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Two goats are each tied at corners _A_ and _B_ of a fenced rectangular field of area 50. The goat's tethers allow them to graze within circular regions of radius 5, centered at _A_ and _B_, respectively. The two grazing areas meet at point _G_ but do not overlap. If the goats graze the entire shaded area within their respective reach by the end of the day, what is the area of the ungrazed, unshaded region?
Why did you choose this answer choice?
Good thinking, bad implementation. Indeed you learned that since $$\pi$$ doesn't cancel out in subtraction, you expect the format $$\text{number}-\pi$$ and can __POE__ any answer that is not in that format. That's a good reason why you can __POE__ answers D and E, but the remaining ones all meet that criteria. So when you implement a criterion for __POE__, remember—don't jump on the first answer that is not eliminated. Check all the others for the same criterion.
Correct. [[snippet]] Since $$\pi \approx 3$$, answer choice B is definitely in the __Ballpark__ of >B) $$25(2-\frac{\pi}{2}) = 25(2 - \frac{3}{2}) = 25 \times 0.5 = 12.5$$. After checking all of the other answer choices, you discover that none of the others are in the __Ballpark__. >A) $$50-25 \pi \approx 50-25\times 3 = 50-75$$. >>This answer choice results in a negative area! >C) $$25(2-\frac{\pi}{4}) \approx 25(2-\frac{3}{4}) = 25\times 1.25$$. >>No need to calculate since this is already greater than 25. >D) $$\frac{25}{2}\pi \approx 12.5\times 3 = 37.5$$. >>This answer choice is greater than 25, so this is also too big. >E) $$25 \pi \approx 25\times 3 = 75$$. >>This answer choice is greater than 25 (actually, it's greater than 50, which is the area of the entire rectangle), so this is also too big. Thus, B remains the only answer choice available.
Incorrect. [[snippet]] When you replace $$\pi$$ with 3, this answer choice comes out as >$$25(2-\frac{\pi}{4}) \approx 25(2-\frac{3}{4}) = 25\times 1.25$$. No need to calculate this. It's already greater than 25, so answer choice C can't be right. The unshaded region is *less* than half of the rectangle.
Why did you choose this answer choice?
Incorrect. [[snippet]] When you replace $$\pi$$ with 3, this answer choice comes out as >$$25\pi \approx 25\times 3 =75$$. This is not only greater than 25 (half of the rectangle), but also greater than 50, which is the area of the rectangle itself!
Review the summary for useful numbers in __Ballparking__ figures. Knowing these values by heart can save time and increase accuracy.
This is a classic mistake for the formula-minded, textbook-oriented test taker. Using formulas and solving equations is a technical, algebraic exercise that forces you to switch off your brain and focus on the formula. By the time you're done finding the area of the circular regions, you've forgotten what the question asked in the first place. On the other hand, __Ballparking__ forces you to ask yourself first, "what is the answer?" instead of "what is the area of a circle?" If you focus your mind on a simple concept such as "the answer is less than 25," there is no way you will choose a trap answer such as D.
Indeed.
Incorrect. [[snippet]] If you're __Ballparking__ correctly, this answer choice is the first to go. Since $$\pi \approx 3$$, this answer choice is actually equal to >$$50-{25\pi} \approx 50-25\times 3 = 50-75 = -25$$. A negative area? Not in this universe. __POE__ and move on.
Incorrect. [[snippet]] When you replace $$\pi$$ with 3, this answer choice comes out as >$$\frac{25}{2}\pi \approx 12.5\times 3 = 37$$. Since 37.5 is greater than 25, answer choice D can't be right. The unshaded region is *less* than half of the rectangle.
$$50-{25\pi}$$
$$25(2-\frac{\pi}{2})$$
$$25(2-\frac{\pi}{4})$$
$$\frac{25}{2}\pi$$
$$25\pi$$
The unshaded region is calculated by the formula >$$\text{Rectangle Area} - \text{Circular Regions}$$. This was the only answer choice that fit the "$$50 - \text{something}$$" structure.
Ah, I made some silly calculation mistake along the way. I should be more careful.
I calculated the area of the circular regions using $$A=\pi r^2$$ but didn't notice that the question asked for the _un_shaded region.
I used __Ballparking__ but forgot that $$\pi \approx 3$$.