IB Applications and Interpretation SL · Trigonometry · Section 4 of 4

Area of a triangle

3 worked examples · By Álvaro López

Key idea

The formula “base times height divided by two” requires knowing a height, and on a plot of land or a drawing you almost never have one. If you know two sides and the angle between them, you can work out the area directly with the sine. In the IB exam it appears in problems about plots of land, gardens, courts or panels, almost always after using the cosine rule or the sine rule to find the missing data.

From “base times height” to the sine formula

You already know the area of a triangle: half of a base times its height. The problem is the height. Look at the figure: from $B$ we draw the height $h$ onto the side $b$ .

Triangle with sides a and b forming the angle C, and the height h onto side b

In the right-angled triangle on the left, the leg $h$ is opposite the angle $C$ and the hypotenuse is $a$ , so $sin C = \frac{h}{a}$ and, therefore, $h = a sin C$ . Substituting into the usual formula:

$Area = \frac{1}{2} b h = \frac{1}{2} b (a sin C)$

This formula also works if $C$ is obtuse, because $sin (18 0^{\circ} - C) = sin C$ : the sine of an obtuse angle is positive, so the area comes out correctly. If $C = 9 0^{\circ}$ , $sin C = 1$ and you recover “base times height divided by two” using the two legs.

When the included angle is missing

The formula needs a specific angle: the one between the two sides. If the question gives you other data, you must first obtain it.

If you know all three sides, work out the angle you need with the cosine rule, for example $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ , and then apply the area formula with $a$ , $b$ and that $C$ .
If you know two angles, the third one is $18 0^{\circ}$ minus their sum. With the sine rule you can find any sides you are missing.
If the angle you are given is not between the two sides, use the sine rule or the cosine rule to find the included angle, or change to a different pair of sides.

Plots of land and compound figures

A plot with four or more sides has no direct formula, but it can be divided into triangles. You draw a diagonal (or several from the same vertex), calculate the area of each triangle and add them together.

For the split to work, each triangle must have enough data: two sides and the included angle, or all three sides. Often the first triangle is solved with the data in the question, and the diagonal that appears (calculated with the cosine rule) becomes a side of the second triangle.

Worked examples

Example 1 Basic

A triangular solar panel

A solar panel is triangular. Two of its sides measure $7.4$ m and $5.2$ m, and the angle they form is $3 8^{\circ}$ .

a) Calculate the area of the panel.

b) Calculate the height of the triangle relative to the side of $7.4$ m.

Solution

a) The angle of $3 8^{\circ}$ lies between the two given sides, so the formula applies directly: $A = \frac{1}{2} \cdot 7.4 \cdot 5.2 \cdot sin 3 8^{\circ} = 11.8 m^{2}$

b) Taking the side of $7.4$ m as the base, $A = \frac{1}{2} \cdot 7.4 \cdot h$ . Solving for $h$ , $h = \frac{2 A}{7.4} = \frac{2 \cdot 11.845}{7.4} = 3.20 m$ This agrees with $h = 5.2 sin 3 8^{\circ}$ , which is the leg opposite the $3 8^{\circ}$ angle in the right-angled triangle.

Example 2 Medium

A plot with three known sides

A triangular plot of land has sides of $9.5$ m, $12$ m and $14$ m.

a) Find the angle formed by the sides of $9.5$ m and $12$ m.

b) Calculate the area of the plot.

Solution

a) The angle $C$ formed by the sides of $9.5$ m and $12$ m is opposite the side of $14$ m. By the cosine rule, $1 4^{2} = 9. 5^{2} + 1 2^{2} - 2 \cdot 9.5 \cdot 12 cos C \Rightarrow cos C = \frac{9. 5 ^{2} + 1 2 ^{2} - 1 4 ^{2}}{2 \cdot 9.5 \cdot 12} = \frac{38.25}{228}$ So $C = 80. 3^{\circ}$ .

b) We now have two sides and the included angle: $A = \frac{1}{2} \cdot 9.5 \cdot 12 \cdot sin 80. 3^{\circ} = 56.2 m^{2}$ To avoid carrying rounding errors, on the GDC use the full value of $C$ (the result agrees to 3 significant figures).

Example 3 Hard

A four-sided plot

A plot of land is in the shape of a quadrilateral $A B C D$ . We know that $A B = 80$ m, $B C = 65$ m, $C D = 50$ m, $D A = 70$ m and that the angle $A B C = 7 2^{\circ}$ . The diagonal $A C$ is drawn.

a) Calculate the length of the diagonal $A C$ .

b) Calculate the area of the triangle $A B C$ .

c) Calculate the angle $A D C$ .

d) Calculate the area of the triangle $A C D$ and the total area of the plot.

Solution

a) In triangle $A B C$ we know two sides and the included angle, so we use the cosine rule: $A C^{2} = 8 0^{2} + 6 5^{2} - 2 \cdot 80 \cdot 65 cos 7 2^{\circ} \Rightarrow A C = 86.1 m$

b) With the same data, $A_{A B C} = \frac{1}{2} \cdot 80 \cdot 65 \cdot sin 7 2^{\circ} = 2470 m^{2}$ .

c) In triangle $A C D$ we know all three sides: $A C = 86.1$ , $C D = 50$ and $D A = 70$ . The angle $D$ is opposite $A C$ : $cos D = \frac{5 0 ^{2} + 7 0 ^{2} - 86. 1 ^{2}}{2 \cdot 50 \cdot 70} \Rightarrow D = 90. 1^{\circ}$

d) With $C D$ , $D A$ and the included angle $D$ : $A_{A C D} = \frac{1}{2} \cdot 50 \cdot 70 \cdot sin 90. 1^{\circ} = 1750 m^{2}$ The total area is the sum of the two triangles: $2470 + 1750 = 4220 m^{2}$ .

In the exam: in a problem like this, each step earns marks: the diagonal with the cosine rule, the correct included angle in each triangle, and the final sum. Keep the unrounded values in the GDC (the diagonal and the angle $D$ ) to avoid carrying error, and write the unit, $m^{2}$ .

Common mistakes

Summary

If you know two sides and the included angle, $A = \frac{1}{2} a b sin C$ ; it is the “base times height divided by two” formula with $h = a sin C$ .
The angle must be between the two sides used. With any pair of sides it holds: $\frac{1}{2} b c sin A = \frac{1}{2} a c sin B = \frac{1}{2} ab sin C$ .
If you do not have that angle, find it: with the cosine rule if you have all three sides, or with the fact that the angles sum to $18 0^{\circ}$ and the sine rule.
A figure with more than three sides is divided into triangles with a diagonal, and the areas are added.
For three sides: first the angle with the cosine rule, then the area formula.
The GDC in degree mode, without rounding intermediate values; give the result to 3 significant figures and its unit of area.

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