Today on her Mathematics Calendar 2018, Theoni Pappas writes:
P is the circle's center. 9.9sqrt(2) is the length of the hypotenuse. Find the area of the shaded region to the nearest whole #.
(The shaded region is a segment of a circle -- the region between a chord of a circle and the arc subtended by the chord. Angle P is both an angle of the triangle and the central angle that also subtends this same arc. The word "hypotenuse" is a giveaway that this is a right triangle.)
To find the area of the segment, we subtract the area of the triangle from that of the sector. Let's find the area of the triangle first, as this is easier. This is indeed a right triangle -- indeed an isosceles right triangle, since its two legs are congruent radii. Thus it is a special right triangle, a 45-45-90 triangle -- its hypotenuse is 9.9sqrt(2), so its legs, the radius of the circle, are 9.9. Either leg of a right triangle is its base with the other leg the height, and so its area is 9.9^2/2 = 98.01/2 = 49.005 square units.
Now to find the area of the sector, we note that the radius is 9.9, so the area of the entire circle must be 9.9^2 pi, or 98.01pi square units. Since the arc (like the central angle) is 90 degrees, we conclude that the area of the circle is 1/4 that of the circle. So it is 98.01pi/4 = 24.5025pi square units.
So the exact area of the segment is 24.5025pi - 49.005 square units. Of course, we use a calculator to round this off. I obtain about 27.97 square units, which we round off to 28. Therefore the area is 28 square units -- and of course, today's date is the 28th.
It's possible to estimate this answer without a calculator. Notice that the area is (9.9^2/4)(pi - 2), and if we approximate 9.9^2 by 98 and pi by 3.14, we must then multiply (24.5)(1.14). This product works out to be 27.93, which is close enough to the correct answer.
This is a strange question -- we're given an exact length 9.9sqrt(2), yet we must round off the answer in the end. Perhaps we might as well have specified the hypotenuse as an integer, such as 14. (Notice that 9.9sqrt(2) is about 14.0007.)
This problem uses a special 45-45-90 triangle. Special triangles don't appear until Lesson 14-1 of the U of Chicago text. Glencoe students are currently learning about this in Chapter 8. Our students can still use the Pythagorean theorem to find the radius, but it's tricky.
Today I subbed in a high school chemistry class. Since this is not a math class, there will be no "Day in the Life" today.
Meanwhile, this is the second day of a two-day Performance Task on the blog, so there's no worksheet for me to post today. Hopefully our students can figure out the answer. As a hint, students should first try proving that the figure is a rectangle (which requires four slope calculations). Two distance calculations (length and width) are need to confirm that the rectangle is a square. The Midpoint Formula given on this page is a red herring, as it's not needed.
By the way, I hear that PBS is showing a math-related episode of NOVA tonight. Unfortunately, I don't have time to watch it tonight. Whenever I do get around to seeing this episode, I'll report on it on the blog.
And so thus ends this rather short (at least by my standards) post!