Use the diameter where the radius is needed in a circle formula, or vice versa.
An error-pattern drill: practice spotting and avoiding this specific SHSAT trap.
Loading questions…
A circle has a diameter of 10. What is its area? If you put 10 into \(\pi r^2\), you get \(100\pi\). But \(r\) is the radius, which is half the diameter: \(r = 5\). The area is \(\pi(5)^2 = 25\pi\). Using the diameter makes the area 4 times too big.
The trap works because the question gives one length, and you want to use it right away. Circle and cylinder formulas need \(r\), but some questions give \(d\). The trap choice is often exactly 4 times the right area or volume.
Check habits:
Radius \(= 26 \div 2 = 13\), so area \(= \pi r^2 = \pi(13)^2 = 169\pi\) sq in. (Squaring the diameter instead of the radius gives \(676\pi\). Using the circumference formula \(\pi d\) gives \(26\pi\). Squaring the diameter first and then dividing by 2 gives \(338\pi\).)
The sector is \(\dfrac{240}{360}\) of the circle. Whole area \(= \pi r^2 = 144\pi\). Sector area \(= \dfrac{240}{360} \times 144\pi = 96\pi\) square m. (Stopping at the whole circle's area gives \(144\pi\). Using the circumference \(2\pi r\) instead of the area gives the arc length, \(16\pi\). Using the diameter, 24, in place of the radius gives \(384\pi\).)
Volume \(= \pi r^2 h = \pi(13)^2(15) = 2535\pi\) cubic in. (Using the diameter, 26, in place of the radius gives \(10140\pi\). Forgetting to square the radius gives \(195\pi\). Doubling the radius instead of squaring it gives \(390\pi\).)