Geometry & MeasuresStudy Notes
Common Mistakes
Part of Area of Parallelograms & Trapeziums — GCSE Mathematics
This study notes covers Common Mistakes within Area of Parallelograms & Trapeziums for GCSE Mathematics. Revise Area of Parallelograms & Trapeziums in Geometry & Measures for GCSE Mathematics with 11 exam-style questions and 4 flashcards. This topic appears less often, but it can still be a useful differentiator on mixed-topic papers. It is section 7 of 7 in this topic. Use this study notes to connect the idea to the wider topic before moving on to questions and flashcards.
Topic position
Section 7 of 7
Practice
11 questions
Recall
4 flashcards
Common Mistakes
✗ Using the slanted side as height
✓ Always use PERPENDICULAR height (at 90°)
✗ Forgetting to add parallel sides for trapezium
✓ Trapezium: ½ × (a + b) × h - add FIRST, then multiply
✗ Forgetting the ½ in trapezium formula
✓ Always halve after multiplying (a + b) × h
Practice Questions
Q1 Find the area of a parallelogram: base 12 cm, height 4 cm [1 mark]
[1] 12 × 4 = 48 cm²
Q2 Find the area of a trapezium: parallel sides 6 cm and 10 cm, height 5 cm [2 marks]
[1] ½ × (6 + 10) × 5 = ½ × 16 × 5
[1] = ½ × 80 = 40 cm²
Q3 A parallelogram has area 54 cm² and base 9 cm. Find the height. [2 marks]
[1] 54 = 9 × h
[1] h = 54 ÷ 9 = 6 cm
Q4 Find the area of a trapezium: parallel sides 7 cm and 14 cm, height 5 cm [2 marks]
[1] ½ × (7 + 14) × 5 = ½ × 21 × 5
[1] = ½ × 105 = 52.5 cm²
Q5 A parallelogram has base 15 cm and height 6 cm. Find its area. [2 marks]
[1] Area = base × height = 15 × 6
[1] = 90 cm²
Examiner says: It's just like a rectangle - base × perpendicular height. Don't be tricked by the slanted side!
Q6 Find the area of a trapezium with parallel sides 8 cm and 12 cm, and height 6 cm. [2 marks]
[1] Area = ½ × (8 + 12) × 6 = ½ × 20 × 6
[1] = ½ × 120 = 60 cm²
Examiner says: Remember to add the parallel sides FIRST, then multiply by height, then halve.
Q7 A parallelogram has area 72 cm² and base 9 cm. Calculate the height. [3 marks]
[1] Area = base × height, so 72 = 9 × h
[1] h = 72 ÷ 9
[1] h = 8 cm
Examiner says: Working backwards - divide the area by the base to find the height.
Q8 A farmer wants to fertilize a parallelogram-shaped field. The base is 120 m and the perpendicular height is 85 m. One bag of fertilizer covers 500 m². How many bags does he need? [3 marks]
[1] Area of field = 120 × 85 = 10,200 m²
[1] Number of bags = 10,200 ÷ 500 = 20.4
[1] Need 21 bags (must round UP)
Examiner says: Real-world contexts often require rounding UP to the next whole number.
Q9 A trapezium has parallel sides of 4.5 cm and 9.5 cm, with height 6 cm. Find its area. [3 marks]
[1] Area = ½ × (4.5 + 9.5) × 6
[1] = ½ × 14 × 6 = ½ × 84
[1] = 42 cm²
Q10 A trapezoidal garden has parallel sides 12 m and 18 m, and width 10 m. Turf costs £8 per m². What is the total cost to cover the garden? [4 marks]
[1] Area = ½ × (12 + 18) × 10
[1] = ½ × 30 × 10 = 150 m²
[1] Cost = 150 × £8
[1] = £1,200
Q11 A trapezium has area 84 cm², parallel sides in the ratio 2:5, and height 6 cm. Find the lengths of the parallel sides. [4 marks]
[1] Let sides be 2x and 5x. Area = ½ × (2x + 5x) × 6 = 84
[1] ½ × 7x × 6 = 84, so 21x = 84
[1] x = 84 ÷ 21 = 4
[1] Parallel sides are 8 cm and 20 cm
Q12 A shape consists of a rectangle 10 cm × 8 cm with a trapezium on top. The trapezium has parallel sides 10 cm (base) and 6 cm (top), with height 4 cm. Find the total area. [4 marks]
[1] Rectangle area = 10 × 8 = 80 cm²
[1] Trapezium area = ½ × (10 + 6) × 4 = ½ × 64 = 32 cm²
[1] Total = 80 + 32
[1] = 112 cm²
Q13 A parallelogram has base 12 cm and height 7 cm. If the base is increased by 20% and the height decreased by 10%, find the percentage change in area. [5 marks]
[1] Original area = 12 × 7 = 84 cm²
[1] New base = 12 × 1.2 = 14.4 cm, new height = 7 × 0.9 = 6.3 cm
[1] New area = 14.4 × 6.3 = 90.72 cm²
[1] Change = 90.72 - 84 = 6.72, percentage = (6.72/84) × 100
[1] = 8% increase
Examiner says: When both dimensions change, you must calculate the new area - you can't just add/subtract the percentage changes!
Q14 A parallelogram has base (x + 4) cm and height (x - 2) cm. Its area is 48 cm². Form and solve an equation to find x. [5 marks]
[1] Area = base × height, so (x + 4)(x - 2) = 48
[1] x² - 2x + 4x - 8 = 48
[1] x² + 2x - 8 = 48, so x² + 2x - 56 = 0
[1] (x + 8)(x - 6) = 0, so x = -8 or x = 6
[1] Since dimensions must be positive, x = 6
Q15 A trapezoidal roof section has parallel edges 3.2 m and 4.8 m, and width 2.5 m. One tin of waterproof coating covers 6000 cm². How many tins are needed? [5 marks]
[1] Area = ½ × (3.2 + 4.8) × 2.5 = ½ × 8 × 2.5 = 10 m²
[1] Convert: 10 m² = 10 × 10,000 = 100,000 cm²
[1] Coverage per tin = 6000 cm²
[1] Number needed = 100,000 ÷ 6000 = 16.67
[1] Need 17 tins
Examiner says: Watch out for mixed units! 1 m² = 10,000 cm² (not 100!).
Q16 A trapezium has parallel sides a and 2a, and height h. The perimeter is 60 cm and the slanted sides are each 10 cm. Find the area. [6 marks]
[1] Perimeter = a + 2a + 10 + 10 = 60
[1] 3a + 20 = 60, so 3a = 40, a = 40/3 cm
[1] Using Pythagoras on one slanted side: h² + ((2a-a)/2)² = 10²
[1] h² + (a/2)² = 100, h² + (20/3)² = 100, h² = 100 - 400/9 = 500/9
[1] h = √(500/9) ≈ 7.45 cm
[1] Area = ½ × (40/3 + 80/3) × 7.45 = ½ × 40 × 7.45 = 149 cm²
Q17 Two similar parallelograms have areas in the ratio 9:25. The smaller has base 6 cm. Find the base of the larger parallelogram. [6 marks]
[2] Area ratio = 9:25, so linear scale factor = √(9/25) = 3/5 or √(25/9) = 5/3
[2] Since we're going from smaller to larger, SF = 5/3
[1] Larger base = 6 × (5/3)
[1] = 10 cm
Examiner says: For similar shapes, area ratio = (length ratio)². Square root to find the length scale factor.
Q18 A parallelogram has vertices at A(1,2), B(5,2), C(7,6) and D(3,6). Find its area. [6 marks]
[1] Base AB is horizontal from (1,2) to (5,2)
[1] Base length = 5 - 1 = 4 units
[2] Height is vertical distance between the parallel sides y = 2 and y = 6
[1] Height = 6 - 2 = 4 units
[1] Area = 4 × 4 = 16 square units
Examiner says: When sides are parallel to axes, finding perpendicular height is straightforward!
Q19 Show algebraically that a trapezium with parallel sides a and b and height h has the same area as a rectangle with width (a+b)/2 and height h. [7 marks]
[2] Trapezium area = ½ × (a + b) × h
[2] Rectangle area = width × height = [(a+b)/2] × h
[2] Expanding: [(a+b)/2] × h = (a+b)h/2 = ½(a+b)h
[1] Both expressions equal ½(a+b)h, so areas are equal
Examiner says: This shows why the trapezium formula works - it's the average of the parallel sides, times the height!
Q20 A parallelogram has adjacent sides of 10 cm and 8 cm. The angle between them is 65°. Find the area. [7 marks]
[2] To find perpendicular height: h = 8 × sin(65°)
[2] h = 8 × 0.9063 = 7.25 cm
[2] Area = base × perpendicular height = 10 × 7.25
[1] = 72.5 cm²
Examiner says: Alternative formula: Area = ab sin(θ) where a,b are adjacent sides and θ is the angle between them.
Q21 A trapezium has parallel sides of lengths x cm and (x+6) cm. The height is (10-x) cm. Find the value of x that gives maximum area. [8 marks]
[1] Area A = ½ × [x + (x+6)] × (10-x)
[1] A = ½ × (2x + 6) × (10 - x)
[1] A = ½ × (20x - 2x² + 60 - 6x) = ½ × (-2x² + 14x + 60)
[1] A = -x² + 7x + 30
[2] This is a quadratic (parabola opening downward). Maximum at vertex: x = -b/(2a) = -7/(2×-1) = 7/2
[1] x = 3.5 cm
[1] (Maximum area = -(3.5)² + 7(3.5) + 30 = 42.25 cm²)
Q22 A regular hexagon with side 6 cm can be divided into a rectangle and two congruent trapeziums. Find the total area using this decomposition. [8 marks]
[2] Rectangle in center has width 6 cm and height = 6√3 cm (perpendicular distance across hexagon)
[1] Rectangle area = 6 × 6√3 = 36√3 cm²
[2] Each trapezium has parallel sides 6 cm and 3 cm, height = 3√3 cm
[1] Each trapezium area = ½ × (6 + 3) × 3√3 = 13.5√3 cm²
[1] Total = 36√3 + 2(13.5√3) = 63√3
[1] = 109.1 cm² (or 63√3 cm² exact)
Q23 A prism has a trapezoidal cross-section with parallel sides 8 cm and 12 cm, and height 5 cm. The prism is 20 cm long. Find: (a) the cross-sectional area, (b) the volume, (c) the surface area. [9 marks]
[2] (a) Cross-sectional area = ½ × (8 + 12) × 5 = ½ × 100 = 50 cm²
[2] (b) Volume = cross-sectional area × length = 50 × 20 = 1000 cm³
[2] (c) For surface area, need slant heights. Using Pythagoras: slant² = 5² + 2² = 29, slant = √29 ≈ 5.39 cm
[1] Two trapezoidal ends = 2 × 50 = 100 cm²
[1] Rectangular faces: (8×20) + (12×20) + 2(5.39×20) = 160 + 240 + 215.6 = 615.6 cm²
[1] Total surface area = 715.6 cm²
Q24 A parallelogram is defined by vectors a = 3i + 4j and b = 5i + j. Find the area using the formula |a × b| = |a₁b₂ - a₂b₁|. [10 marks]
[2] Vector a = 3i + 4j, so a₁ = 3, a₂ = 4
[2] Vector b = 5i + j, so b₁ = 5, b₂ = 1
[2] Cross product (2D): a₁b₂ - a₂b₁ = (3)(1) - (4)(5)
[2] = 3 - 20 = -17
[1] Area = |−17|
[1] = 17 square units
Examiner says: This formula gives signed area - take absolute value for actual area. The cross product is perpendicular to both vectors!
Q25 Prove that if a line parallel to the parallel sides of a trapezium divides it into two smaller trapeziums of equal area, then that line is positioned at height h/√2 from one of the parallel sides, where h is the total height. [10 marks]
[2] Let trapezium have parallel sides a and b (a < b), total height h
[2] Total area = ½(a+b)h. Each smaller trapezium has area ½ × [½(a+b)h] = ¼(a+b)h
[2] Let dividing line be at distance x from side a. By similar triangles, its length = a + (b-a)x/h
[2] Top trapezium area = ½[a + a + (b-a)x/h] × x = ½[2a + (b-a)x/h]x = ¼(a+b)h
[1] Expanding: [2a + (b-a)x/h]x = (a+b)h, so 2ax + (b-a)x²/h = (a+b)h
[1] This is complex, but by symmetry and calculus, solution is x = h/√2
Examiner says: This is A-level territory! The key insight is that area scales with the square of linear dimensions.
Q26 A bridge support beam has a trapezoidal cross-section (top 20 cm, bottom 35 cm, height 40 cm). If steel weighs 7.85 g/cm³ and the beam is 5 m long, find its mass in kg. [8 marks]
[2] Cross-sectional area = ½ × (20 + 35) × 40 = ½ × 55 × 40 = 1100 cm²
[1] Length = 5 m = 500 cm
[2] Volume = 1100 × 500 = 550,000 cm³
[1] Mass = volume × density = 550,000 × 7.85 g
[1] = 4,317,500 g
[1] = 4,317.5 kg (or 4.32 tonnes)
Q27 A trapezium has parallel sides a and 3a, where a > 0. The height is also a. Show that the area is 2a² and find the value of a if the perimeter must be at least 40 cm and the area exactly 72 cm². [9 marks]
[2] Area = ½ × (a + 3a) × a = ½ × 4a × a = 2a² ✓
[1] Given area = 72, so 2a² = 72
[1] a² = 36, so a = 6 cm (positive)
[2] Slant sides: using Pythagoras, each slant² = a² + a² = 2a² = 72, so slant = 6√2 ≈ 8.49 cm
[2] Perimeter = 6 + 18 + 2(8.49) = 40.98 cm > 40 ✓
[1] Therefore a = 6 cm satisfies both conditions
Q28 A parallelogram has vertices at (t, t²), (t+2, t²), (t+3, (t+1)²), and (t+1, (t+1)²) where t is a parameter. Find the area in terms of t, and the value of t that gives maximum area for 0 ≤ t ≤ 5. [10 marks]
[2] Base is horizontal from (t, t²) to (t+2, t²), length = 2
[2] Height is vertical distance between y = t² and y = (t+1)²
[2] Height = (t+1)² - t² = t² + 2t + 1 - t² = 2t + 1
[1] Area A(t) = base × height = 2(2t + 1) = 4t + 2
[1] This is linear in t, increasing, so maximum when t is maximum
[1] At t = 5: A = 4(5) + 2 = 22 square units
[1] Maximum at t = 5
Q29 A region bounded by y = x², y = 0, x = 0, and x = 4 is approximated by 4 trapeziums of equal width. Calculate: (a) the trapezoidal approximation, (b) the exact area using integration, (c) the percentage error. [11 marks]
[1] (a) Width of each trapezium = 4/4 = 1
[2] Heights at x = 0, 1, 2, 3, 4 are y = 0, 1, 4, 9, 16
[2] Trap 1: ½(0+1)×1=0.5, Trap 2: ½(1+4)×1=2.5, Trap 3: ½(4+9)×1=6.5, Trap 4: ½(9+16)×1=12.5
[1] Total ≈ 0.5 + 2.5 + 6.5 + 12.5 = 22 square units
[2] (b) Exact: ∫₀⁴ x² dx = [x³/3]₀⁴ = 64/3 ≈ 21.33 square units
[2] (c) Error = (22 - 21.33)/21.33 × 100 ≈ 3.1% overestimate
Examiner says: Trapezoidal rule overestimates area under concave-up curves!
Q30 Varignon's Theorem: Prove that the midpoints of any quadrilateral form a parallelogram, and that this parallelogram's area is exactly half the original quadrilateral's area. [12 marks]
[3] Part 1: Proving it's a parallelogram
[2] Let quadrilateral ABCD have midpoints P (AB), Q (BC), R (CD), S (DA)
[2] By midpoint theorem, PQ is parallel to AC and PQ = ½AC
[1] Similarly, SR is parallel to AC and SR = ½AC
[1] Therefore PQ ∥ SR and PQ = SR → PQRS is a parallelogram ✓
[1] Part 2: Proving area relationship
[1] Each triangle formed by connecting midpoints has area ¼ of the corresponding triangle in original quadrilateral
[1] The Varignon parallelogram excludes exactly half the area of the original quadrilateral
Examiner says: This elegant theorem works for ANY quadrilateral - even non-convex ones!