I know it's counterintuitive, but millions of devoted readers have expressed interest in my 6th-grader's math homework, and comparatively few in, say, Israeli electoral politics. Whoever pitched "Are You Smarter than a Fifth Grader" as a game show idea may have been onto something. As it happens, I've been consulted again regarding (b) in the picture. The context—"learning objective," I think they say in ed-speak—concerns the geometry of circles. This mysterious number pi (π) is the ratio of the circle's circumference to its diameter. Half the diameter is the radius. The area of the circle is the product of π and the square of the radius. From the definition of π it follows that the circumference is equal to π times the diameter. I don't think you can see in the picture that they are given the information that the squares are 1 unit long on a side. The problem is to find the area and perimeter of the figures.

Lydia could do (a) and (c). Maybe you can see enough of them to agree, with me, that (b) is indeed a little harder. To find the area, you have to satisfy yourself that there are two partial circles of the same size, plus the square, and no overlap between the three—a little tricky, perhaps, since the circles, if whole, would overlap. The radii of the circles are the same as the side of the square, length 1 unit, and, since it's a square, the missing portion from each circle is a 90-degree sector. Then you just have to put all that together. The individual circles, if whole, would have an area of π square units. Since 90 of the 360 degrees (one-fourth) are missing, they both have an area of 0.75 π square units, and together, an area of 1.5 π square units. Plus 1 square unit for the square containing the two missing sectors: 1 + 1.5 π, or about 5.71 square units.

I beg Lydia to consider, once she has an answer, whether it makes any sense, a habit that tends to guard against errors in number grinding or keying on a calculator. I might as well beg Kyle Gibson, through the tv screen, to "work quickly and throw strikes." In this case though, it seems that a useful observation might be that, since the circles if completed would slightly overlap, there is a portion of the square within both circles. Therefore the final answer should be a little less than the sum of the areas of two full circles, which would be 2π, or about 6.28, and 5.71 satisfies that check.

Regarding the perimeter: it's three-fourths of the way around a circle with circumference of 2π, and then the same trip again, so (1.5)(π)(2), or 3π, about 9.42.

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