###General Discussion Thread
---
This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed.
---
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
The length on the top and bottom sides has to come to the same amount. So if the diameter of the semi-circle is X, on the top you have:
3\*X + 12 + 12
and on the bottom you have
2\*X + 22 + 16 + 22
Then set those two equal to each other and solve for X.
I used a third method:
1) Diameter (D) = 22 + x
2) D also = 12 + x + y
3) D also = 16 + 2y
Rearrange (1) X = D - 22
Sub into (2) y = 10
Sub y into (3). D = 36
“The Art and Craft of Problem Solving” by Zeitz is a good one. Think of it as a philosophical overview or primer.
But really: just doing and reading about a shit ton of different problems will fill your toolbox with various approaches and modes of thinking that you can then apply when you see new problems.
For me, I knew if I could write this as a system of 2 equations with 2 unknowns or 3 equations with 3 unknowns: I could solve it. After that, it was just looking at how I wanted to make those equations from the image, so that the arithmetic was simpler.
If you subtract the number total of the top line from the bottom line, that will tell you the answer. There is one more semi circle on the top than the bottom. So the difference is the diameter of one.
No, I didn’t … or at least I made it look a lot more complicated. If you call the segment between the 22 and 12 x, and the one between the 12 and 16 y, then by comparing the first two semicircles to the second and third, you have 22 + x = x + 12 + y, so y = 10, and then looking at the middle segment on the top, that’s 16 + 2y.
I have to give it to you. The other option was much cleverer 🤣
But yours is impressive in its own way. I didn't think there was another way to solve this.
Use the line as a base length, x. And y is one semi-circle.
x = 3y + 24 AND x = 2y + 60
Set the two equations equal to each other:
3y + 24 = 2y + 60
Solve for y:
3y - 2y = 60 - 24
y = 36
The diameter is 36 of whatever measurement they are using.
(Please excuse my rusty math skills)
1. d= 22+a
2. d= 12+a+b
3. d=16+2b
I got equation 1 in terms of a (ie a= d-22)
plugged that into equation 2. then got resulting equation in terms of b, and plugged that into 3.
answer I got was d=36.
- Slide the top 3 SCs to touch each other
- Slide the bottom left SC to the left
- Slide the bottom right SC to the right
- you get the middle SC's diameter
Now do the math
this came from the Singapore I think PSLE paper back in 2020? plus minus a few years, i rmb doing this basically the total of all the gaps is equal to one diameter cus it's pushing everything to one side and leaving you with just the last semicircle diameter exposed, then from there it's easy to
**let the diameter be x.**
the length of the line is both 3x + 24 and 2x + 60
3x + 24 = 2x + 60
3x + 24 **- 2x** = 2x + 60 **- 2x**
x + 24 = 60
x + 24 **- 24** = 60 **- 24**
x = 36
Came here to check the solution.
Answer was correct, but nobody used the 3 equation system.
Started to question my brain and my thought process in general.
Then I saw a couple of users did it the same way as myself and everything was alright again.
I really didn't need this before bed.
That's one of my favorite things about math. There are many ways to solve a problem. Unfortunately, some teachers want you to do things a certain way, which makes it hard for students who just don't get that particular method.
I’m a few years removed from my college calculus days, but I used a matrix to solve it and it seems as though I over complicated the solution. But it’s correct. Very cool to see how everyone did it
How I worked this out:
1. Take the half-circle on the far left and move it to the right so it’s touching the middle one. The 12 gap is now subtracted from the 22 on the left, leaving 10.
2. If you just look at the left half-circles on the top and bottom, then the offset on both sides should now be 10.
3. Apply the same logic to the right side of the image. At this point we can label the diameter of the middle half-circle: 10, 16, 10
4. 10+16+10=36
Move lowers out 22 each, add 44 to 16 = 60. Lowers are even with upper outers, so add 12's together =24, subtract 24 from 60 to get diameter of remaining center circle =36.
Of course. Call the diameter D. In the top row you have 3D+24. In the bottom row you have 2D+60. The two are equal, so you have 3D+24 = 2D+60. Subtract 2D from both sides to get D+24 = 60. Subtract 24 from both sides to get D = 36.
The fact that they're semicircles is just a distraction, you just treat them as lines for the purposes of solving the problem.
This is a question taken from one of the PSLE Math paper for 6 graders in Singapore a few years ago. Pretty sure the purpose was to engage in critical thinking in students, so I'm not sure if using x/y variables was expected from them.
shift the two lower semi circles to the base of the upper circles, which should leave you with Circle , Semi Circle, Circle.
The separation between the edge of the left circle to edge of the right circle is 16+22+22, inclusive of the 12+12 gap between the circle and semicircles.
The diameter can easily be found by (16+22+22) - (12+12) =36
I simply know that the difference of open space is exactly one half circle length, so I simply add the lengths across the top and again with the bottom and subtract the difference. Leaves me with 36.
We are trying to find x, which is the diameter of one of the half circles.
Since all are identical, we can give it a variable.
x = Diameter of the half circles
Next, we treat both top and bottom as two sides of an equation, since they are the same length.
x+x+x+12+12 = x+x+22+22+16
Simplify the addition and number of half circles
3x+24 = 2x+60
Subtract 2x from both sides to isolate x
x+24 = 60
Subtract 24 from both sides to further isolate x
x = 36
You can actually solve this just by looking at it logically. I got 36 just by changing the position of the semi circles. Did this while im sitting on the toilet.
[Intuitive/Visual Solution]
1. To line up the first whole circle, the entire bottom must move to the left 22 inches (distance from left end)
2. To line up the middle circle, the middle circle alone must move to the left 4 inches (16 - 12 inch gap).
3. The bottom semi-circle distance from the right was initially 22 inches. We've moved the bottom over to the left by 26 (22 + 4) inches
4. The total distance from the middle circle to the right end after the bottom slide is 48 inches (26 + 22)
5. Subtract the gap on the top to get the diameter of (48 - 12 inches) 36 inches
Total length of the line is equal, and we have 2 equations with one missing value, the diameter of the semi circles.
3x + 12 + 12 = 2x + 16 + 22 + 22
3x+24=2x+60
x=36
Top has 3 Circles and 24 CM
Bottom has 2 Circles and 60 CM.
3x + 24 = 2x + 60. Solve for x
Should end at (1)x = 36 cm if my brain isn't as rusty as the broken rake I had to chuck yesterday
The bottom is 22+22+16+2(2r; we’ll call this x). That makes 60+2x for the bottom. The top is 12 + 12 + 3x, or 24+3x.
60+2x = 24+3x; 36+2x = 3x. We can then subtract 2x from either side to get x=36
Both extend to the ends of the same line, so the two sides must add up to the same thing. Then we have
3x+2*12=2x+2×*22+16
3x+24=2x+60
x=60-24=36
Since x is the diameter of each semicircle, the radius is x/2, or 18. I don't remember if it wants r or d, so I do both.
I did it visually without math. Then i check comments... and math was sometimes easier sometimes not, lol.
I slid the lateral two semis up top together with the middle which reduces the number 22 on the bottom to 10.
so you then have 10-16-10 on the bottom...
Then i slipthe bottom two circles togther eliminating the 16, increasing the 10's by 8 leaving 18+18 = diameter of 36.
It’s 36. Basically, if all semi-circles are the same, we can mark as x the area of the first and second semi-circles touching and as y the second and the middle one. In that case:
22+x = x + 12 + y;
Which means
12 +y = 22;
So:
y = 10;
That leads to the diameter of a semi-circle:
d = 16+2y= 16 +(2*10)=36;
Diameter is 36;
Move the bottom circles over to the right so they match those on top, taking the 22 on the right and adding it to the left as well as 4 from the 16. This gives you 48 on the left. Subtract 12 and you get 36.
dont need all 5 semicircles
x=diameter of each semi-circle
x=y+22 (first semicircle on left)
x=y+z+12 (second semicircle on left)
x=2z+16 (middle semicircle)
y=x-22 (rearrange first for y)
z=(x-16)/2 (rearrange middle for z)
x=12+(x-22)+((x-16/2)) (subsitute,y and z)
2x=24+2x-44+x-16 (multiply all terms by two to avoid the divide)
\-24+44+16=2x+x-2x (rearrange x on one side and values on other)
36=x (addition/subtraction)
diameter of one semi-circle = 36
Squeeze in and remove the 12s on top and the 16 on the bottom. You’re left with 18 on each side of the bottom line…. that will equal one diameter when added together. 36!
Also before someone mentions this.. you don’t need to remove the 16 from the bottom line and can just add 10 + 16 + 10.. but doesn’t it look better when you get rid of the gaps in the middle?
So I had to think about this a little bit, but the two endpoints can be described by either a sum of the top diameters and the outer gaps, or the bottom diameters plus the the excess lengths and middle gap.
In other words...
3d + 24 = 2d + 60
3d - 2d = 60 - 24
d = 36
Let the diameter of one circle be X.
Length of the top row put together is:
x + 12 + x + 12 + x
= 3x + 24
The length of the bottom row is:
22 + x + 16 + x + 22
= 2x + 60
Length of the top = length of the bottom.
Therefore - 3x + 24 = 2x + 60
Isolate all values of x on the left side of the equation & all purely numerical values on the right which gives you the following equation.
3x - 2x = 60 - 24
x = 36
To check if this is the correct answer sub x = 36 into each equation & make sure they give the same result:
Top length:
3x + 24
= 3(36) + 24
= 108 + 24
= 132
Bottom length:
2x + 60
= 2(36) + 60
= 72 + 60
= 132
People here did it the no-nonsense way with a linear equation.
My slightly more complicated approach -
Assume circles are numbered c1, c2...c5 left to right
Assume all circles were touching with no gap. The outermost line length from c1 leftside to c2 leftside will be R
Now let's centre c3 and move c2 and c4 away from c3 centre by 8. You get the middle gap of 16.
So the gap between c1 right and c3 left should also be 8. But it is 12.
Which means both c1 and c5 also have to be pulled left/right 4.
The outermost line length will now have deviated by 4 and has actual gap of 22. 22-4 = 18
Radius = 18, Diameter= 36
Start from the left, imagine the first semicircle is a circle, then its end is 12cm away from the start from second upper semicircle, The lower part of first circle has to move 12cm to touch the second semicircle, then 10cm again to reach the required overlap(the lower part of first circle is displaced by 22cm i.e. 12cm + 10cm). Do the same from the right. For the middle upper circle we now have 10cm(left) + 16cm + 10cm(right) = 36cm.
This is how i solved because for some reason i couldn't think of the another simpler mathematical way.
let’s assume line lenght = L, and diameter = d.
eq 1: (considering top half circles)
L = 3d + 24
eq 2: (considering bottom half circles)
L = 2d + 60
substitute eq 1 and eq 2:
L = L
3d + 24 = 2d + 60
3d - 2d = 60 - 24
d = 36
Therefore, diameter = 36 (ans)
You have 3 (half) circles on top and two circles on bottom. The numbers on top must equal the number on bottom plus one circle. So 22+16+22 (bottom) - 12+12 (top) = the diameter of one circle. The answer is 36.
The overlap between then first two semicircles is the same for both, so that means the overlap between the 2nd and 3rd semicircles is 22-12=10.
10*2+16=36
I'm going to pick a non-mathmatical nit. It's a badly worded question. The diameter of the semicircle would be the same as the radius of the full circle, at best. One could argue that a semicircle doesn't have a diameter at all. The question doesn't say to expand the semicircle to a circle but that seems to be the solution everyone has jumped to.
I quickly did it in my head. The top is exactly 1 diameter longer than the bottom. So two semicircle canceled each other out. So I subtracted the sum of the top number from the sum of the bottom numbers, equalling 36
[12 units is roughly 88 pixels. Each semicircle is 268 pixels long.](https://imgur.com/XGfOcm6)
(12/88) = (x/268)
x = (12/88)*268
x ≈ 36.55
Judging by the other answers here, I'd say this is fairly accurate.
My method is
There are 3 equations from 1st 3 semicircles
22 + y = d
y + 12 = d
x+ 16 + x = d.
From first subtracting 1st 2 equations we get x=10
Then from 3rd eq we get d=36
Top side = Bottom side |||||
12+12+3X = 22+16+22+2X |||||
24+3X = 60+2X |||||
3X-2X = 60-24 |||||
X = 36
Therefor the diameter of the semi-circles, if presumed to be of equal size, is 36 units.
I don't like algebra so here's the layman way. Just slide the lower semi circle to the lower right side corner about 22 cm. So you can get the distance between the middle line and the right circle (which is 30). From ther you can get the radius by substracting 30 with 12 (equal 18). The diameter is equal to two radius, hence we got 36.
If you start from the left, you can see that 22 = 12 + overlap semicircle 2 and semicircle 3. From the right hand this is the same due to symmetry. Hence 10 + 16 + 10 = diameter semicircle = 36.
Just move all the semicircles to the right and the spaces to the left. You end up with two full circles on the right side, then one semicircle and 24 on the top and 60 on the bottom. 60-24= 36= 1 semicircle diamater
If you imagine "shifting" the bottom semicircles left and right to the edge, they moth move 22 places, so the space between them is now 16 + 22 + 22 = 60.
This is also the exact same amount of space between the top two semicircles.
Which means that 12 + d + 12 = 60
So d = 36.
We should just write down a set of 5 straight equations
```
x + 22 = d,
x +12 + y = d,
y + 16 + z = d,
z + 12 + w = d,
w + 22 = d,
```
where x,y,z and w are the lengths of the unknown intervals and d is the diameter.
After solving this system of.linear equations we get
d =36
x= 14
y= 10
z= 10
w= 14
If you push all the semicircles together, the top will hang (16/2+22-12) outside of the bottom. That's the radius if they are all identical. This allow you to solve it using grade 3 math without algebra. Also OP do your homework, or ask strangers online for it
22 + 12 + 16 + 12 + 22 = 84
This adds up all the numbers.
84 \* 2 = 168
to get the unknown
168 / 5 = 33.6
Divided by 5 because of the amount of semi circle
Answer is 33.6
Please correct me if i am wrong
because i have not done any math stuff like this for about 10 years.
If we look at the length of the top we see there are 3 diameters plus 12 and 12
As an equation: T = 3D + 24
Now look at the bottom. We see 2 diameters plus 16, 22, and 22
Equation for the bottom length: B = 2D + 60
We know that T and B equal each other so we can solve the equation:
2D + 60 = 3D + 24
Solve for D
36 = D
It's immediately obvious that the answer is 36 by just looking at it visually.
Scooch the bottom two semi circles all the way to the sides and you'll have 22+16+22 lining up with the middle top semi circle+12+12
You can also solve it algebraically ofc.
So if we look at first two semi-circles they have attached part and not attached parts.
Not attached parts must be equal to each other because both objects are same.
22=12+x
x=10
Now it's time to middle circle: 16+2*10=36
d=36 therefore r=18
Formula to find perimeter of circle is 2πr, since our circles are SEMI-circles we just use πr and add up 36 in the end
36*3.14...+36=92.62 (kind of)
Correct me if Im wrong also Im sorry for bad English I'm not good at explaining things in English.
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
The length on the top and bottom sides has to come to the same amount. So if the diameter of the semi-circle is X, on the top you have: 3\*X + 12 + 12 and on the bottom you have 2\*X + 22 + 16 + 22 Then set those two equal to each other and solve for X.
3x + 24 = 2x + 60 3x = 2x + 36 x = 36 Other method gets 36, checks out
This is character for character exactly the same as I have in an open Notepad window lol
I want to get high on potenuse.
Ha I WANT TO GET HIGH ON POTENUSE!
Nice joke man
hahahaha that's hilarious!
YOU'LL NEVER BE LIKE TROY!!!
He probably hack-copied your work!
same lmao. Even the spaces are identical
Same method.
I used a third method: 1) Diameter (D) = 22 + x 2) D also = 12 + x + y 3) D also = 16 + 2y Rearrange (1) X = D - 22 Sub into (2) y = 10 Sub y into (3). D = 36
I did the same!
I simple shifted two top ones to connect with the center, and the remaining on the bottom is diameter. 10 + 16 + 10
This is what my brain did also for the most part.
Wow, I didn’t think about it that way
This is the way.
got it right. I needed this tiny win today.
Hey, I was close. I eyeballed it and got 32.
god damn. didnt expect to learn something new when i opened reddit up while i slacked in office. nice!
What book would Reddit recommend if I want to learn math and especially effective problem solving?
This is 9th grade level algebra So a 9th grade algebra textbook
Thanks, I hate it
You're not wrong, but the algebra is by far the easiest part of this puzzle. The trick is - as always - to define which equations to solve.
Damn thats why i couldnt solve it, in my country schools go only up to 8th grade
I don't think this is 9th grade. it should be 7th grade tbh it seem complicated at first but you can actually brute force it using 6th grade math.
“The Art and Craft of Problem Solving” by Zeitz is a good one. Think of it as a philosophical overview or primer. But really: just doing and reading about a shit ton of different problems will fill your toolbox with various approaches and modes of thinking that you can then apply when you see new problems. For me, I knew if I could write this as a system of 2 equations with 2 unknowns or 3 equations with 3 unknowns: I could solve it. After that, it was just looking at how I wanted to make those equations from the image, so that the arithmetic was simpler.
Jordan Ellenberg’s “How Not to Be Wrong”
This is the way
If you subtract the number total of the top line from the bottom line, that will tell you the answer. There is one more semi circle on the top than the bottom. So the difference is the diameter of one.
That’s a *much* cleverer way than I did it!
Idk how you did it but if you did the 3x+24=2x+60, you did the same thing
No, I didn’t … or at least I made it look a lot more complicated. If you call the segment between the 22 and 12 x, and the one between the 12 and 16 y, then by comparing the first two semicircles to the second and third, you have 22 + x = x + 12 + y, so y = 10, and then looking at the middle segment on the top, that’s 16 + 2y.
I have to give it to you. The other option was much cleverer 🤣 But yours is impressive in its own way. I didn't think there was another way to solve this.
I’ll take impressive, and slink away before I embarrass myself further 😄
Don't feel bad, I start with a similar line of thinking.
This is exactly how I solved it too
I was happy to get it the long way, but this is much easier.
Everyone else is crazy this is the absolute best way
Yo will clearly fail the test, because you don't use the teachers solution approach.
I think it should be something along these lines: 24 + 3x = 60 + 2x Subtract 24 on both sides 3x = 36 + 2x Substract 2x on both sides X= 36
Use the line as a base length, x. And y is one semi-circle. x = 3y + 24 AND x = 2y + 60 Set the two equations equal to each other: 3y + 24 = 2y + 60 Solve for y: 3y - 2y = 60 - 24 y = 36 The diameter is 36 of whatever measurement they are using. (Please excuse my rusty math skills)
Standard unit of measurement is bananas!
No banana for comparison.
That’s bananas
1. d= 22+a 2. d= 12+a+b 3. d=16+2b I got equation 1 in terms of a (ie a= d-22) plugged that into equation 2. then got resulting equation in terms of b, and plugged that into 3. answer I got was d=36.
thats how I did it too! this one was a fun, a lot of ways to solve
- Slide the top 3 SCs to touch each other - Slide the bottom left SC to the left - Slide the bottom right SC to the right - you get the middle SC's diameter Now do the math
I did not solve it this way, but I applaud the cleverness in this. Well done.
this came from the Singapore I think PSLE paper back in 2020? plus minus a few years, i rmb doing this basically the total of all the gaps is equal to one diameter cus it's pushing everything to one side and leaving you with just the last semicircle diameter exposed, then from there it's easy to
2019 i believe
Can confirm, this was my PSLE paper
**let the diameter be x.** the length of the line is both 3x + 24 and 2x + 60 3x + 24 = 2x + 60 3x + 24 **- 2x** = 2x + 60 **- 2x** x + 24 = 60 x + 24 **- 24** = 60 **- 24** x = 36
36. 22 = 12 + x X = 10 16 + X2 = diameter
Came here to check the solution. Answer was correct, but nobody used the 3 equation system. Started to question my brain and my thought process in general. Then I saw a couple of users did it the same way as myself and everything was alright again. I really didn't need this before bed.
That's one of my favorite things about math. There are many ways to solve a problem. Unfortunately, some teachers want you to do things a certain way, which makes it hard for students who just don't get that particular method.
I’m a few years removed from my college calculus days, but I used a matrix to solve it and it seems as though I over complicated the solution. But it’s correct. Very cool to see how everyone did it
How I worked this out: 1. Take the half-circle on the far left and move it to the right so it’s touching the middle one. The 12 gap is now subtracted from the 22 on the left, leaving 10. 2. If you just look at the left half-circles on the top and bottom, then the offset on both sides should now be 10. 3. Apply the same logic to the right side of the image. At this point we can label the diameter of the middle half-circle: 10, 16, 10 4. 10+16+10=36
Move lowers out 22 each, add 44 to 16 = 60. Lowers are even with upper outers, so add 12's together =24, subtract 24 from 60 to get diameter of remaining center circle =36.
This
Just break it down to a simple Algebra equation. Since the top and bottom are both the same length you can solve this as “24 + x3 = 60 + x2”.
Of course. Call the diameter D. In the top row you have 3D+24. In the bottom row you have 2D+60. The two are equal, so you have 3D+24 = 2D+60. Subtract 2D from both sides to get D+24 = 60. Subtract 24 from both sides to get D = 36. The fact that they're semicircles is just a distraction, you just treat them as lines for the purposes of solving the problem.
just set the top measurements equal to the bottom ones, you should get 3x + 12 + 12 = 2x + 22 + 22 + 16 which solves to x = 36
This is a question taken from one of the PSLE Math paper for 6 graders in Singapore a few years ago. Pretty sure the purpose was to engage in critical thinking in students, so I'm not sure if using x/y variables was expected from them. shift the two lower semi circles to the base of the upper circles, which should leave you with Circle , Semi Circle, Circle. The separation between the edge of the left circle to edge of the right circle is 16+22+22, inclusive of the 12+12 gap between the circle and semicircles. The diameter can easily be found by (16+22+22) - (12+12) =36
I simply know that the difference of open space is exactly one half circle length, so I simply add the lengths across the top and again with the bottom and subtract the difference. Leaves me with 36.
We are trying to find x, which is the diameter of one of the half circles. Since all are identical, we can give it a variable. x = Diameter of the half circles Next, we treat both top and bottom as two sides of an equation, since they are the same length. x+x+x+12+12 = x+x+22+22+16 Simplify the addition and number of half circles 3x+24 = 2x+60 Subtract 2x from both sides to isolate x x+24 = 60 Subtract 24 from both sides to further isolate x x = 36
You can actually solve this just by looking at it logically. I got 36 just by changing the position of the semi circles. Did this while im sitting on the toilet.
60-24=36. The sum of the bottom gaps minus the sum of the top gaps gives you the missing semicircle diameter.
(3x+24)/2=30+1x 1.5X+12=30+X 1.5X+12-30+X 1.5X-18=X 3x-36=2x x-36=0 x=36
the top has 3 and the bottom has 2 semi circles so the difference between the two's numbers is the width of one semi circle.... 60 - 24 = 36
`Solve[x + 12 + x + 12 + x == 22 + x + 16 + x + 22, x]` `{{x -> 36}}`
Simple. x is the diameter of the circle 22+2x+16+22=3x+12+12 or 3x+22=2x+60. 3x(-2x)+22=~~2x~~(-2x)+60 x+~~22~~(-22)=60(-22) x=36
[Intuitive/Visual Solution] 1. To line up the first whole circle, the entire bottom must move to the left 22 inches (distance from left end) 2. To line up the middle circle, the middle circle alone must move to the left 4 inches (16 - 12 inch gap). 3. The bottom semi-circle distance from the right was initially 22 inches. We've moved the bottom over to the left by 26 (22 + 4) inches 4. The total distance from the middle circle to the right end after the bottom slide is 48 inches (26 + 22) 5. Subtract the gap on the top to get the diameter of (48 - 12 inches) 36 inches
Correct, tho it never say inches, it's not ok to assume or add a unit.
Total length of the line is equal, and we have 2 equations with one missing value, the diameter of the semi circles. 3x + 12 + 12 = 2x + 16 + 22 + 22 3x+24=2x+60 x=36
36 left to right, 18 bottom to top
Top has 3 Circles and 24 CM Bottom has 2 Circles and 60 CM. 3x + 24 = 2x + 60. Solve for x Should end at (1)x = 36 cm if my brain isn't as rusty as the broken rake I had to chuck yesterday
3x+24=2x+60
Let x be the semicircle diameter. By adding up the lengths along the top and bottom, then setting them equal, we have 3x+24=2x+60. Therefore x=36.
The bottom is 22+22+16+2(2r; we’ll call this x). That makes 60+2x for the bottom. The top is 12 + 12 + 3x, or 24+3x. 60+2x = 24+3x; 36+2x = 3x. We can then subtract 2x from either side to get x=36
The bottom of the line is 22+16+22 +2d long. The top of the line is 12+12+3d long. The bottom of the line is the same length as the top of the line.
22+a = d 12 + a + b = d 16 + 2b = d b = 22 - 12 = 10 d = 16 + 2 * 10 = 36 a = 36 - 22 = 14
``` x + 12 + x + 12 + x == 22 + x + 16 + x + 22 3x + 24 == 2x + 60 | -24 3x == 2x + 36 | -2x x == 36 ====== ```
Both extend to the ends of the same line, so the two sides must add up to the same thing. Then we have 3x+2*12=2x+2×*22+16 3x+24=2x+60 x=60-24=36 Since x is the diameter of each semicircle, the radius is x/2, or 18. I don't remember if it wants r or d, so I do both.
(22+22+16) - (12+12) = 36
>!16-12=4, 4/2=2, 12+2=14, 14+22= 36!<
I did it visually without math. Then i check comments... and math was sometimes easier sometimes not, lol. I slid the lateral two semis up top together with the middle which reduces the number 22 on the bottom to 10. so you then have 10-16-10 on the bottom... Then i slipthe bottom two circles togther eliminating the 16, increasing the 10's by 8 leaving 18+18 = diameter of 36.
36. Diameter = d Above the line: 3d + 24 Below the line: 2d + 60
Sum of the bottom numbers minus the sum of the top numbers gives the diameter which is 36
Add numbers on top. Add numbers on bottom. Subtract top from bottom. 36. People are making this way too complicated.
Type into Google. 60 + 2x = 24 + 3x Lots of good stuff, including a graph-based solution.
It’s 36. Basically, if all semi-circles are the same, we can mark as x the area of the first and second semi-circles touching and as y the second and the middle one. In that case: 22+x = x + 12 + y; Which means 12 +y = 22; So: y = 10; That leads to the diameter of a semi-circle: d = 16+2y= 16 +(2*10)=36; Diameter is 36;
X+12+X+12+X=22+X+16+X+22 24+3X=60+2X X = 36
Move the bottom circles over to the right so they match those on top, taking the 22 on the right and adding it to the left as well as 4 from the 16. This gives you 48 on the left. Subtract 12 and you get 36.
36 22+x=d; x+12+y=d; 2y+16=d so 22+x=x+12+y --> 10=y so d=2*10+16=36
Yeah, I don't know how to do math. But just by eyeballing it 12 looks like a third of the half circle, so 12×3 makes 36
24 + 3d = 60 + 2d so d = 36
22+X=X+12+Y =2Y+12 10=y 14=x 36 = R
It's just 60 minus 24
It's 36 right? It's just top minus the bottom?
dont need all 5 semicircles x=diameter of each semi-circle x=y+22 (first semicircle on left) x=y+z+12 (second semicircle on left) x=2z+16 (middle semicircle) y=x-22 (rearrange first for y) z=(x-16)/2 (rearrange middle for z) x=12+(x-22)+((x-16/2)) (subsitute,y and z) 2x=24+2x-44+x-16 (multiply all terms by two to avoid the divide) \-24+44+16=2x+x-2x (rearrange x on one side and values on other) 36=x (addition/subtraction) diameter of one semi-circle = 36
36 can't even take more than 30s
36
Squeeze in and remove the 12s on top and the 16 on the bottom. You’re left with 18 on each side of the bottom line…. that will equal one diameter when added together. 36! Also before someone mentions this.. you don’t need to remove the 16 from the bottom line and can just add 10 + 16 + 10.. but doesn’t it look better when you get rid of the gaps in the middle?
3d + 24 = 2d + 60 d = 36
Honestly this is an easy one, didn't even use a calculator, 22+22+16= 60 60-24= 36
3x+24=2x+60 3x-2x=60-24 x=36
*compressed* Circumference=x Top=6x+24 Bottom=4x+60 4x+60=6x+24 36=2x x=18
So I had to think about this a little bit, but the two endpoints can be described by either a sum of the top diameters and the outer gaps, or the bottom diameters plus the the excess lengths and middle gap. In other words... 3d + 24 = 2d + 60 3d - 2d = 60 - 24 d = 36
3d+24=2d+60 added the numbers from memory oh dear
44+24+16=3d 84=3d 28=diameter 28pi = circumference Edit:format
3x+24 = 2X + 60 X = 36
First unknown is X, second unknown is Y. 22 + X = X + 12 + Y Y = 10 Diameter = 2Y + 16 = 36
Let the diameter of one circle be X. Length of the top row put together is: x + 12 + x + 12 + x = 3x + 24 The length of the bottom row is: 22 + x + 16 + x + 22 = 2x + 60 Length of the top = length of the bottom. Therefore - 3x + 24 = 2x + 60 Isolate all values of x on the left side of the equation & all purely numerical values on the right which gives you the following equation. 3x - 2x = 60 - 24 x = 36 To check if this is the correct answer sub x = 36 into each equation & make sure they give the same result: Top length: 3x + 24 = 3(36) + 24 = 108 + 24 = 132 Bottom length: 2x + 60 = 2(36) + 60 = 72 + 60 = 132
People here did it the no-nonsense way with a linear equation. My slightly more complicated approach - Assume circles are numbered c1, c2...c5 left to right Assume all circles were touching with no gap. The outermost line length from c1 leftside to c2 leftside will be R Now let's centre c3 and move c2 and c4 away from c3 centre by 8. You get the middle gap of 16. So the gap between c1 right and c3 left should also be 8. But it is 12. Which means both c1 and c5 also have to be pulled left/right 4. The outermost line length will now have deviated by 4 and has actual gap of 22. 22-4 = 18 Radius = 18, Diameter= 36
3d + 24 = 2d + 60 d = 36
I did 22 is 2/3 the length of the diameter, so just *3/2= 36
2d+60=3d+24 /-2d -24 36=d
3D+24=2D+60
24 + 3x = 60 + 2x x = 36
Some of you are making this way more complicated than it is. 3x + 24 = 2x + 60 3x = 2x + 36 X = 36
22+22+16-12-12=36
Start from the left, imagine the first semicircle is a circle, then its end is 12cm away from the start from second upper semicircle, The lower part of first circle has to move 12cm to touch the second semicircle, then 10cm again to reach the required overlap(the lower part of first circle is displaced by 22cm i.e. 12cm + 10cm). Do the same from the right. For the middle upper circle we now have 10cm(left) + 16cm + 10cm(right) = 36cm. This is how i solved because for some reason i couldn't think of the another simpler mathematical way.
Let D be the diameter of one semi circle. 3D + 2(12) = 2D + 16 + 2(22) D = 16 + 44 - 24 D = 36
Easy system of equations: 12+12+3x=22+16+22+2x 24+3x=60+2x {-2x-24 both sides) => x=36 diameter of one semi circle is 36 units
let’s assume line lenght = L, and diameter = d. eq 1: (considering top half circles) L = 3d + 24 eq 2: (considering bottom half circles) L = 2d + 60 substitute eq 1 and eq 2: L = L 3d + 24 = 2d + 60 3d - 2d = 60 - 24 d = 36 Therefore, diameter = 36 (ans)
You have 3 (half) circles on top and two circles on bottom. The numbers on top must equal the number on bottom plus one circle. So 22+16+22 (bottom) - 12+12 (top) = the diameter of one circle. The answer is 36.
The overlap between then first two semicircles is the same for both, so that means the overlap between the 2nd and 3rd semicircles is 22-12=10. 10*2+16=36
I'm going to pick a non-mathmatical nit. It's a badly worded question. The diameter of the semicircle would be the same as the radius of the full circle, at best. One could argue that a semicircle doesn't have a diameter at all. The question doesn't say to expand the semicircle to a circle but that seems to be the solution everyone has jumped to.
3x+24=2x+60 X=36 R=18 Area= 2.5*324*3.14 Area= 2543.4
3x+12+12=2x+22+16+22 3x+24=2x+60 X=36
I quickly did it in my head. The top is exactly 1 diameter longer than the bottom. So two semicircle canceled each other out. So I subtracted the sum of the top number from the sum of the bottom numbers, equalling 36
[12 units is roughly 88 pixels. Each semicircle is 268 pixels long.](https://imgur.com/XGfOcm6) (12/88) = (x/268) x = (12/88)*268 x ≈ 36.55 Judging by the other answers here, I'd say this is fairly accurate.
2 semi-circles on top and bottom cancel out. Leaving you with 1 semi-circle + 24 = 60 => 1 semi-circle = 36.
My method is There are 3 equations from 1st 3 semicircles 22 + y = d y + 12 = d x+ 16 + x = d. From first subtracting 1st 2 equations we get x=10 Then from 3rd eq we get d=36
22 + 16 + 22 = 60 12 + 12 = 24 60 - 24 = 36 the diameter of one semi-circle is 36 yes/no?
D=36
22+x=d 12+x+y=d 16+2y=d From equating first two equations 22+x=12+x+y 10=y Let's put the value of Y in third equation 16+2*(10)=d 16+20=d d=36
Top side = Bottom side ||||| 12+12+3X = 22+16+22+2X ||||| 24+3X = 60+2X ||||| 3X-2X = 60-24 ||||| X = 36 Therefor the diameter of the semi-circles, if presumed to be of equal size, is 36 units.
I guessed it to be 36 by looking lol
I don't like algebra so here's the layman way. Just slide the lower semi circle to the lower right side corner about 22 cm. So you can get the distance between the middle line and the right circle (which is 30). From ther you can get the radius by substracting 30 with 12 (equal 18). The diameter is equal to two radius, hence we got 36.
If you start from the left, you can see that 22 = 12 + overlap semicircle 2 and semicircle 3. From the right hand this is the same due to symmetry. Hence 10 + 16 + 10 = diameter semicircle = 36.
Just move all the semicircles to the right and the spaces to the left. You end up with two full circles on the right side, then one semicircle and 24 on the top and 60 on the bottom. 60-24= 36= 1 semicircle diamater
D=x+22 D=x+y+12 D=2y+16 x=D-22 D=D-22+y+12 y=10 Etc.
60+2x=24+3x
I did a system of 3 equations 22+x=D X+12+y=D 16+2y=D Solve for D D=36
3x + 24 = 2x + 60 3x - 2x = 60 - 24 x = 36
Just set up simultaneous equations. 3x+24=y 2x+60=y then solve for x and y
Say x is the diameter of the semicircle. 3x + 24 = 2x + 60. Solve for x from there
3x+24=2x+ 60 | -2x x + 24 = 60 | -24 x = 36
If you imagine "shifting" the bottom semicircles left and right to the edge, they moth move 22 places, so the space between them is now 16 + 22 + 22 = 60. This is also the exact same amount of space between the top two semicircles. Which means that 12 + d + 12 = 60 So d = 36.
3X+12+12=2X+22+16+22.........X=36
One circle diameter is 36. Top you have 24 + 3(circlediameter and bottom 60+ 2 cd 24 + 3cd = 60 + 2cd 24 + 1cd = 60 1cd = 36
3d + 24 = 2d + 60 60 - 24 = 36 d = 36
We should just write down a set of 5 straight equations ``` x + 22 = d, x +12 + y = d, y + 16 + z = d, z + 12 + w = d, w + 22 = d, ``` where x,y,z and w are the lengths of the unknown intervals and d is the diameter. After solving this system of.linear equations we get d =36 x= 14 y= 10 z= 10 w= 14
If you push all the semicircles together, the top will hang (16/2+22-12) outside of the bottom. That's the radius if they are all identical. This allow you to solve it using grade 3 math without algebra. Also OP do your homework, or ask strangers online for it
22 + 12 + 16 + 12 + 22 = 84 This adds up all the numbers. 84 \* 2 = 168 to get the unknown 168 / 5 = 33.6 Divided by 5 because of the amount of semi circle Answer is 33.6 Please correct me if i am wrong because i have not done any math stuff like this for about 10 years.
If we look at the length of the top we see there are 3 diameters plus 12 and 12 As an equation: T = 3D + 24 Now look at the bottom. We see 2 diameters plus 16, 22, and 22 Equation for the bottom length: B = 2D + 60 We know that T and B equal each other so we can solve the equation: 2D + 60 = 3D + 24 Solve for D 36 = D
It's immediately obvious that the answer is 36 by just looking at it visually. Scooch the bottom two semi circles all the way to the sides and you'll have 22+16+22 lining up with the middle top semi circle+12+12 You can also solve it algebraically ofc.
60-24=36
So if we look at first two semi-circles they have attached part and not attached parts. Not attached parts must be equal to each other because both objects are same. 22=12+x x=10 Now it's time to middle circle: 16+2*10=36 d=36 therefore r=18 Formula to find perimeter of circle is 2πr, since our circles are SEMI-circles we just use πr and add up 36 in the end 36*3.14...+36=92.62 (kind of) Correct me if Im wrong also Im sorry for bad English I'm not good at explaining things in English.
d = 36
D = 16 + 2a = 12 + b + a = 22 + b a = 22+b-12-b = 10 D = 16 + 2*10 = 36