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9.5 Solve Geometry Applications: Circles and Irregular Figures

3 min readjune 25, 2024

Circles are all about curves and measurements. We'll look at how to find the distance around a circle and the space inside it. These formulas use special numbers like and to calculate what we need.

are shapes that don't fit neat categories. We'll learn to break them into smaller, familiar shapes to find their areas. This skill helps solve real-world problems, like figuring out how much carpet you need for an oddly-shaped room.

Circles

Circumference and area of circles

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  • measures distance around a circle's perimeter
    • Formula C=2πrC = 2\pi r uses radius rr (distance from center to edge)
    • Alternative formula C=πdC = \pi d uses dd (line segment through center with endpoints on circle)
    • π\pi is a constant approximately equal to 3.14159
  • quantifies space inside a circle
    • Formula A=πr2A = \pi r^2 uses radius rr squared
    • π\pi is a constant approximately equal to 3.14159
  • Radius spans from circle's center to any point on its edge
    • Radius is half the diameter (radius of 5 cm means diameter of 10 cm)
  • Diameter is a line segment passing through circle's center with endpoints on the circle
    • Diameter is twice the radius (diameter of 10 cm means radius of 5 cm)
  • is a line segment with endpoints on the circle (e.g., diameter is the longest chord)

Circle Components

  • is formed by two radii and has its vertex at the circle's center
  • is a portion of the circle's circumference
  • is the region bounded by two radii and an arc
  • is a line that touches the circle at exactly one point

Irregular Figures

Area of irregular figures

  • Irregular figures have non-standard shapes unlike circles, triangles, or rectangles
  • Break irregular figure into smaller, recognizable shapes to calculate total area
    • Rectangles have area A=lwA = lw (length ll times width ww)
    • Triangles have area A=12bhA = \frac{1}{2}bh (one-half base bb times height hh)
    • Circles have area A=πr2A = \pi r^2 (pi times radius rr squared)
  • Find each shape's individual area and sum them for the irregular figure's total area
    • Irregular figure composed of a rectangle (4 cm by 6 cm) and (radius 4 cm) has total area 65.13 cm² (rectangle 24 cm² + semicircle 25.13 cm²)

Applications of geometric formulas

  • Identify relevant information in problem (measurements, shapes)
  • Choose appropriate formulas based on given info and shape(s)
    • Circles use circumference C=2πrC = 2\pi r or area A=πr2A = \pi r^2 formulas
    • Irregular figures break into familiar shapes (rectangles A=lwA = lw, triangles A=12bhA = \frac{1}{2}bh, circles A=πr2A = \pi r^2)
  • Plug given values into formulas
  • Calculate to find desired measurement (circumference, area)
  • Verify answer is in correct units and reasonable for the problem
    • Circular garden with 10 ft diameter needs 31.42 ft of fencing (circumference C=πd=3.1415910=31.42C = \pi d = 3.14159 * 10 = 31.42)
    • Irregularly shaped room composed of a 12 ft by 16 ft rectangle and a triangle with base 12 ft and height 8 ft has area 240 ft² (rectangle 192 ft² + triangle 48 ft²)
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© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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