math-2017:G.SRT.9
math-2017:G.SRT.9 Derive the formula A = 1/2 ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side.
Learning goal: Derive the formula A = ½ab·sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side, and apply the formula to acute, right and obtuse triangles.
Scope note. G.SRT.9 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full as part of the Geometry course, as extension content for students on that path.
Notation. In triangle ABC, side a is opposite vertex A, side b is opposite vertex B and side c is opposite vertex C. So a = BC, b = CA, c = AB, and the angle C sits between the sides a and b. The formula A = ½ab·sin(C) says that the area is half the product of two sides times the sine of the angle between them. Any vertex can play the role of C; the three versions ½ab·sin C, ½bc·sin A and ½ca·sin B all give the same area.
The derivation for an acute angle C. Start from the ordinary area formula, area = ½ × base × height. Take side b = CA as the base. The height is the distance from the opposite vertex B to the line CA, and it is not one of the sides, so draw it in: the auxiliary line from B perpendicular to CA, meeting CA at a point D. Now triangle BDC is a right triangle with the right angle at D, hypotenuse CB = a, and acute angle C. The height BD is the side opposite angle C, so by the definition of sine, sin C = BD/a, which gives BD = a·sin C. Substitute into the area formula: area = ½ × b × BD = ½ × b × a·sin C = ½ab·sin C. The auxiliary perpendicular converted a height, which was not given, into a side and an angle, which were.
The model with numbers. A triangle has a = 8 cm, b = 5 cm and included angle C = 60°. Draw the perpendicular from B to CA. In the right triangle BDC, BD = 8 sin 60° = 8 × (√3/2) = 4√3 ≈ 6.93 cm, and CD = 8 cos 60° = 4 cm, so D lies on the segment CA, 4 cm from C and 1 cm from A. Area = ½ × 5 × 4√3 = 10√3 ≈ 17.32 cm². Check by another route: the third side, by the Law of Cosines from the next lesson or by the Pythagorean Theorem in triangle BDA, is √(BD² + DA²) = √(48 + 1) = 7 cm; Heron's formula with sides 5, 7, 8 and half-perimeter 10 gives √(10 × 5 × 3 × 2) = √300 = 10√3 cm². The two methods agree exactly.
The right-angle case. If C = 90°, then sin C = 1 and the formula reads A = ½ab, the familiar half the product of the legs. The perpendicular from B to CA is the side BC itself, so BD = a directly, consistent with a·sin 90° = a.
The obtuse case. If C is obtuse, the foot D of the perpendicular from B falls on the extension of CA beyond C, outside the segment. Triangle BDC is still a right triangle with hypotenuse a, but the acute angle at C inside it is the supplement 180° − C. So BD = a·sin(180° − C). The sine of an obtuse angle is defined to equal the sine of its supplement, sin(180° − C) = sin C, which is exactly what makes the formula A = ½ab·sin C hold for every angle between 0° and 180°. A model: a = 10, b = 6, C = 120°. Then BD = 10 sin 60° = 5√3 ≈ 8.66 and area = ½ × 6 × 5√3 = 15√3 ≈ 25.98 square units. Check: the foot D is 10 cos 60° = 5 units beyond C, so DA = 11, and the third side is √(75 + 121) = √196 = 14; Heron with sides 6, 10, 14 and half-perimeter 15 gives √(15 × 9 × 5 × 1) = √675 = 15√3. Again exact agreement.
Why the derivation is a proof. Nothing in it depended on the particular lengths 8 and 5 or the angle 60°. The auxiliary perpendicular exists for every triangle, the right triangle it creates always has hypotenuse a and an acute angle equal to C or to 180° − C, and the definition of sine always gives the height. So the formula holds for every triangle. The one convention it needs is the extension of sine to obtuse angles by sin(180° − C) = sin C, which is a definition designed to make the height formula work without separate cases.
Using the formula. It computes an area from two sides and the included angle, which is a common way to measure a triangular plot of land: two boundary lengths and the angle between them from a corner. It also runs backwards: an area and two sides give sin C, and hence C. A triangle with a = 8, b = 10 and area 20 has sin C = 20/(½ × 8 × 10) = 0.5, so C = 30° or C = 150°. Both are possible, since both angles have sine 0.5, and they give different triangles with the same area; more information is needed to decide between them. The formula also gives the area of a parallelogram with sides a, b and angle θ as ab·sin θ, since a diagonal cuts it into two congruent triangles, and the area of a regular n-gon with circumradius r as n × ½r²·sin(360°/n).
The misconception. The angle in the formula must be the included angle, the one between the two sides used. With sides 8 and 5 and an angle of 60° at vertex A instead of C, the formula ½ × 8 × 5 × sin 60° is wrong, because A is not between BC = 8 and CA = 5; the side b = 5 does touch A but a = 8 does not. Label the triangle, identify which two sides meet at the known angle, and use those.
Worked, acute: a = 7 cm, b = 9 cm, C = 40°. Area = ½ × 7 × 9 × sin 40° ≈ 31.5 × 0.6428 ≈ 20.25 cm². The height from B to line CA is 7 sin 40° ≈ 4.50 cm, and ½ × 9 × 4.50 ≈ 20.25 cm² confirms the base-and-height reading of the same computation.
Worked, equilateral: An equilateral triangle with side 6 has every angle 60°. Area = ½ × 6 × 6 × sin 60° = 18 × (√3/2) = 9√3 ≈ 15.59 square units. Check with the altitude: the altitude of an equilateral triangle of side 6 is 3√3, and ½ × 6 × 3√3 = 9√3.
Worked, obtuse: a = 10, b = 6, C = 120°. sin 120° = sin 60° = √3/2, so area = ½ × 10 × 6 × √3/2 = 15√3 ≈ 25.98. The perpendicular from B lands 5 units beyond C on the extension of AC, so the height is 10 sin 60° = 5√3, and ½ × 6 × 5√3 = 15√3, the same value; the obtuse angle did not change the height formula.
Worked, backwards: A triangle has sides 8 and 10 meeting at an unknown angle, and area 20. Then 20 = ½ × 8 × 10 × sin C, so sin C = 0.5 and C = 30° or 150°. With C = 30° the third side is √(64 + 100 − 160 cos 30°) ≈ √(164 − 138.6) ≈ 5.04; with C = 150° it is √(164 + 138.6) ≈ 17.39. Two different triangles, same area. The area alone does not decide.
Recording task: Draw triangle ABC with CA = 5 cm, CB = 8 cm and angle C = 60°. Draw the perpendicular from B to line CA with a set square and mark its foot D. Measure BD and CD and record them beside the computed values 4√3 ≈ 6.93 cm and 4 cm. Record the area two ways: ½ × 5 × 6.93 ≈ 17.3 cm² and ½ × 8 × 5 × sin 60° ≈ 17.3 cm². Then draw a triangle with CA = 6 cm, CB = 10 cm and angle C = 120°, draw the perpendicular from B, and record that its foot lies outside segment CA, 5 cm beyond C, with height ≈ 8.66 cm and area ≈ 26.0 cm².
Independent performance: Write the derivation of A = ½ab·sin C in your own words, with a labeled diagram showing the auxiliary perpendicular, for an acute angle C, and then a second paragraph explaining what changes when C is obtuse and why the same formula still holds. Apply the formula to three triangles: sides 7 and 9 with included angle 40°; an equilateral triangle of side 6; and sides 10 and 6 with included angle 120°; check each by a second method (base and height, or Heron’s formula). Finally solve one backwards problem, area 20 with sides 8 and 10, and explain why there are two answers. Success criteria: the derivation names the right triangle created, its hypotenuse and the sine definition used; the obtuse paragraph states sin(180° − C) = sin C; every area is confirmed by a second method to within rounding; and the backwards problem gives both 30° and 150° with a reason.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Derive the formula A = ½ab·sin C for a triangle with acute angle C, naming the auxiliary line you draw.
Take side b = CA as the base and draw the auxiliary line from B perpendicular to CA, meeting it at D. Triangle BDC is a right triangle with hypotenuse CB = a and acute angle C, and BD is the side opposite C, so sin C = BD/a and BD = a·sin C. Then area = ½ × base × height = ½ × b × a·sin C = ½ab·sin C.
A triangle has sides 8 cm and 5 cm with an included angle of 60°. Find its area exactly and to two decimal places, and confirm by a second method.
Area = ½ × 8 × 5 × sin 60° = 20 × √3/2 = 10√3 ≈ 17.32 cm². Second method: call the included vertex C, with CB = 8 and CA = 5, and draw BD perpendicular to CA. Then BD = 8 sin 60° = 4√3 and CD = 8 cos 60° = 4 cm, so DA = 1 cm and the third side is √(48 + 1) = 7 cm; Heron’s formula with half-perimeter 10 gives √(10 × 5 × 3 × 2) = √300 = 10√3 cm².
Explain what happens to the derivation when angle C is obtuse, using a = 10, b = 6, C = 120°, and show the formula still gives the area.
The foot of the perpendicular from B falls on the extension of CA beyond C, and the acute angle of the right triangle at C is 180° − 120° = 60°, so the height is 10 sin 60° = 5√3 ≈ 8.66. Since sin 120° is defined as sin 60°, the formula ½ × 10 × 6 × sin 120° = 15√3 ≈ 25.98 gives the same value as ½ × 6 × 5√3. Check: the third side is √(75 + 11²) = 14 and Heron with 6, 10, 14 gives √(15 × 9 × 5 × 1) = 15√3.
Find the area of an equilateral triangle with side 6 using A = ½ab·sin C, and check with the altitude.
Any two sides are 6 and 6 with included angle 60°: area = ½ × 6 × 6 × sin 60° = 18 × √3/2 = 9√3 ≈ 15.59. The altitude of an equilateral triangle of side 6 is 6 sin 60° = 3√3, and ½ × 6 × 3√3 = 9√3, the same. The formula is the altitude computation packaged as a product.
A triangle has two sides 8 and 10 and area 20. Find the included angle, and explain why there are two possibilities.
20 = ½ × 8 × 10 × sin C gives sin C = 0.5, so C = 30° or C = 150°, because an angle and its supplement have the same sine. Both produce a triangle of area 20: with C = 30° the third side is about 5.04, with C = 150° about 17.39. The area does not distinguish them; another measurement, such as the third side, would.
A student computes the area of a triangle with BC = 8, CA = 5 and angle A = 60° as ½ × 8 × 5 × sin 60°. What is wrong?
The formula needs the angle included between the two sides used. Angle A is between sides AB and AC, not between BC and CA; BC does not touch A. So either use the two sides that meet at A, which requires knowing AB, or use the angle at C, which is between BC and CA. Applying the formula with a non-included angle gives a number that is not the area.
Use the triangle area formula to find the area of a parallelogram with sides 7 and 4 and an angle of 35° between them.
A diagonal divides the parallelogram into two congruent triangles, each with sides 7 and 4 and included angle 35° (or its supplement 145°, which has the same sine). Each triangle has area ½ × 7 × 4 × sin 35° ≈ 14 × 0.5736 ≈ 8.03, so the parallelogram has area ≈ 16.06 square units, which is 7 × 4 × sin 35°.