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Question of the Week for September 25, 2017

Jonathan EvansJonathan Evans PowerScore StaffJoined: 10/31/2016

On your own as a challenge, attempt the question of the week above and enter your solution in comments below. Answer explanation and multiple approaches to follow. 


  • Jonathan EvansJonathan Evans PowerScore Staff Joined: 10/31/2016
    ...and now the explanation. First, two hints:

    These two figures, noted in the problem, are key. 

    We're looking at the areas of circles and squares, so we need to get those formulas down on our scratch paper, stat!

    Circle area = πr²

    Square area = s²

    Can we use these equations right now to make any calculations? Sure. The big circle has radius 4, so:

    Area Big Circle = π · 4² = 16π

    Also, the square has sides of 4, so:

    Area Square = 4² = 16

    Now observe how the large circle and the square are related. 

    The square takes up part of the big circle. It covers up one quarter of the big circle. In order to calculate the shaded region, we cannot double count this area.

    So, ask yourself, if we don't want to double count the area of the circle covered up by the square, how could we get rid of it?

    We could start by finding out how much of the circle is left. If one quarter is covered up, that leaves us with 3/4 left.

    In other words, three quarters of the area of the big circle is:

    (3/4) · 16π = 12π

    So, forgetting about the little circles for a second, what's the total area of the uncovered portion of the big circle and the square?

    12π + 16 

    Now it's time to take away the area of the little circles. Both have radii of 2. Thus:

    Area Small Circle = π · 2² = 4π

    We have to take this away twice from the total area, 12π + 16, noted above.

    12π + 16 - 4π - 4π = 4π + 16 
    Answer Choice (D) is correct

    Alternative Approach

    Let's see if we can estimate the area. We'd need to calculate perhaps the square's area, 16. Look at the figure with the blue square above. If we had to take that square and fit its area into the shaded region, how many times might we be able to do that? Perhaps twice. I can see about a square's worth of area on the left hand side of the figure. I can also see maybe a square's worth of area in the rest of the shaded region. Thus, the shaded region looks like it has roughly an area of 32, or perhaps a little less. 

    Since we know that:

    π ≈ 3.14

    We could use that to estimate in our answer choices.

    (A) this comes out way too small
    (B) again, way too small; don't bother
    (C) still too small
    (D) looks like around 29; pretty good
    (E) closer to 40; too big

    We still come out with (D).
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