Choosing a triangle-area formula starts with the evidence. A sloping side is not automatically a height, and a supplied angle is not automatically the angle between the two known sides. The best first method uses measurements whose relationships are actually stated.
The original cases below deliberately supply different information. One has an external height, one has two sides with their included angle, and one gives three sides. Before calculating, write what each measurement belongs to and express all lengths in compatible units.
Three methods, three evidence checks
- A = ½bh: b is the chosen base and h is the perpendicular distance from the opposite vertex to the line containing that base.
- A = ½ab sin C: a and b are the two supplied sides and C is the angle between them. Use degree mode when angles are stated in degrees.
- A = √[s(s − a)(s − b)(s − c)]: a, b and c are all three side lengths and s = (a + b + c)/2 is the semiperimeter.
For three positive sides, the largest must be strictly shorter than the sum of the other two. Equality produces a flattened, zero-area configuration rather than an ordinary triangle. A negative Heron factor is a reason to check the givens, not an instruction to remove a minus sign.
Text decision route: use base and perpendicular height if that pair is known; use two sides and their included angle if that package is known; use Heron when all three valid sides are known. If no package is complete, derive missing information using justified geometry or state what is missing.
Worked case 1: a height outside an obtuse triangle
Triangle ABC has base AB = 14 cm. The perpendicular from C meets the extended base line at H, outside the segment AB, with CH = 9 cm. Find its area. The required base-height pair is still AB and CH, so A = ½ × 14 × 9 = 63 cm².
To specify the diagram completely, use centimetre coordinates A(0, 0), B(14, 0), C(−4, 9) and H(−4, 0). AB and AH are horizontal, while CH is vertical, so CH is perpendicular to the base line. The drawing’s sloping side AC is not the height.
A second-method check subtracts two right-triangle areas. Triangle HBC has base HB = 18 cm and area ½ × 18 × 9 = 81 cm². Triangle HAC has base HA = 4 cm and area 18 cm². Their difference is 81 − 18 = 63 cm², confirming why the external height works.
Diagram data in words: A shows AB = 14 cm and outside height CH = 9 cm. B shows DE = 18 cm, DF = 13 cm and the included angle EDF = 40°. C shows JL = 9 cm, KL = 10 cm and JK = 17 cm. The displayed drawings are aids; calculate from these stated relationships.
Worked case 2: the included angle replaces a supplied height
In triangle DEF, DE = 18 cm, DF = 13 cm and angle EDF = 40°. The angle is at D, exactly where the two known sides meet. Therefore A = ½ × 18 × 13 × sin 40° = 117 sin 40° ≈ 75.21 cm², rounded to two decimal places.
For an independent route, choose DE as the base. The height from F is 13 sin 40° ≈ 8.3562389 cm. Half the base times that unrounded height gives 9 × 13 sin 40° = 117 sin 40°, the same area. The horizontal projection of DF is 13 cos 40°, approximately 9.96 cm, which lies within the 18 cm base.
The factor sin 40° is below one, so the area is below ½ × 18 × 13 = 117 cm². That upper bound corresponds to those sides meeting at a right angle. Using 40 as a multiplier, or interpreting it in radians, would not describe this triangle.
Worked case 3: use three sides without guessing a height
Triangle JKL has side lengths 9 cm, 10 cm and 17 cm. The largest side passes the feasibility check because 17 < 9 + 10. The semiperimeter is s = (9 + 10 + 17)/2 = 18 cm, not 36 cm.
Heron’s formula gives A = √[18(18 − 9)(18 − 10)(18 − 17)] = √(18 × 9 × 8 × 1) = √1296 = 36 cm². Because the product is exact, no decimal approximation is needed.
Check by recovering a compatible height to base JK = 17 cm. The area implies h = 2A/b = 72/17 cm. A consistent coordinate placement is J(0, 0), K(17, 0), L(135/17, 72/17), in centimetres. The squared distances JL² and KL² are respectively 81 and 100. This independently confirms both supplied shorter sides and the area ½ × 17 × (72/17) = 36 cm².
Three solved practice tasks
Practice 1: reject an impossible side triple
Can lengths 5 cm, 7 cm and 13 cm form a triangle for Heron’s formula? No: 5 + 7 = 12 < 13. Their semiperimeter would be 12.5 cm and the factor s − 13 would be negative. No such triangle exists; there is no real triangle area to report.
Practice 2: identify the missing information
A triangle has sides AB = 11 cm and AC = 15 cm. Nothing else is supplied. A proposed solution uses ½ × 11 × 15. Explain the problem. This silently assumes the two sides are perpendicular. The area is not uniquely determined. It is 82.5 sin A cm², so a known included angle, a matching height or the third side would provide the missing information. An angle elsewhere cannot simply be substituted for A.
Practice 3: compare two justified methods
A right triangle has perpendicular legs 8 cm and 15 cm. Find its area, then check using all three sides. The direct method gives ½ × 8 × 15 = 60 cm². Pythagoras gives hypotenuse 17 cm, so s = 20 cm. Heron gives √(20 × 12 × 5 × 3) = √3600 = 60 cm². Both methods use the stated right angle; neither assumes it from appearance.
Keep the unit attached to the result
Area has squared units because it combines two length dimensions. If a base is 0.8 m and its height is 35 cm, first write 0.8 m as 80 cm or write 35 cm as 0.35 m. The area is 1400 cm² or 0.14 m². These agree because one square metre contains 10,000 square centimetres. Round only the final result, particularly when an intermediate height comes from a sine.
Choose area methods for quadrilaterals
The same measurement-first approach works for four-sided figures. Use a rectangle’s two perpendicular side lengths, a parallelogram’s base and perpendicular height, a trapezium’s parallel sides and their perpendicular separation, or a justified split into triangles. Four side lengths alone do not normally fix the area of a general quadrilateral.
Worked trapezium: use the parallel pair
A trapezium has parallel sides 12 cm and 20 cm, separated perpendicularly by 7 cm. Area = ½(12 + 20) × 7 = 112 cm². A coordinate model is A(0,0), B(20,0), C(16,7), D(4,7), in that boundary order. The middle rectangle contributes 12 × 7 = 84 cm² and the two side triangles together contribute ½(4 + 4) × 7 = 28 cm². Their sum confirms 112 cm².
Worked parallelogram: a sloping side is not its height
A parallelogram has base 0.8 m and perpendicular height 35 cm. Convert 35 cm to 0.35 m: area = 0.8 × 0.35 = 0.28 m². Equivalently 80 × 35 = 2,800 cm², and 2,800/10,000 = 0.28 m². Multiplying 0.8 by 35 while labelling the result m² mixes units. The triangle formula’s factor ½ does not apply to the whole parallelogram.
Worked irregular quadrilateral: split along a diagonal
A convex quadrilateral ABCD has diagonal AC = 10 cm. B and D are on opposite sides of line AC; their perpendicular distances to it are respectively 4 cm and 7 cm. Triangle ABC has area ½ × 10 × 4 = 20 cm², and triangle ACD has area ½ × 10 × 7 = 35 cm². They share only the diagonal, so the quadrilateral area is 55 cm².
For a fully specified model take A(0,0), B(4,4), C(10,0), D(6,−7), in centimetres. The heights meet AC at (4,0) and (6,0). The shape is not assumed to be a kite or trapezium; the supplied diagonal and two perpendicular heights are sufficient. The compact expression ½ × AC × (4 + 7) uses that justified split.
Independent method selection
- A trapezium has parallel sides 9 cm and 15 cm and perpendicular separation 8 cm. A sloping side is 10 cm. Choose the required measurements and find its area.
- A convex quadrilateral has diagonal 1.2 m. The other vertices are on opposite sides of its line at perpendicular distances 25 cm and 40 cm. Find its area in m² and cm².
Explained answers
- Use the parallel sides and their separation: ½(9 + 15) × 8 = 96 cm². The 10 cm sloping side is not the perpendicular height and is unnecessary.
- Split along the diagonal. Area = ½ × 1.2 × (0.25 + 0.40) = 0.39 m². In centimetres, ½ × 120 × (25 + 40) = 3,900 cm². Both routes agree after multiplying square metres by 10,000.
To see why side lengths alone can be insufficient, compare a square with side 5 cm (area 25 cm²) with a rhombus of side 5 cm and included angle 30° (area 25 sin 30° = 12.5 cm²). Both have four sides of length 5 cm. A right angle, a height, a diagonal with supporting information or another sufficient constraint must be supplied.
Frequently asked questions
Can any triangle side be called the base?
Yes, but each chosen base needs its own perpendicular height from the opposite vertex. Changing the base without changing the height generally changes the calculation incorrectly.
Must the height lie inside the triangle?
No. In an obtuse triangle, a perpendicular may meet an extension of the base. Its length still measures the distance to the base line.
Which angle belongs in the sine-area formula?
The angle between the two sides being multiplied. Identify their shared vertex before entering the angle.
Can two sides alone determine the area?
No. Their included angle can vary, changing the area. More information is needed unless a right angle or another constraint is explicitly given.
Why does Heron use half the perimeter?
The formula is defined with s = (a + b + c)/2. Substituting the full perimeter changes every factor and gives a different, incorrect calculation.
What if the two shorter sides add to the longest?
The shape is degenerate: its vertices are collinear and its area is zero. It is not a non-degenerate triangle.
Which method should I use when two methods work?
Choose the direct method using the clearest supplied measurements. A second valid method can provide a useful check without replacing the original justification.
Subject references
The original questions, diagrams and solutions above use standard mathematical definitions and methods. These source-authored references support the underlying facts.
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