Have you seen this type of question in QRT?
Started by mashimlro
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I would make an educated guess, mark it and come back.
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Find the area of a regular pentagon (n-gon) whose side length is 2 (or others)?
Only triangles and half of the quadrilaterals. If they give you anything pentagon or above, they would have to break it down enough that you'd have triangles and quadrilaterals with known side lengths.
Since it's a regular (meaning all sides are equal) pentagon or w/e-gon, it can be calculated, not so easily though. You'd break it up into 5 triangles, you know the base of each triangle, and you find the height (the perpendicular distance between each side and the center) by knowing that the height would half the triangle, and divide the central angle lol. I gotta start taking this full-length test, I can show how later.
Ok I got caught up with some shiznet, so imma start the test a bit later, but here it is:
Break it up to 5 triangles, each triangle is equilateral. The top angle of each triangle (on the end that is at the center) is 360/5 = 72 degrees. Now if you split each triangle in 2 by drawing a perpendicular line from center to middle of side, you will have a right triangle, where the top angle is 72/2=36. Now since the bottom of that right triangle is half of the side (a/2), you have tan(36)=(a/2)/h so h = (a/2)/tan(36). The area of each triangle is then 1/2*b*h = 1/2*a*[(a/2)/tan(36)], or a^2/(4*tan(36)). Now just multiply it by 5. so you have A= 5*a^2/(4tan(36)). I can't see any reason why this can't be generalized for an normal n-gon. It would be A = n* a^2/(4tan(360/2n)).
that works for a square at least. I might have made a mistake somewhere, didn't do on paper, check my work, but seems to work. (And it's probably confusing without any figures, but can't do much about that)
Break it up to 5 triangles, each triangle is equilateral. The top angle of each triangle (on the end that is at the center) is 360/5 = 72 degrees. Now if you split each triangle in 2 by drawing a perpendicular line from center to middle of side, you will have a right triangle, where the top angle is 72/2=36. Now since the bottom of that right triangle is half of the side (a/2), you have tan(36)=(a/2)/h so h = (a/2)/tan(36). The area of each triangle is then 1/2*b*h = 1/2*a*[(a/2)/tan(36)], or a^2/(4*tan(36)). Now just multiply it by 5. so you have A= 5*a^2/(4tan(36)). I can't see any reason why this can't be generalized for an normal n-gon. It would be A = n* a^2/(4tan(360/2n)).
that works for a square at least. I might have made a mistake somewhere, didn't do on paper, check my work, but seems to work. (And it's probably confusing without any figures, but can't do much about that)
are you guys serious?
s^2 (1.72)
you're welcome😎
(not 100% exact, but extremely close)
s^2 (1.72)
you're welcome😎
(not 100% exact, but extremely close)
Wikipedia? Some other site? We can all search for that. The exam could give an EXACT answer (with square roots and such). The 1.72 s^2 wouldn't cut it.
And to repeat, I highly doubt you have to find the area of any figure unless it is broken down into 3- or 4-sided shapes (or circles) with clearly marked sides (or sides you can figure out the lengths).
The DAT isn't math for math majors.
middleschool mathlete actually lol... but seriously, 1.72 is the 3.14 of regular pentagons... if you know that other way to do it, there is no reason why you shouldn't know how to do the shortcut... Plus, when I said not 100%, I meant it'll be off by ~0.01 or less (not unlike using 3.14 instead of 1200 sigfigs for pi).
I wasn't hating on the people who don't know that little formula... It's obviously not as well known as the formula for the area of a square... Just that if you can write a whole dissertaion on how to use the 10 triangles (impressive btw), I don't know how your geometry class could have missed this.
Also, I don't think it's absurd for them to ask the area of a regular pentagon and only give you one side. Altough you're right, if they have the answer in the other form, you'd be screwed.. then again, if it's in numerical form, this will save you 5mins 😉
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oh. and since you're the one saying thee dat isn't "math for math majors"... do you honestly think that they would expect you to know that lengthy formula over one that consists of a few numbers and a letter? doubt it.
I haven't taken DAT but from the samples I've done, I agree, they probably won't expect you to derive something like that. Nevertheless, it was asked and I like doing math anyways lol.
Last time I took an actual geometry class was about 6 years ago, but the basic shapes have been revisited many times in my other classes (calc, etc), so I remember them. That formula, along with most of the random math formulas, I tend to not memorize, cause I'd rather take like 20-30 seconds to derive it.
And in a math test (probably not on DAT), it's usually more important to be able to derive/utilize stuff and not just have the memorized.
Nevertheless, if that question came up on DAT (not in 50 years), you'd have 20 seconds on me ;p
Last time I took an actual geometry class was about 6 years ago, but the basic shapes have been revisited many times in my other classes (calc, etc), so I remember them. That formula, along with most of the random math formulas, I tend to not memorize, cause I'd rather take like 20-30 seconds to derive it.
And in a math test (probably not on DAT), it's usually more important to be able to derive/utilize stuff and not just have the memorized.
Nevertheless, if that question came up on DAT (not in 50 years), you'd have 20 seconds on me ;p
Just curious, Mathcounts?
Hey I said not to expect pentagon problems 🙂
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