I think it’s always good to have 0 in your back pocket as well. Sometimes 0 takes care of that one anomalous case, or can be used as a quick counterexample. In this case it leads to a simple and useful equation.
So i have this math question it consists of 4 overlapping semicircles the biggest one has a diameter of 4 and then we have two smaller semicircles drawn by connecting the center of the biggest circle and the endpoint of the diameter and then we have a semicircle drawn by connecting the center of the two semicircles with an equal area to the two drawn before
and the question is what is the value of the remaining area of the bigger semicircle after cutting these 3 smaller circles
It was kind of hard for me to visualize the ending there, with the last semicircle, but I’ll give it a shot.
Usually when asked to find “what’s left” of the area of a figure, subtraction is involved, so…
A(area of biggest figure) - B(area of next biggest figure) - … Z(area of smallest figure) = Remaning area of A.
Edit: After rereading it, I managed to draw it out, hahaha, hold on a sec.
Boy, what a question. You actually got that from the SAT? I’m assuming not due to the lack of alternatives. I’d go with an approximation method. Look for the answer closest to (2pi - 4)/2 .
How did I get there? I drew a rectangle spanning from the center of the circle to the left to the center of the circle to the right. The length of that rectangle happens to be the diameter of the middle circle whose ends touch the centers of the left and right circles. I divided the rectangle into 2 squares, each with area of 1. Remaining should be the following, in order
A quarter circle of area pi/2 ; a square with area 1 ; another identical square with area 1 ; another quarter circle of area pi/2 .
Adding them together and substracting them from the biggest semicircle, would equal the answer I gave above. Now, what you’ve probably noticed, and I couldn’t solve for the life of me, is that little extra piece remaining on both squares. If there is some way to figure out the area of the little piece of fluff, then, you’ll have your precise answer. If not, you’ll be safe with the approximation above, granted you have alternatives.
some1 help pls? Hahaha, good luck pal.
@kareem5511 @AGoodFloridian I think what kareem’s asking is we have a semicircle of diameter 4 or radius 2 (assume center is at (0,0) and its diameter is on the x-axis), and there are three semicircles of radius 1 with centers (-1,0), (0,0), (1,0) completely contained inside the large semicircle, and we want the area of the region inside the big semicircle but outside the smaller ones.
This is quite difficult for an SAT problem, but I have seen much more difficult geometry problems, and this problem, while harder, is solvable by anyone who’s taken geometry.
Many ways to solve, but the way I did it was taking the area of the big semicircle (which is 2π), subtracting the area of the two non-intersecting semicircles (which is 2*(π/2) = π) so you have 2π-π = π.
Now you need to subtract the area inside the small middle semicircle that is outside the other two. This is equal to the area of a 60-degree arc, minus 2 times the area of the small region that is contained in the intersection of the semicircles. This equals (π/6) - 2*(π/6 - sqrt(3)/4) = (sqrt(3)/2) - (π/6).
Area = π - ((sqrt(3)/2) - (π/6)) = 7π/6 - (sqrt(3))/2 which is about 2.8.
Let me know if I made any mistakes or computation errors (which I sometimes do).
i think your answer is wrong cuz the area of the large semicircle as a whole is 2π so how could it equal 2.8π when we take parts of it your method is correct but you calculated it wrong at the end @MITer94
so i tried to solve it and i got this 0.3006412629
by dividing the non-shaded area into 5 equal sectors and two equilateral triangles
the area of each sector would be 1/6 times 5 = 5/6 and the triangles would have a total area of sqrt3/2 so
2π-(5/6+sqrt3/2) = 0.3006412629 @MITer94 @AGoodFloridian
@kareem5511 I said 2.8, not 2.8*pi.
How is the area of each sector 1/6? How are you defining those sectors? Btw, the expression you gave evaluates to roughly 4.583.
Unless I misinterpreted what your question was asking, I’m fairly certain that my solution is correct.
am i allowed to post photos pr it is not possible
@kareem5511 You can upload the image to a free photo sharing site, then post the link here.
the question and my answer
@kareem5511 sorry I forgot, URL domain name is blocked out
I think since most questions can be solved without a calculator this is being overthought? Could be my mistake though.
@Sophiclees I didn’t use a calculator here (except for computing the value at the end). But I don’t know if this is an actual SAT problem.
Here, I got one for ya’. I haven’t seen any question like this on the SAT, mainly because I just made it up, but it’s certainly doable.
A rectangle is inscribed within a circle with radius 3, with its 4 corners touching the circle itself. The left side of the rectangle is increased outward by 50 percent, and so is the right side.(Imagine getting two squares, each with an area equal to half of the rectangle inside the circle, and putting one on each side of the rectangle.)
What is the total area of the two protrusions outside of the circle?
I hope I worded that right. Motivation for this question came from the last guy who made that circle-inception question. Thanks bud.
Just for the sake of clarity up there… The length of the rectangle inscribed in the circle is two times the height. I really want to emphasize the fact that it’s actually just four squares in a row. But I’ll hush hush now.
forgot to add alternatives.
A) approx. 16.4
B) approx. 14
C) approx. 17.7
D) 18
E) 20^cos(7pi) - ln(7)
Yup, just like the SAT. It’d suck if I didn’t get my own question right and just wrote down all wrong alternatives.
Hahaha, my third SAT is in four days from now, wish me luck!
@AGoodFloridian sorry to say, but I don’t think any of those choices are right, assuming I interpreted your question correctly.
I think you meant to say “increased by 100%” since an increase of 50% is like going from 10 to 15, or x --> 1.5x.
Assuming the “squares” have a side length of s, we have (s/2)^2 + s^2 = 3^2, or s^2 = 36/5. Then 2s^2 = 72/5 = 14.4, but since the area of the two regions is less than the sum of the areas of the two squares, the answer must be less than 14.4. I can determine the exact area using some trig, but the computations look somewhat messy so I’ll skip that for now.
Here’s a math problem I made up (not SAT level, or perhaps a hard level 5, but solvable with an algebra II and geometry background):
Q: Square ABCD has side length 1. Point P is on segment BC such that the circle with diameter AP is tangent to line CD. Find the length of BP.
A) 1/4
B) 1/2
C) sqrt(2)/2
D) 3/4
E) 2(sqrt(2) - 1)
Note: Part of writing multiple choice tests is creating good answer choices. I am, just like everyone else, human; feel free to point out any ambiguities or mistakes.
Congrats on you’re 3000th post @MITer94 !
Ah… I’d figured I messed something up, I feel like showing you the drawing I made, but it seems like links aren’t too well received here. I’m definitely going to try out yours in the morning, after midnight my brain goes from mathematical to philosophical, so I’ll spare you that for now.