Geometry and Area: TEAS Practice Question with Answer & Explanation

Understanding the Relationship Between Rectangles and Circles

To solve this problem, you must visualize how a circle fits inside a rectangle. The "limiting factor" is always the shorter side of the rectangle.

In this scenario, we have a rectangle that is 12 inches (width) by 18 inches (length). If we try to draw a circle inside it:

• The circle can grow until it touches the top and bottom edges (12 inches apart).
• If we tried to make the circle 18 inches wide to match the length, the top and bottom of the circle would "bleed" off the edges of the banner because the banner is only 12 inches tall.

Therefore, the diameter ($d$) of the largest possible circle is equal to the shorter side of the rectangle.

• Diameter ($d$) = 12 inches.

Step-by-Step Calculation

1. Find the Radius ($r$): The radius is half of the diameter.

   • $r = d / 2$
   • $r = 12 / 2 = 6$ inches.

2. Calculate the Area ($A$): The formula for the area of a circle is $A = \pi r^2$.

   • $A = \pi \times (6)^2$
   • $A = \pi \times 36$
   • $A = 36\pi$ square inches.

Analyzing the Distractors

• Choice A (12π): This error often occurs if a student uses the diameter in place of the radius or forgets to square the number (using $A = \pi \times d$ or $A = \pi \times 2r$ incorrectly).
• Choice C (144π): This occurs if the student uses the diameter (12) as the radius ($12^2 = 144$). Always remember to divide the diameter by 2 before squaring!
• Choice D (144): This is the area of a square with 12-inch sides ($12 \times 12$), and it lacks the $\pi$ symbol required for circular area.

TEAS Math Concept: Geometry and Measurement

On the TEAS, you are expected to know basic geometric formulas (Area, Perimeter, Circumference) and how shapes interact. A common "trick" is providing the diameter when the formula requires the radius.

Memory Tip: Think of a hallway. If a hallway is 4 feet wide, you can't carry a 5-foot wide hula hoop flat through it. The "width" of your object (diameter) is limited by the "width" of the space. Always use the smaller dimension as your diameter.