SAT Question of the Day Explained – January 13, 2014 – Circles, Semicircles, Area

 

Image copyright The College Board

Today’s official SAT question of the day is about a shaded circle inscribed in a semicircle. Given only the radius of the semicircle, we are asked to find the area of the shaded circle.

How to start:
From the diagram and the given information, we must notice that the circle’s diameter is also a radius of the semicircle; this segment is labeled CD.

We are told that the area of the semicircle is 24. We must recall the formula for the area of a circle (pi • the radius squared) and see that the semicircle is only half of a circle, so we have to take just half that area formula and set it equal to 24.

24 = (pi • radius squared)/2

From this point, we can arrive at our answer in two different ways.

Way #1: Algebra
48 = pi • radius squared
15.28 = radius squared
3.9 = radius (of the semicircle)

The radius of the semicircle is twice the radius of the small circle, so 3.9/2 gives us 1.95, the radius of the small circle. Using the formula for the area of a circle again, we arrive at 11.99, or 12.

Way #2: Algebraic Reasoning

Let s stand for the radius of the shaded circle. The radius of the semicircle (the big circle) is equal to 2s. Substituting this large radius back into the equation for the area of the semicircle which we left back above way #1, we get
24 = (pi • (2s)^2) / 2

which simplifies to

24 = (pi • 4s^2)/2

and then to

24 = pi • 2s^2.

The area of our small circle, which is still what we are trying to find, is equal to pi • s^2. If 24 is pi • 2s^2, dividing both sides by 2 gives us what we’re looking for. 24/2 = 12.

How to solve all questions like this one:

  • Remember your circle formulas (area and circumference), as well as the number of degrees in a circle
  • Look for relationships within the figure. If we hadn’t noted that the shaded circle’s radius was half the radius of the semicircle, we wouldn’t have unlocked this problem with the information given.
  • Keep your quantities straight. Another common issue with problems like this one is losing the idea that the area of a semicircle is half the area of a whole circle. Work neatly and carefully to avoid dropping terms or operators.

Key terms:

  • tangent
  • perpendicular
  • semicircle

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