11.1 Areas of triangles and quadrilaterals

Areas of Triangles and Quadrilaterals

Triangles and quadrilaterals are the building blocks for nearly every polygon area problem you'll encounter. Each shape has its own area formula, but they all share a common thread: every formula connects some measure of width to some measure of height. Mastering these formulas (and knowing when to use each one) sets you up for the more complex area problems later in this unit.

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Areas of Triangles

Triangle area with base and height

The area of any triangle is given by:

$A = \frac{1}{2}bh$

• $b$ is the base (you can choose any side as the base)
• $h$ is the height, the perpendicular distance from the base to the opposite vertex

The height must be perpendicular to whichever side you pick as the base. In acute triangles, the height falls inside the triangle. In obtuse triangles, the height can fall outside the triangle, which sometimes trips people up on diagrams.

If the height isn't given directly, you'll often need to find it using the Pythagorean theorem or trigonometry.

Example 1: An equilateral triangle with side length 6. Dropping an altitude splits the base in half, creating a right triangle with hypotenuse 6 and one leg 3. By the Pythagorean theorem:

$h = \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3} \approx 5.2 \text{ units}$

So the area is $A = \frac{1}{2}(6)(3\sqrt{3}) = 9\sqrt{3} \approx 15.6 \text{ square units}$.

Example 2: A right triangle with legs 3 and 4 (the classic 3-4-5 Pythagorean triple). The two legs are the base and height since they're already perpendicular:

$A = \frac{1}{2}(4)(3) = 6 \text{ square units}$

Areas of Quadrilaterals

Parallelogram area calculation

A parallelogram's area works just like a rectangle's, because you can "slide" the slanted part to form a rectangle with the same base and height:

$A = bh$

• $b$ is the base (either pair of parallel sides)
• $h$ is the height, the perpendicular distance between the two parallel sides the base belongs to

Notice this formula does not use the slant side length. The slant side only matters if you need it to calculate the height.

Example 1: Base 8 units, height 5 units → $A = (8)(5) = 40 \text{ square units}$.

Example 2: Side lengths 6 and 10, with a 60° angle between them. The height drops from the top side perpendicular to the base of length 10. Using the side of length 6 and the given angle:

$h = 6 \sin 60° = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \approx 5.2 \text{ units}$

$A = 10 \cdot 3\sqrt{3} = 30\sqrt{3} \approx 52.0 \text{ square units}$

Trapezoid area using parallel sides

A trapezoid has exactly one pair of parallel sides. Its area formula averages those two parallel sides and multiplies by the height:

$A = \frac{1}{2}(a + b)h$

• $a$ and $b$ are the lengths of the two parallel sides (called bases)
• $h$ is the perpendicular distance between those parallel sides

Think of $\frac{1}{2}(a + b)$ as the "average base." You're finding the area of a rectangle whose width equals the average of the two parallel sides.

Example 1: Parallel sides 6 and 10, height 4 → $A = \frac{1}{2}(6 + 10)(4) = \frac{1}{2}(16)(4) = 32 \text{ square units}$.

Example 2: An isosceles trapezoid with parallel sides 8 and 12, and non-parallel sides of length 5. To find the height, notice the difference in the parallel sides is $12 - 8 = 4$. That extra length splits equally on both sides (because it's isosceles), so each right triangle at the base has a horizontal leg of $\frac{4}{2} = 2$ and a hypotenuse of 5:

$h = \sqrt{5^2 - 2^2} = \sqrt{21} \approx 4.6 \text{ units}$

$A = \frac{1}{2}(8 + 12)(\sqrt{21}) = 10\sqrt{21} \approx 45.8 \text{ square units}$

Rhombus and kite area from diagonals

Both rhombuses and kites have perpendicular diagonals, which gives them the same clean area formula:

$A = \frac{1}{2}d_1 d_2$

• $d_1$ and $d_2$ are the lengths of the two diagonals

This formula works because the perpendicular diagonals divide the shape into four right triangles whose combined area simplifies to $\frac{1}{2}d_1 d_2$.

Key diagonal properties to remember:

• Rhombus: Diagonals bisect each other at right angles. Each diagonal cuts the other into two equal halves.
• Kite: Diagonals are perpendicular, but only the main diagonal (the one connecting the two vertices where unequal sides meet) bisects the other diagonal. The shorter diagonal does not bisect the longer one.

These properties come up constantly when you're given half-diagonal lengths and need to reconstruct the full diagonals before applying the formula.

Example 1: A rhombus with diagonals 8 and 6 → $A = \frac{1}{2}(8)(6) = 24 \text{ square units}$.

Example 2: A kite with diagonals 12 and 16 → $A = \frac{1}{2}(12)(16) = 96 \text{ square units}$.