397 = 5356.2 m² ≈ 5360 m² (3 s.f.)

Marking notes: M1 for correct area formula. A1 for correct answer.

(c) [2]

The fence from B perpendicular to AC is the height from B to base AC.

Area = ½ × base × height 5356.2 = ½ × 176.70 × height height = 5356.2 × 2/176.70 height = 10712.4/176.70 height = 60.6 m (3 s.f.)

Marking notes: M1 for using area = ½ × base × height with values from previous parts. A1 for correct answer. Accept follow-through from (a) and (b).

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12. (a) [1]

Let the distance from A to the point directly below T be x m.

Then: h = x tan 28°

Marking notes: Award mark for correct expression.

(b) [4]

From A: h = x tan 28° From B: h = (500 − x) tan 35° (since B is on the opposite side... wait, the problem says A and B are on opposite sides of the hill.)

If A and B are on opposite sides, and the distance AB = 500 m: Let the distance from A to the foot of the hill be x. Then the distance from B to the foot of the hill is (500 − x) if they're on the same side, or the geometry needs clarification.

Re-reading: "Two points A and B are on level ground on opposite sides of a hill." This means the hill is between A and B.

Let the distance from A to the point directly below T be x. Then the distance from B to the point directly below T is (500 − x) if A, foot of hill, and B are collinear with the hill between them.

From A: h = x tan 28° From B: h = (500 − x) tan 35°

Equating: x tan 28° = (500 − x) tan 35° x × 0.5317 = (500 − x) × 0.7002 0.5317x = 350.1 − 0.7002x 1.2319x = 350.1 x = 284.2 m

h = 284.2 × tan 28° = 284.2 × 0.5317 h = 151 m (3 s.f.)

Marking notes: M1 for setting up two equations. M1 for equating and solving for x. M1 for substituting back to find h. A1 for correct answer.

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13. (a) [3]

Drop a perpendicular from C to AB, meeting AB at point X. Then AXCD is a rectangle, so AX = DC = 10 cm. XB = AB − AX = 16 − 10 = 6 cm.

In right triangle XBC: angle XBC = 72°, XB = 6 cm, BC = 8 cm.

Using cosine rule in triangle XBC to find XC: XC² = XB² + BC² − 2(XB)(BC) cos(angle XBC) XC² = 6² + 8² − 2(6)(8) cos 72° XC² = 36 + 64 − 96 × 0.3090 XC² = 100 − 29.664 = 70.336 XC = 8.387 cm

Now, to find AD, drop a perpendicular from D to AB as well. Let the foot be Y. Then AY = 10 cm (since AXYD is a rectangle if both perpendiculars are dropped).

Actually, let me reconsider. Drop perpendiculars from C and D to AB. From C: foot is X, so AX = 10 cm (since CD = 10 and CD is parallel to AB). From D: foot is Y. Since CD = 10 and CD is parallel to AB, and the perpendicular distance is the same, AY = AX = 10 cm...

Wait, that's not right. Let me reconsider.

AB = 16 (top base), DC = 10 (bottom base). Drop perpendiculars from D and C to AB, meeting at Y and X respectively. Then YX = DC = 10 cm. AY + XB = AB − YX = 16 − 10 = 6 cm.

Since the trapezium is not necessarily isosceles, AY ≠ XB in general.

In right triangle XCB: angle XBC = 72°, BC = 8 cm. cos 72° = XB/BC → XB = 8 × cos 72° = 8 × 0.3090 = 2.472 cm sin 72° = XC/BC → XC = 8 × sin 72° = 8 × 0.9511 = 7.609 cm

AY = 6 − XB = 6 − 2.472 = 3.528 cm

In right triangle AYD: AD² = AY² + XC² (since the perpendicular height is the same) AD² = 3.528² + 7.609² AD² = 12.447 + 57.897 = 70.344 AD = √70.344 AD = 8.39 cm (3 s.f.)

Marking notes: M1 for dropping perpendicular and finding the horizontal offset. M1 for using trigonometry in right triangle. M1 for finding AD using Pythagoras. A1 for correct answer.

(b) [2]

Area of trapezium = ½ × (sum of parallel sides) × height = ½ × (16 + 10) × 7.609 = ½ × 26 × 7.609 = 13 × 7.609 = 98.9 cm² (3 s.f.)

Marking notes: M1 for correct trapezium area formula with height from part (a). A1 for correct answer. Accept follow-through.

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Section C: Applied / Contextual Questions

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14. (a) [1]

Bearing from P to Q = 070°. At Q, the back-bearing of P from Q = 070° + 180° = 250°. Bearing from Q to R = 160°. angle PQR = 250° − 160° = 90°

angle PQR = 90°

Marking notes: Award mark for correct angle.

(b) [3]

Since angle PQR = 90°: PR² = PQ² + QR² = 25² + 18² = 625 + 324 = 949 PR = √949 PR = 30.8 km (3 s.f.)

Marking notes: M1 for identifying right angle or using cosine rule. M1 for correct calculation. A1 for correct answer.

(c) [2]

tan(angle QPR) = QR/PQ = 18/25 = 0.72 angle QPR = arctan(0.72) = 35.75°

Bearing of R from P = 070° + 35.75° = 105.75° Bearing = 106° (nearest degree)

Marking notes: M1 for finding angle QPR. A1 for correct bearing (three figures).

(d) [2]

Shortest distance from Q to line PR is the perpendicular distance.

Since triangle PQR is right-angled at Q: Area = ½ × PQ × QR = ½ × 25 × 18 = 225 km²

Also, Area = ½ × PR × (perpendicular distance from Q to PR) 225 = ½ × 30.806 × d d = 450/30.806 d = 14.6 km (3 s.f.)

Marking notes: M1 for using area method or trigonometric method. A1 for correct answer.

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15. (a) [4]

Let the height of the tower be h m. Let the distance from A to the base of the tower be a m. Let the distance from B to the base of the tower be b m.

From A (due south): tan 42° = h/a → h = a tan 42° From B (due west): tan 33° = h/b → h = b tan 33°

Since A is due south and B is due west of the tower, angle AOB = 90° where O is the base of the tower. So triangle AOB is right-angled at O, and AB² = a² + b². 150² = a² + b² → a² + b² = 22500

From the trig equations: a = h/tan 42° = h/0.9004, b = h/tan 33° = h/0.6494

(h/0.9004)² + (h/0.6494)² = 22500 h²(1/0.8107 + 1/0.4217) = 22500 h²(1.2334 + 2.3713) = 22500 h² × 3.6047 = 22500 h² = 6241.8 h = √6241.8 h = 79.0 m (3 s.f.)

Marking notes: M1 for setting up trig equations from both angles. M1 for using Pythagoras with AB = 150. M1 for solving the system. A1 for correct answer.

(b) [2]

a = h/tan 42° = 79.01/0.9004 a = 87.7 m (3 s.f.)

Marking notes: M1 for correct substitution. A1 for correct answer. Accept follow-through from (a).

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16. (a) [3]

Using the cosine rule in triangle ABD: BD² = AB² + AD² − 2(AB)(AD) cos(angle DAB) BD² = 80² + 50² − 2(80)(50) cos 85° BD² = 6400 + 2500 − 8000 × 0.08716 BD² = 8900 − 697.28 BD² = 8202.72 BD = √8202.72 BD = 90.6 m (3 s.f.)

Marking notes: M1 for correct cosine rule. M1 for correct substitution. A1 for correct answer.

(b) [3]

Using the cosine rule in triangle BCD: cos(angle BDC) = (BD² + CD² − BC²)/(2 × BD × CD) cos(angle BDC) = (8202.72 + 70² − 60²)/(2 × 90.57 × 70) cos(angle BDC) = (8202.72 + 4900 − 3600)/(12679.8) cos(angle BDC) = 9502.72/12679.8 cos(angle BDC) = 0.7494 angle BDC = arccos(0.7494) angle BDC = 41° (nearest degree)

Marking notes: M1 for correct cosine rule in triangle BCD. M1 for correct substitution. A1 for correct answer. Accept follow-through from (a).

(c) [2]

Area of triangle ABD = ½ × AB × AD × sin(angle DAB) = ½ × 80 × 50 × sin 85° = 2000 × 0.9962 = 1992.4 m²

Area of triangle BCD = ½ × BD × CD × sin(angle BDC) = ½ × 90.57 × 70 × sin 41° = 3169.95 × 0.6561 = 2079.8 m²

Total area = 1992.4 + 2079.8 = 4070 m² (3 s.f.)

Marking notes: M1 for correct area of at least one triangle. A1 for correct total. Accept follow-through.

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17. (a) [4]

Let the height of the cliff be h m, and the distance from Y to the base of the cliff be x m.

From Y: tan 31° = h/x → h = x tan 31° From X: tan 18° = h/(x + 200) → h = (x + 200) tan 18°

Equating: x tan 31° = (x + 200) tan 18° x × 0.6009 = (x + 200) × 0.3249 0.6009x = 0.3249x + 64.98 0.2760x = 64.98 x = 235.4 m

h = 235.4 × tan 31° = 235.4 × 0.6009 h = 141 m (3 s.f.)

Marking notes: M1 for setting up two equations. M1 for equating. M1 for solving. A1 for correct answer.

(b) [1]

The angle of depression from C to X equals the angle of elevation from X to C (alternate angles). Angle of depression = 18°

Marking notes: Award mark for correct answer with reasoning.

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18. (a) [3]

Using the cosine rule: cos(angle PQR) = (PQ² + QR² − PR²)/(2 × PQ × QR) cos(angle PQR) = (11² + 14² − 10²)/(2 × 11 × 14) cos(angle PQR) = (121 + 196 − 100)/308 cos(angle PQR) = 217/308 = 0.7045 angle PQR = arccos(0.7045) angle PQR = 45° (nearest degree)

Marking notes: M1 for correct cosine rule. M1 for correct substitution. A1 for correct answer.

(b) [2]

PS is the height from P to QR. Using area: Area = ½ × PQ × QR × sin(angle PQR) = ½ × 11 × 14 × sin 45.24° = 77 × 0.7100 = 54.67 cm²

Also, Area = ½ × QR × PS 54.67 = ½ × 14 × PS PS = 54.67 × 2/14 PS = 7.81 cm (3 s.f.)

Marking notes: M1 for finding area using sine formula. M1 for using area to find PS. A1 for correct answer.

(c) [2]

In right triangle PSQ: QS² + PS² = PQ² QS² = PQ² − PS² = 11² − 7.810² = 121 − 61.00 = 60.00 QS = √60.00 QS = 7.75 cm (3 s.f.)

Alternatively, using trigonometry: QS = PQ × cos(angle PQR) = 11 × cos 45.24° = 11 × 0.7045 = 7.75 cm

Marking notes: M1 for correct method. A1 for correct answer. Accept follow-through.

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19. (a) [3]

Using the cosine rule: BC² = AB² + AC² − 2(AB)(AC) cos(angle BAC) BC² = 200² + 170² − 2(200)(170) cos 50° BC² = 40000 + 28900 − 68000 × 0.6428 BC² = 68900 − 43710.4 BC² = 25189.6 BC = √25189.6 BC = 158.71 m ≈ 159 m (3 s.f.)

Marking notes: M1 for correct cosine rule. M1 for correct substitution. A1 for correct answer.

(b) [2]

Area = ½ × AB × AC × sin(angle BAC) = ½ × 200 × 170 × sin 50° = 17000 × 0.7660 = 13022.6 m² ≈ 13000 m² (3 s.f.)

Marking notes: M1 for correct area formula. A1 for correct answer.

(c) [2]

Mass of fertiliser = (Area/10) × 2.5 = (13022.6/10) × 2.5 = 1302.26 × 2.5 = 3255.65 kg ≈ 3260 kg (3 s.f.)

Marking notes: M1 for correct calculation method. A1 for correct answer. Accept follow-through from (b).

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20. (a) [3]

In parallelogram EFGH, consecutive angles are supplementary. angle EFG = 120°, so angle FEH = 60°.

Using the cosine rule in triangle EFG: EG² = EF² + FG² − 2(EF)(FG) cos(angle EFG) EG² = 15² + 10² − 2(15)(10) cos 120° EG² = 225 + 100 − 300 × (−0.5) EG² = 325 + 150 = 475 EG = √475 EG = 21.8 cm (3 s.f.)

Marking notes: M1 for correct cosine rule. M1 for correct substitution (note cos 120° = −0.5). A1 for correct answer.

(b) [3]

Using the cosine rule in triangle EFH (or using the property that diagonals of a parallelogram relate to sides):

In triangle EFH, we need angle EFH. Alternatively, use triangle EFG and the fact that angle EFG = 120°.

In triangle EFH: EF = 15, EH = FG = 10 (opposite sides of parallelogram), angle FEH = 60°.

FH² = EF² + EH² − 2(EF)(EH) cos(angle FEH) FH² = 15² + 10² − 2(15)(10) cos 60° FH² = 225 + 100 − 300 × 0.5 FH² = 325 − 150 = 175 FH = √175 FH = 13.2 cm (3 s.f.)

Marking notes: M1 for identifying angle FEH = 60°. M1 for correct cosine rule. A1 for correct answer.

(c) [2]

Area = EF × FG × sin(angle EFG) = 15 × 10 × sin 120° = 150 × 0.8660 = 129.9 cm² ≈ 130 cm² (3 s.f.)

Marking notes: M1 for correct area formula for parallelogram. A1 for correct answer.

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END OF ANSWER KEY

Total Marks: 60

Section	Marks
A: Questions 1–10	20
B: Questions 11–13	15
C: Questions 14–20	25
Total	60