Composite & shaded regions drills

Find area or perimeter of composite figures and shaded regions.

Free, timed SHSAT practice sets on composite & shaded regions— every question includes a full worked explanation, and there's no account required.

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How to solve composite & shaded region problems

Break the figure into shapes you know, then add or subtract.

Common traps: squaring the difference of the radii (\(\pi(5 - 3)^2 = 4\pi\) is wrong), counting a hidden inside edge in the perimeter, using a full circle's circumference for a semicircle, and adding a cut-out instead of subtracting it.

Worked examples

Example 1.
A square with side 8 yd has four quarter-circles, each of radius 4 yd, cut out from its four corners. Using 3.14 for \(\pi\), what is the area of the shaded (remaining) region?
Show answer and explanation
Answer:
A

The four quarter-circles together make one full circle of radius 4, with area \(3.14(4)^2 = 50.24\). Shaded area \(= 64 - 50.24 = 13.76\) sq yd. (Subtracting only one quarter-circle, 12.56, gives 51.44. Adding the circle instead of subtracting it gives 114.24. Forgetting to subtract the corners gives the square's area, 64.)

Example 2.
A washer-shaped ring is formed by cutting a circle of radius 8 cm out of the center of a larger circle of radius 14 cm. What is the area of the ring, in terms of \(\pi\)?
Show answer and explanation
Answer:
C

Ring area \(= \pi R^2 - \pi r^2 = \pi(14^2 - 8^2) = 132\pi\) sq cm. (Subtracting the radii first and then squaring, \(\pi(14 - 8)^2\), gives \(36\pi\). Adding the two circles instead of subtracting gives \(260\pi\). Forgetting to take out the hole gives the large circle's area, \(196\pi\).)

Example 3.
A running track's edge is shaped like a rectangle 46 in by 20 in with a semicircle (of diameter 20 in) attached to one short side, replacing that side. Using 3.14 for \(\pi\), what is the total perimeter of the shape (the two long sides, the one remaining short side, and the semicircular arc)?
Show answer and explanation
Answer:
A

Perimeter \(= 2(46) + 20 + 3.14(10) = 112 + 31.4 = 143.4\) in. The semicircle's straight edge is not part of the boundary. (Also counting the short side that the semicircle replaces, 20, gives 163.4. Using a full circle's circumference, \(2(3.14)(10)\), gives 174.8. Finding the area, \(46 \times 20 + \dfrac{1}{2}(3.14)(10)^2\), gives 1077.)