I didn’t do a Weekend Challenge last weekend. It’s not like I didn’t want to, y’all. Things just got totally cray-cray. My use of “cray-cray” in the previous sentence should be enough of an indication to you that things remain squarely thus.

I actually wrote two questions for this weekend, but I’m only going to use one of them. The other one is going into the strategic reserve. For emergencies.

If you’re the first to answer this correctly (and not anonymously), I’ll bestow upon you coveted access to the PWN the SAT Math Guide Beta.

Other ways you can gain access to my Magnum Opus:

  • Buy it. It’s $5. You get about 300 pages of useful SAT math help. I get a footlong meatball sub at Subway.
  • Send me a question of your own making. Added bonus: this is a fantastic way to solidify your knowledge of the test.
On to the question:

In the figure above A is the center of the circle, A and D lie on BF and CE, respectively, and B, D, F, and G lie on the circle. If BC = 3, and DG (not shown) bisects BF, what is the total area of the shaded regions?

Good luck, and have a great weekend. I’ll post the solution early next week.

UPDATE: Commenter Katieluvgold got it first. Nice work Katie!  Solution posted below the cut.

This is a shaded region question, and the usual shaded region technique applies, but with a twist. Since only parts of the top half of the circle are shaded, let’s just look use the top semicircle as Awhole.

We know AD is perpendicular to CE because it’s a radius connecting to a tangent line. Those are always perpendicular. It’s a rule.  And since A and D are both on rectangle BCEF, and BC = 3, AD must also equal 3.

So the radius of the circle is 3, so the area of the circle is 9π. That means the area of the semicircle is half that. Awhole = 4.5π.

Point A is the center of the circle, so it’s obviously the center of diameter BF. When the question tells us that DG bisects BF, it’s telling us that DG is also a diameter because it also passes through the center of the circle. That means ΔCEG has both a base and a height of 6.

Because the rectangle’s top and bottom are parallel, that means the top triangle (which is our Aunshaded) is similar to the large triangle, because all the angles of the two triangles are the same (the top one is shared, and the bottom ones are corresponding angles across parallel lines). We don’t have a way of knowing what the base is automatically, but we DO know that the height of it is 3, because the height of the little triangle is a radius. So the small triangle has a base of 3 and a height of 3.

The area of a triangle is (1/2)bh, so Aunshaded = 4.5.

To calculate Ashaded, we just subtract Aunshaded from Awhole:

AwholeAunshaded = Ashaded
4.5π – 4.5 = Ashaded

Comments (6)

Semicircle: 9pi/2 = 4.5pi —> 3 = radius
Triangle = 9/2 = 4.5 —> height = 3, base = 3 –> using similar triangles property
Shaded Region = 4.5pi – 4.5 —> Area of semicircule – Area of Triangle

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