Our latest book, 200 Challenging ACT Math Problems, is in the proofreading stage and should be ready for publication in a couple of weeks. In the meantime, here’s a taste of the type of problem you’ll find in the book. See if you can figure out the answer to this challenging shaded region problem (Scroll down for the solution).
The large square in the diagram below has an area of 64 square units. A circle is inscribed in the large square and then a smaller square is inscribed in the circle. What is the total area, rounded to the nearest tenth of a square unit, of the shaded regions?
Give it a try and then scroll down for the solution!
Your strategy for a shaded region problem is usually to find the area of the region that encloses the shaded parts and then subtract out the unshaded area(s). In this case, we want to do (Area of the Circle) – (Area of the Small Square).
The area of the large square is 64 square units, so each of its sides must have a length of 8 units (because A = s²). The length of the side of the square is also the diameter of the circle.
Now you can compute the area of the circle. If its diameter is 8 units, its radius is 4 units.
A = πr²
A = 16π
Now let’s move on to the small square. The diameter of the circle, which was 8 units long, is also the diagonal of the small square. This diagonal, along with two sides of the square, forms a 45-45-90 special right triangle. If you know the length of the hypotenuse in that triangle, you can find the length of each leg by dividing by √2. Dividing 8 by √2 and then rationalizing gives you a length for each side of the small square of 4√2 units.
So, the area of the smaller square is:
A = s²
A = (4√2)²
A = 32
The area of the smaller square is 32 square units. Now you can find the total area of the four shaded regions. That area is 16π – 32 ≈ 18.3 square units. The correct answer is Choice C.
If you have questions about this problem, shaded region problems in general, or anything at all to do with the ACT, send us an email at info@cardinalec. com