Area of 2D Shapes

Calculate areas of rectangles, triangles, parallelograms, trapezoids, and composite figures for the Digital SAT.

Area calculations are fundamental to the Geometry and Trigonometry section of the Digital SAT. You need to know the standard area formulas and be able to apply them to composite figures — shapes made by combining or subtracting simpler shapes.

Core Concepts

Key Area Formulas

Shape Formula
Rectangle A=lwA = lw
Square A=s2A = s^2
Triangle A=12bhA = \frac{1}{2}bh
Parallelogram A=bhA = bh
Trapezoid A=12(b1+b2)hA = \frac{1}{2}(b_1 + b_2)h
Circle A=πr2A = \pi r^2

All heights (hh) must be perpendicular to the base.

Composite Figures

Break complex shapes into simpler parts:

  • Add areas of parts that make up the shape.
  • Subtract areas of parts that are removed.

Example: An L-shaped room = large rectangle − small rectangle cut from corner.

Shaded Region Problems

A common SAT pattern: find the area of a shaded region by subtracting the unshaded part from the whole.

Ashaded=AouterAinnerA_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}}

Example: A circle inscribed in a square. Shaded area = square area − circle area.

If the square has side ss and the circle has radius r=s/2r = s/2:

Ashaded=s2π(s/2)2=s2πs24A_{\text{shaded}} = s^2 - \pi(s/2)^2 = s^2 - \frac{\pi s^2}{4}

Strategy Tips

Tip 1: The SAT Provides Formulas

Area formulas for standard shapes are on the SAT reference sheet. But knowing them saves time.

Tip 2: Height Must Be Perpendicular

For triangles and parallelograms, don't use the slant side — use the perpendicular height.

Tip 3: Draw Helper Lines

For composite shapes, draw dashed lines to split the shape into rectangles and triangles.

Tip 4: Label Everything

Write dimensions on the figure to keep track.

Worked Example: Example 1

Problem

Find the area of a triangle with base 12 and height 8.

A=12(12)(8)=48A = \frac{1}{2}(12)(8) = 48

Solution

Worked Example: Example 2

Problem

A trapezoid has parallel sides 10 and 14, with height 6. Find the area.

A=12(10+14)(6)=12(24)(6)=72A = \frac{1}{2}(10 + 14)(6) = \frac{1}{2}(24)(6) = 72

Solution

Worked Example: SAT-Style Shaded Region

Problem

A square of side 10 has a circle of radius 5 inscribed in it. Find the area of the shaded region outside the circle.

A=102π(5)2=10025π10078.5=21.5A = 10^2 - \pi(5)^2 = 100 - 25\pi \approx 100 - 78.5 = 21.5

Solution

Worked Example: Example 4

Problem

An L-shaped figure: outer rectangle 12 × 8, with a 4 × 3 rectangle removed from one corner. Find the area.

A=9612=84A = 96 - 12 = 84

Solution

Worked Example: Example 5

Problem

A parallelogram has base 15 and slant side 10. The perpendicular height is 8. What is the area?

A=15×8=120A = 15 \times 8 = 120 (NOT 15×1015 \times 10).

Solution

Practice Problems

  1. Problem 1

    Area of a triangle: base = 7, height = 10.

    Problem 2

    Area of a circle with diameter 12.

    Problem 3

    A rectangle has area 72 and width 6. What is the length?

    Problem 4

    Shaded region: rectangle 20 × 15 with two circles of radius 3 cut out.

    Problem 5

    A hexagonal room is split into 6 equilateral triangles of side 4. Find total area.

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Common Mistakes

  • Using slant height instead of perpendicular height. This gives the wrong answer for triangles and parallelograms.
  • Forgetting to halve for triangles. Triangle area = 12bh\frac{1}{2}bh, not bhbh.
  • Using diameter instead of radius. Area of circle: πr2\pi r^2, not πd2\pi d^2.
  • Not reading the question. Does it ask for the shaded or unshaded area?

Key Takeaways

  • Know the standard area formulas — rectangle, triangle, circle, trapezoid, parallelogram.

  • Height must be perpendicular to the base.

  • Composite shapes: add or subtract simpler areas.

  • Shaded regions: outer area − inner area.

  • Formulas are on the SAT reference sheet, but knowing them saves time.

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