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Area of polygons (triangles, quadrilaterals)

20 practice questions 0 video lessons Theory + worked examples

Area of Polygons

Texas Geometry (TEKS) • Standard G.11(B) • Two-Dimensional Measurement

Area of Polygons is the opening topic of Two-Dimensional Measurement in the Texas Essential Knowledge and Skills (Geometry, §111.41). It is aligned to Standard G.11(B), which requires students to apply the formulas for the area of two-dimensional figures to solve problems.

Polygon area formulas include the triangle \(\dfrac12 bh\), the parallelogram \(bh\), and the trapezoid \(\dfrac12(b_1+b_2)h\).

Texas Geometry (TEKS) › Two-Dimensional Measurement › Area of Polygons  —  Standard G.11(B)

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Theory

Each polygon has an area formula based on a base and a perpendicular height:

  • Triangle: \(A=\dfrac12 bh\).
  • Rectangle / parallelogram: \(A=bh\).
  • Trapezoid: \(A=\dfrac12(b_1+b_2)h\), where \(b_1,b_2\) are the parallel sides.
Height is always perpendicular to the base — not a slanted side.
Area of a triangle The area of a triangle is one half base times height. base b height h triangle: A = ½ b h
Triangle: \(A=\dfrac12 bh\), with height perpendicular to the base.
Area of a trapezoid A trapezoid's area is one half the sum of the parallel sides times the height. b₁ b₂ h trapezoid: A = ½(b₁+b₂)h
Trapezoid: \(A=\dfrac12(b_1+b_2)h\).

The area formulas:

\[\text{triangle } \dfrac12 bh,\quad \text{parallelogram } bh,\quad \text{trapezoid } \dfrac12(b_1+b_2)h\]
triangle area is one half base times height; parallelogram is base times height; trapezoid is one half the sum of parallel sides times height
The trapezoid uses the average of the two parallel sides times the height.

How to find a polygon's area

  1. Identify the shape and its base(s) and height.
  2. Use the perpendicular height, not a slant side.
  3. Substitute into the formula.
  4. Rearrange to find a missing dimension if the area is given.
Example 1 — Triangle area
Find the area of a triangle with base \(10\) and height \(6\).
Solution

Use \(A=\dfrac12 bh\).

\(A\)\(=\)\(\dfrac12(10)(6)=30\)
the area is 30 square units
Example 2 — Parallelogram area
Find the area of a parallelogram with base \(8\) and height \(5\).
Solution

A parallelogram's area is base times height.

\(A\)\(=\)\(8\times 5=40\)
the area is 40 square units
Example 3 — Trapezoid area
Find the area of a trapezoid with parallel sides \(6\) and \(10\) and height \(4\).
Solution

Use \(A=\dfrac12(b_1+b_2)h\).

\(A\)\(=\)\(\dfrac12(6+10)(4)\)
\(=\)\(\dfrac12(16)(4)=32\)
the area is 32 square units
Example 4 — Find a missing dimension
A triangle has area \(24\) and base \(8\). Find its height.
Solution

Solve \(\dfrac12 bh=24\).

\(\dfrac12(8)h\)\(=\)\(24\)
\(4h\)\(=\)\(24\)
\(h\)\(=\)\(6\)
the height is 6

Common pitfalls

Use the perpendicular height, not the slanted side length.
The trapezoid formula averages the two parallel sides; don't forget the \(\dfrac12\).
Area is in square units.

Frequently asked questions

What is the area of a triangle?

\(A=\dfrac12 bh\): one half the base times the perpendicular height.

What is the area of a trapezoid?

\(A=\dfrac12(b_1+b_2)h\): the average of the two parallel sides times the height.

What is the area of a parallelogram?

Base times perpendicular height, \(A=bh\).

What height do you use in area formulas?

Always the perpendicular height to the chosen base, not a slanted side.