if it were drawn to scale, simple cheat: just take your ruler and measure on the picture itself, then cross-multiply the 18 and X by the real measures on the picture, basically ignore the whole point of using the circles.
@@anotherpromotor Porque es una competencia, duh Y también, se evalúa el procedimiento con el que llegaste al resultado, si solo tienes el resultado y no demuestras ni cómo lo sacaste (no puedes decir que lo hiciste midiendo con una regla jajaja) no tienes los puntos
I was the one that sent to you. Thank you so much for solving it. I had seen that somewhere in a video and figured out it was 25 cm but I wasn't sure because the guy in the video gave the wrong answer. I'm glad I did it correctly. Exciting video.
I love that it's not drawn to scale because for my work as a draftsman, i work all the time with hand drawn sketches that are never to scale, but with correct measurements. This question is exactly the kind of thing I would encounter at my job!
Fooled me! I figured that the diagonal of the square Y would be 5 + 8 + √50 + √128, meaning X would be Y² - 18², but the circle center are not on the diagonal line because the figure isn't a square.
I remember something I studied a long time ago, and I'm not sure how true this is, but bear with me for a second... 1. What if we divide the square/rectangle into two triangles, based on the angles, we will have 2 right triangles (a 90° angle each) sharing the same Hypotenuse. 2. Now if you look at the circles you may notice that the combined diameters of the circles are equal to the Hypotenuse mentioned earlier...that's 26cm. 3. Now that we know the length of the Hypotenuse we should use it as well as the 18cm rectangle side (also triangle base)...we'll work with one of the triangles: Chard²=base²+x² which means x²=Hypotenuse² - base² x²= 26² - 18² x≈ 18.76cm
Grade 2 mafs: You always re-draw and re-write the problem. NEVER work with default schematics and text Aka: never trust "The last Guy"s work, the one that couldn't solve it. If he could solve it, he wouldn't ask you to solve it 👍
good advice, but here you can't redraw it correctly until you start solving it, but if you're on the right path you don't need to redraw it anymore, it's straightforward
Scale is off, but definitely possible. The radii of the two circles are 5 and 8, so their centers are 13cm apart in a straight line. This also means that their centers are a combined 13cm from the left and right sides of the rectangle (5 on left, 8 on right) and top and bottom sides of the rectangle (5 from top, 8 from bottom). This means that the distance left to right (∆x) of their centers is 18-13 = 5. 5-12-13 is a Pythagorean triple, so that means that the top to bottom distance (∆y) of their centers is 12, so x = 12+13 = 25cm.
Solution: Move the sides inward, until they 'hit' the first centerpoint on their path. You end up with a rectangle that is 5 cm wide (18 - 10/2 - 16/2 = 18 - 5 - 8 = 5) and x - 13 cm high (x - 10/2 - 16/2 = x - 5 - 8 = x - 13). Its diagonal is a combination of both circle radii and therefore 13 cm long (10/2 + 16/2 = 5 + 8 = 13) As such, we can use pythagoras to get: 5² + (x - 13)² = 13² 5² + x² - 26x + 13² = 13² |-13² x² - 26x + 25 = 0 (x - 1)(x - 25) = 0 So x is either 1 - which can be discarded, as it is not big enough - or it is 25. Therefore x = 25 cm
The practical application of this is in three dimensions. Imagine the image is a frontal view of a box with a slope and with two pipes in. It, also cut at an angle. The actual dimensions of the box are the rectangle etc but from your pov it looks like a square. @andymath, you should draw this.
Looks pretty simple at first glance. Circle centres to left/right edges are 5 and 8 respectively, same to top/bottom. Distance between centers is 5+8=13. Right triangle between two centres: bottom edge is 18-(5+8)=5. Hypotenuse is 13. Left edge is √(13²-5²) = √(169-25) = √(144) = 12 So x = 12+5+8 = 25.
My first thought when I saw the problem on the screen was "it's not drawn to scale. This would be confusing if it were given during an exam." Great solution, nonetheless.
The line joining the centers of the two circles makes an angle Ɵ with the vertical. Then by construction, 13 (1 + sin Ɵ) = 18 , so sin Ɵ = 5/13, so cos Ɵ = 12/13 Then X = 13 * (1 + cos Ɵ) = 25
To be completely correct, there is another solution from sqrt 144, which is -12. Of course this doesn‘t make any sense. But when I remember back in school, it was considered incorrect when we forgot tje negtive solution.
It's intentionally not drawn to scale as a trap for the unwary. And the plural of radius is radii, because _radius_ is a second-declension noun in Latin.
It was probably compacted in the book to conserve space. The extra space required represent 2-3 lines of text. Tho I guess they could have rotate the figure 90°...
I love to solve this kind of stuff when I was elementary and highschool but since I do programming as my academic and gaming when I resting, I have no more time for this.
Pretty easy one, I could solve the entire thing in my head within a couple of seconds. I just imagined the radius of each circle building a 5x5 and 8x8 square eliminating the big square with the ugly "void spots" which were preventing me to use pyth.t., got. a neat square in the middle, plugged in the remaining values and done.
I feel like you should explain where you got the 5 at 2:50. I get how you got there but people might forget and not be able to replicate it themselves. I got there by finding that the 5cm circle had 2cm overlap to the left of the 8cm circle (18-16). The 5cm circle has 3cm to the center of it from that point. The 3cm are then subtracted from the 8cm circle's radius => 5cm horizontaly between the 2 centers.
This one is an easy task. Call M₁ center of big circle with radius r₁ = 8 and M₂ center of small circle with radius r₂ = 5. The distance M₁M₂ is then r₁ + r₂ = 13. Since the base of the rectangle is 18, the horizontal distance of M₁ and M₂ is 18 - 8 - 5 = 5. Therefore, the vertical distance z of M₁ and M₂ is Pythagoras with z = √(13² - 5²) = √(169 - 25) = √144 = 12. Then we have height x = r₁ + z + r₂ = 8 + 12 + 5 = 25. Funny sketch BTW, 8cm < 5cm ;)
You were still able to solve it even with the scale being wrong. I was trying to solve it and then realized something was very off about this picture. 😆
i get a different answer I created a cube whose side length is equal to half the diameter of the large circle. Then I used the Pythagorean theorem to find the diameter and then combined it with the semicircle. I made a square whose side length was half the diameter of the small circle, then I found its hypotenuse and merged it with the radius of the small circle. I added together the longest lines to show me the major square hypotenuse Then I used the Pythagorean theorem on the large square because I have the hypotenuse and the length of one of the two sides. Then I moved the side whose value is 18 to the left side so that it is “X” alone. Then I simplified the calculation, then eliminated the square with X, and then the following result appeared. 33,525....
18/sqrt(2)+8 just from looking at the thumbnail. Square from upper left corner to center of big circle has diagonal 10+16/2, side that/sqrt(2). It's 16/2 above the floor.
I might be doing this the wrong way, but is it possible to transform the circles into squares or rectangles and suntract them from the box? Or is that too inaccurate because pi?
The graphic is misleading. But the centers of the circles are 5 apart (18-8-5=5). The diagonale between the two circles‘ centers is 5+8=13. This is the hypothenuse of of a right triangle formed by the horizontal distance (5) and the vertical distance of the centers. This leads to 13^2-5^2=169-25=144. thus the vertical difference between is 12. Thus the height is 12+8+5=25.
I got ~25.71 units. using the 8cm radii, it creates a square on the bottom right of the rectangle, with hypotenuse of sqrt(128), then did the same with the 5 in top left and had sqrt(50), so I did the hypotenuse of the entire rectangle is equal to sqrt(50)+sqrt(128)+5+8. Then I did 18² + b² is equal to the hypotenuse squared, then square rooted it and got 25.71
When I was in school, they sometimes handed me a test with this picture of a square and it said "the area of this square is 5square kilometers..." I was like "It ain't" and then I measured it and I told them the correct area of the square. And then they failed me. The whole education system failed me...
couldnt you line up the diameters of the circle, making 2 triangles, now you have the the base and the hypotenuse of a right angled triangle, thus you find the height, then since its a rectangle, the height found will be equal to x?
That only works if the line between their centers is in line with the line(s) between their centers and the corners, and if the radii of the circles goes all the way to the corners. In this case neither is true. But you're looking in the right direction. What you do is, you move the corners to where the centers of the circles are. You subtract 5 cm from the left side and top, to reflect where the center of the 10cm circle is, and 8cm from the right side and bottom, to reflect where the center of the 16cm circle is, and then line up the radii with the "new corners", since those are in line and do reach to the centers of the triangles. From that you get a 13cm hypotenuse (8+5) and a 5cm base (18-8-5), which gives you 12cm for the vertical leg. Then you add back the 5cm and 8cm you subtracted from the top and bottom and get x = 25cm.
The first one is a square and rectangle. So a rectangle will have a different area unless the scale is change to equal the same amount of space taken up by the square. I refuse to solve it.
sure. the centers are 13 cm apart and they would lie on the diagonal of the square. so the side is 5 + 8 + 13×(sq.root of 2)/2 cm (upd: fixed a mistake)
Fooled me good on this one--I first thought you were going to get the values between the edges of the circles and corners of the rectangle, and then use Pythagorean theorem on the edges of the rectangle and the diagonal through the rectangle. But as soon as I saw you draw it to scale, I realized my idea wouldn't work. Moral of the story is never trust a drawing that is not to scale. :-)
I'd like this more if the problem was a square and the 18cm wasn't provided. You would then get 2x^2 = A^2, where A = the hypotenuse A = 13 + (√50 - 5) + (√128 - 8) = √50 + √128 This then simplifies to 2x^2 = (√50 + √128)^2 (√50 + √128)(√50 + √128) 50+2(√50 * √128) + 128 50+2(√6400)+128 50+2(80)+128 50+160+128 338 So 2X^2 = 338 X^2 = 169 X = √169 = 13
I got 25 but i just couldn't wrap my head around the fact that the base was 5cm because that would mean it was inside the bigger circle. Never trust images man
Let's have a go before I watch your video .. If you turn the two circles so that the 19cm and the 16cm lines are horizontal you can work out the distance between the centre on the circles ''along the bottom line'' 18 - (5 + 8) = 5 So now using pythagoras you can work out what to add to (5 + 8) along the X line, let's call it A 5^2 + A^2 = (5 + 8)^2 5^2 + A^2 = 13^2 25 + A^2 = 169 A^2 = 144 so A = 12 and 12 + 5 + 8 = X 25CM
To be fair you don't need to have a correct drawing of this problem since you have all you need to solve for x. The correct representation may help, but it's not really necessary.
This puzzle is really engaging and easy to follow! I love how it challenges our dimensional reasoning skills. I remember getting into similar problems and found that resources like SolutionInn helped me think through the tricky geometry topics. It’s interesting how visual challenges can enhance our understanding of shapes and their relationships. What other puzzles do people enjoy that test these concepts?
Okay I’m kind of going insane with my answer cuz it’s wrong but I don’t know why… Here’s what I did: I found the diagonal across the rectangle by making a square with the radius of each circle, and finding the diagonal from the center of each circle to their respective corner. I added up that diagonal and the radius of the circle which SHOULD be the distance from the corner to the edge of the circle, and then I added the same value but with the radius subbed out for the other circle. My current value for the diagonal is 13+13*sqrt(2) We know that we can form a right triangle with the corner of the rectangle, and we have one side length as 18 cm and the hypotenuse being 13+13sqrt(2) so it SHOULD be sqrt((13+13sqrt(2))^2-18^2) but that gets me 25.71 for some reason HELP
Nvm. I figured it out. It's because the centers of both circles do not intersect with the diagonal (when it is drawn to scale). So you are not actually measuring the big diagonal line when you calculate small diagonal lines. Go to 1:21 in the video to see what I mean
The thing is that the line connecting the two centers does not have the same gradient as the line connecting the centers to the corners. When this is not true, you can't add and equate multiple line segments. They must all be on the same line.
I learned, if omniting the unit, and mentioning/ putting it back at the result, thy syntax would be square brackets. So, in this case, [cm]. Anyone with me?
I'm assuming because both circles are tangent to each other so since they make contact there needs to be a single point where they touch, and since a circles radius connects to every point on the circle you are allowed to assume they share the same line.
i first solved for the longer hypotenuse across the whole rectangle by solving the smaller hypotenuses in each corner. And then i ended up with x = sqrt(183+338sqrt(2))cm which roughly equals to 25.71cm. I dont know how i got the extra 0.71, maybe it came from dealing with the sqrt(2)s but if someone wanna check my math on it. Feel Free! 8^2 + 8^2 = C^2 C=8sqrt(2) 5^2 + 5^2 = c^2 c=5sqrt(2) x^2 + 18^2 = (5sqrt(2) + 5 + 8 + 8sqrt(2))^2 x^2 + 18^2 = (13sqrt(2) + 13)^2 x^2 + 18^2 = (2*13^2) + (2*13*13sqrt(2)) + (13^2) x^2 + 324 = 338 + 338sqrt(2) + 169 x^2 = 183 + 338sqrt(2) x = sqrt(183 + 338sqrt(2)) x =~25.71cm
Diagonal is not 26 cm The 2 diameters add upto 26 cm but they dont touch the corners of this quadrilateral, hence not a diagonal "If" x=25 cm then diagonal shall be 30.8 cm by applying Pythagoras
Radiuses has only become accepted due to the overwhelming ignorance of people who don't know that the plural is, and should only have ever been, radii.
Not drawn to scale? Well I'm not solving it then...
if it were drawn to scale, simple cheat: just take your ruler and measure on the picture itself, then cross-multiply the 18 and X by the real measures on the picture, basically ignore the whole point of using the circles.
@@greenheroeswhy not let them do that? It's easier to solve it
You can only know if it's drawn to scale when you're solving it...
@@anotherpromotor Porque es una competencia, duh
Y también, se evalúa el procedimiento con el que llegaste al resultado, si solo tienes el resultado y no demuestras ni cómo lo sacaste (no puedes decir que lo hiciste midiendo con una regla jajaja) no tienes los puntos
@@foldinghomealone Not drawn to scale is fine. "Lets make it a square to mess with them" is cruel.
bro getting wild with the title of the video
no why do you have do that nooo!!!!
bro is cooked
It's remarkable that so many years later people still remember the video.
Generational trauma
While its a well known reference has anyone actually gone out of there way to actually watch it??
At first I was like "wait, if it's inside a SQUARE, shouldn't X be equal to 18cm??"
it has to be a square it seems to me.
That was my first thought 😂
did the youtube title change or did it explicitly say "rectangle" from the start
@@drenzine a square is still a rectangle but he said like thrice that it wasn’t to scale
@@drenzine why draw a square if you mean rectangle? i realize that all squares are a subset of rectangles, but still.
I was the one that sent to you. Thank you so much for solving it.
I had seen that somewhere in a video and figured out it was 25 cm but I wasn't sure because the guy in the video gave the wrong answer. I'm glad I did it correctly.
Exciting video.
Me with a ruler: well sh*t
JAJAJAJA esa es la trampita
I love that it's not drawn to scale because for my work as a draftsman, i work all the time with hand drawn sketches that are never to scale, but with correct measurements.
This question is exactly the kind of thing I would encounter at my job!
Let’s goo, more random geometry questions that I have no idea how to solve if you didn’t make a video about them. Thank you!
Fooled me! I figured that the diagonal of the square Y would be 5 + 8 + √50 + √128, meaning X would be Y² - 18², but the circle center are not on the diagonal line because the figure isn't a square.
I did the same thung
Same! I would have failed this test 😂
Same sir!
Yes it happened with me as well and the answer was 25.70
Yes my answer is 25.70
I remember something I studied a long time ago, and I'm not sure how true this is, but bear with me for a second...
1. What if we divide the square/rectangle into two triangles, based on the angles, we will have 2 right triangles (a 90° angle each) sharing the same Hypotenuse.
2. Now if you look at the circles you may notice that the combined diameters of the circles are equal to the Hypotenuse mentioned earlier...that's 26cm.
3. Now that we know the length of the Hypotenuse we should use it as well as the 18cm rectangle side (also triangle base)...we'll work with one of the triangles:
Chard²=base²+x² which means x²=Hypotenuse² - base²
x²= 26² - 18²
x≈ 18.76cm
Grade 2 mafs: You always re-draw and re-write the problem. NEVER work with default schematics and text
Aka: never trust "The last Guy"s work, the one that couldn't solve it. If he could solve it, he wouldn't ask you to solve it 👍
good advice, but here you can't redraw it correctly until you start solving it, but if you're on the right path you don't need to redraw it anymore, it's straightforward
No bullshit, just to the point and a good demonstration. Thanks!
It’s very well presented and with an elegant solution!
Woot! The little remembered 5-12-13 right triangle! 3:27
I love that every problem circles that are touching each other inside of a different shape boils down to
" Find the correct set of triangles"
HOW EXCITING 🗣🗣🗣❤🔥❤🔥🗣
Scale is off, but definitely possible.
The radii of the two circles are 5 and 8, so their centers are 13cm apart in a straight line. This also means that their centers are a combined 13cm from the left and right sides of the rectangle (5 on left, 8 on right) and top and bottom sides of the rectangle (5 from top, 8 from bottom). This means that the distance left to right (∆x) of their centers is 18-13 = 5. 5-12-13 is a Pythagorean triple, so that means that the top to bottom distance (∆y) of their centers is 12, so x = 12+13 = 25cm.
i love this way of solving it, thanks for sharing
Solution:
Move the sides inward, until they 'hit' the first centerpoint on their path.
You end up with a rectangle that is 5 cm wide (18 - 10/2 - 16/2 = 18 - 5 - 8 = 5) and x - 13 cm high (x - 10/2 - 16/2 = x - 5 - 8 = x - 13).
Its diagonal is a combination of both circle radii and therefore 13 cm long (10/2 + 16/2 = 5 + 8 = 13)
As such, we can use pythagoras to get:
5² + (x - 13)² = 13²
5² + x² - 26x + 13² = 13² |-13²
x² - 26x + 25 = 0
(x - 1)(x - 25) = 0
So x is either 1 - which can be discarded, as it is not big enough - or it is 25.
Therefore x = 25 cm
I saw the thumbnail and decided to try it myself before watching the video. I actually did get it! Love your explanation too 👍
The practical application of this is in three dimensions. Imagine the image is a frontal view of a box with a slope and with two pipes in. It, also cut at an angle. The actual dimensions of the box are the rectangle etc but from your pov it looks like a square. @andymath, you should draw this.
Ok, but the circles will not look like circles if you cut at an angle
Looks pretty simple at first glance.
Circle centres to left/right edges are 5 and 8 respectively, same to top/bottom.
Distance between centers is 5+8=13.
Right triangle between two centres: bottom edge is 18-(5+8)=5. Hypotenuse is 13. Left edge is √(13²-5²) = √(169-25) = √(144) = 12
So x = 12+5+8 = 25.
I would have gotten reduced marks because I forgot my units lol
This is the type of math I like, I just wish it was drawn to scale
For the first time, I could solve it without watching the video, how exciting!
My first thought when I saw the problem on the screen was "it's not drawn to scale. This would be confusing if it were given during an exam."
Great solution, nonetheless.
The line joining the centers of the two circles makes an angle Ɵ with the vertical.
Then by construction, 13 (1 + sin Ɵ) = 18 , so sin Ɵ = 5/13, so cos Ɵ = 12/13
Then X = 13 * (1 + cos Ɵ) = 25
To be completely correct, there is another solution from sqrt 144, which is -12. Of course this doesn‘t make any sense. But when I remember back in school, it was considered incorrect when we forgot tje negtive solution.
i remember my first algebra 1 lesson when all the questions were this easy!! the good ol days.
It's intentionally not drawn to scale as a trap for the unwary. And the plural of radius is radii, because _radius_ is a second-declension noun in Latin.
It was probably compacted in the book to conserve space. The extra space required represent 2-3 lines of text. Tho I guess they could have rotate the figure 90°...
I love to solve this kind of stuff when I was elementary and highschool but since I do programming as my academic and gaming when I resting, I have no more time for this.
Pretty easy one, I could solve the entire thing in my head within a couple of seconds. I just imagined the radius of each circle building a 5x5 and 8x8 square eliminating the big square with the ugly "void spots" which were preventing me to use pyth.t., got. a neat square in the middle, plugged in the remaining values and done.
It truly is exiting
from where do you take those exercices . please give me an answear
I feel like you should explain where you got the 5 at 2:50.
I get how you got there but people might forget and not be able to replicate it themselves.
I got there by finding that the 5cm circle had 2cm overlap to the left of the 8cm circle (18-16). The 5cm circle has 3cm to the center of it from that point. The 3cm are then subtracted from the 8cm circle's radius => 5cm horizontaly between the 2 centers.
I'm getting an X of 25. Cosine of 18, Sine of 25. That is, as to a 90 degree angle at the lower right side of the figure, sets up a hypotenuse of 31.
This one is an easy task. Call M₁ center of big circle with radius r₁ = 8 and M₂ center of small circle with radius r₂ = 5. The distance M₁M₂ is then r₁ + r₂ = 13. Since the base of the rectangle is 18, the horizontal distance of M₁ and M₂ is 18 - 8 - 5 = 5. Therefore, the vertical distance z of M₁ and M₂ is Pythagoras with z = √(13² - 5²) = √(169 - 25) = √144 = 12. Then we have height x = r₁ + z + r₂ = 8 + 12 + 5 = 25. Funny sketch BTW, 8cm < 5cm ;)
How exciting ❤
You were still able to solve it even with the scale being wrong. I was trying to solve it and then realized something was very off about this picture. 😆
i get a different answer
I created a cube whose side length is equal to half the diameter of the large circle. Then I used the Pythagorean theorem to find the diameter and then combined it with the semicircle.
I made a square whose side length was half the diameter of the small circle, then I found its hypotenuse and merged it with the radius of the small circle.
I added together the longest lines to show me the major square hypotenuse
Then I used the Pythagorean theorem on the large square because I have the hypotenuse and the length of one of the two sides. Then I moved the side whose value is 18 to the left side so that it is “X” alone.
Then I simplified the calculation, then eliminated the square with X, and then the following result appeared.
33,525....
18=13+13cos(a), a=arccos(5/13), a=67.38 deg.
x=13+13sin(67.38), therefore x=25
Isnt trig nice? Wonderful set of tools.
18/sqrt(2)+8 just from looking at the thumbnail. Square from upper left corner to center of big circle has diagonal 10+16/2, side that/sqrt(2). It's 16/2 above the floor.
Definitely an Indian Textbook.
I sure had my share of thinking the sides are equal only for them to turn out unequal and many more similar problems.
I might be doing this the wrong way, but is it possible to transform the circles into squares or rectangles and suntract them from the box? Or is that too inaccurate because pi?
Also, this is so much easier if your know that 5, 12, 13 is a Pythagorean triplet.
great video
How exciting 😎
Ahhh, after hours and hours of frustration, i finally have earned the rights to say, "EASIEST QUESTION EVERRR!!!" (andy did a similar question b4 XD)
The graphic is misleading. But the centers of the circles are 5 apart (18-8-5=5). The diagonale between the two circles‘ centers is 5+8=13. This is the hypothenuse of of a right triangle formed by the horizontal distance (5) and the vertical distance of the centers. This leads to 13^2-5^2=169-25=144. thus the vertical difference between is 12. Thus the height is 12+8+5=25.
Imagine a title: ''This problem was not exciting''
We would all be cooked.
I got ~25.71 units.
using the 8cm radii, it creates a square on the bottom right of the rectangle, with hypotenuse of sqrt(128), then did the same with the 5 in top left and had sqrt(50), so I did the hypotenuse of the entire rectangle is equal to sqrt(50)+sqrt(128)+5+8.
Then I did 18² + b² is equal to the hypotenuse squared, then square rooted it and got 25.71
radiuses for the win
When I was in school, they sometimes handed me a test with this picture of a square and it said "the area of this square is 5square kilometers..."
I was like "It ain't" and then I measured it and I told them the correct area of the square.
And then they failed me.
The whole education system failed me...
couldnt you line up the diameters of the circle, making 2 triangles, now you have the the base and the hypotenuse of a right angled triangle, thus you find the height, then since its a rectangle, the height found will be equal to x?
im incredibly fucking stupid i apologise
That only works if the line between their centers is in line with the line(s) between their centers and the corners, and if the radii of the circles goes all the way to the corners. In this case neither is true.
But you're looking in the right direction. What you do is, you move the corners to where the centers of the circles are. You subtract 5 cm from the left side and top, to reflect where the center of the 10cm circle is, and 8cm from the right side and bottom, to reflect where the center of the 16cm circle is, and then line up the radii with the "new corners", since those are in line and do reach to the centers of the triangles. From that you get a 13cm hypotenuse (8+5) and a 5cm base (18-8-5), which gives you 12cm for the vertical leg. Then you add back the 5cm and 8cm you subtracted from the top and bottom and get x = 25cm.
@@quigonkenny yea i realised immediatly after commenting
The first one is a square and rectangle. So a rectangle will have a different area unless the scale is change to equal the same amount of space taken up by the square. I refuse to solve it.
Is it possible to do if the rectangle was actually a square of unknown side?
sure. the centers are 13 cm apart and they would lie on the diagonal of the square. so the side is 5 + 8 + 13×(sq.root of 2)/2 cm (upd: fixed a mistake)
Fooled me good on this one--I first thought you were going to get the values between the edges of the circles and corners of the rectangle, and then use Pythagorean theorem on the edges of the rectangle and the diagonal through the rectangle. But as soon as I saw you draw it to scale, I realized my idea wouldn't work. Moral of the story is never trust a drawing that is not to scale. :-)
No Pythagorean theorem needed when the triple is there…save some time.
How exciting!
Does anybody else find themselves putting boxes around their answers these days? :-) How exciting
As an asian, i rate this at "can't solve without paper" (calculator is optional). I can't solve this until i drew a xOy graph.
I'd like this more if the problem was a square and the 18cm wasn't provided.
You would then get 2x^2 = A^2, where A = the hypotenuse
A = 13 + (√50 - 5) + (√128 - 8) = √50 + √128
This then simplifies to 2x^2 = (√50 + √128)^2
(√50 + √128)(√50 + √128)
50+2(√50 * √128) + 128
50+2(√6400)+128
50+2(80)+128
50+160+128
338
So 2X^2 = 338
X^2 = 169
X = √169 = 13
Very nice
Bravo!
Well there is a identity when 2 tangents are drawn from 1 point to a circle so but I don't remember it right now well, might it help
1:55 I'm sure it was purposefully not drawn to scale so you couldn't tell if 'x' was more, less, or equal to 18 and you would need to do the math.
It's probably intentionally not to scale to prevent students from measuring for the answer.
How exciting
I got 25 but i just couldn't wrap my head around the fact that the base was 5cm because that would mean it was inside the bigger circle. Never trust images man
I was ready to sass you saying "it's radii, not radiuses" but quickly I ate my words 🤐
I went about it by forming the full diagonal using the 8x8 and 5x5 right triangles formed where the circles touch the square and it worked out yay ^^
Let's have a go before I watch your video ..
If you turn the two circles so that the 19cm and the 16cm lines are horizontal you can work out the distance between the centre on the circles ''along the bottom line'' 18 - (5 + 8) = 5
So now using pythagoras you can work out what to add to (5 + 8) along the X line, let's call it A
5^2 + A^2 = (5 + 8)^2
5^2 + A^2 = 13^2
25 + A^2 = 169
A^2 = 144 so A = 12 and 12 + 5 + 8 = X 25CM
seems like it has to be a square, same 2 circles vertically and horizontal.
Me: adding @Andy Math to the list of youtube channels I'll put as subscriptions when my kid is old enough to have an account.
How would you check this answer?
Easiest method to check is probably just a scale drawing, to be honest.
I did it in another way, i used the diagonal (d+D+(r*√2-r)+(R*√2-R)) and than the Pitagora theorem
Im glad i clicked the video instead of assuming it was just a square and x = 18
i have seen diagrams not drawn to scale many times, i thought that was something they do on purpose
To be fair you don't need to have a correct drawing of this problem since you have all you need to solve for x. The correct representation may help, but it's not really necessary.
But can you solve the 2 gurls and 1 cup question that's kinda hard though
This puzzle is really engaging and easy to follow! I love how it challenges our dimensional reasoning skills. I remember getting into similar problems and found that resources like SolutionInn helped me think through the tricky geometry topics. It’s interesting how visual challenges can enhance our understanding of shapes and their relationships. What other puzzles do people enjoy that test these concepts?
Okay I’m kind of going insane with my answer cuz it’s wrong but I don’t know why…
Here’s what I did:
I found the diagonal across the rectangle by making a square with the radius of each circle, and finding the diagonal from the center of each circle to their respective corner. I added up that diagonal and the radius of the circle which SHOULD be the distance from the corner to the edge of the circle, and then I added the same value but with the radius subbed out for the other circle. My current value for the diagonal is 13+13*sqrt(2)
We know that we can form a right triangle with the corner of the rectangle, and we have one side length as 18 cm and the hypotenuse being 13+13sqrt(2) so it SHOULD be sqrt((13+13sqrt(2))^2-18^2) but that gets me 25.71 for some reason
HELP
Likely from rounding
Nvm. I figured it out. It's because the centers of both circles do not intersect with the diagonal (when it is drawn to scale). So you are not actually measuring the big diagonal line when you calculate small diagonal lines. Go to 1:21 in the video to see what I mean
The thing is that the line connecting the two centers does not have the same gradient as the line connecting the centers to the corners. When this is not true, you can't add and equate multiple line segments. They must all be on the same line.
@@Kyle-nm1kh oh my god thank you for catching that
@@redfinance3403 other guy figured it out… the diagonal doesn’t go through the center of both circles
I learned, if omniting the unit,
and mentioning/ putting it back at the result,
thy syntax would be square brackets.
So, in this case, [cm].
Anyone with me?
These would be easier if teachers gave us a second copy of the diagram that was animated. ;)
Can someone explain why he's assuming that the centers of each circle are on the same vector?
I'm assuming because both circles are tangent to each other so since they make contact there needs to be a single point where they touch, and since a circles radius connects to every point on the circle you are allowed to assume they share the same line.
Nice. I paused and did it in another way, and still got around 25 for x
diagonal = hypotenuse = 16 +10 = 36 [36]^2 = 1296 x ^2 + 18^2 =1296 1296 - 348 = x^2 = 972 /972 = 31.17
Couldn't you see immediately that the triangle you constructed was a 5,12,13 triangle?
The inside is bigger than the outside. Doctor who anyone.
Uploaded 40 seconds ago? I must click immediately.
i first solved for the longer hypotenuse across the whole rectangle by solving the smaller hypotenuses in each corner. And then i ended up with x = sqrt(183+338sqrt(2))cm which roughly equals to 25.71cm. I dont know how i got the extra 0.71, maybe it came from dealing with the sqrt(2)s but if someone wanna check my math on it. Feel Free!
8^2 + 8^2 = C^2 C=8sqrt(2)
5^2 + 5^2 = c^2 c=5sqrt(2)
x^2 + 18^2 = (5sqrt(2) + 5 + 8 + 8sqrt(2))^2
x^2 + 18^2 = (13sqrt(2) + 13)^2
x^2 + 18^2 = (2*13^2) + (2*13*13sqrt(2)) + (13^2)
x^2 + 324 = 338 + 338sqrt(2) + 169
x^2 = 183 + 338sqrt(2)
x = sqrt(183 + 338sqrt(2))
x =~25.71cm
The circle centers are not aligned with the square corners.
If it's a rectangle, diagonal is 26, and bottom is 18, so third side is easy to Calc. Sm I missing something?
Diagonal is not 26 cm
The 2 diameters add upto 26 cm but they dont touch the corners of this quadrilateral, hence not a diagonal
"If" x=25 cm then diagonal shall be 30.8 cm by applying Pythagoras
Cool!
Is not y= 13^2 -8^2 as it is not to scale so the left of the big circle would be 2cm instead of 5cm
It'll be better if the rectangle is a square and only diameters of circles are given.
Nice title 😂
can't you do 16 +10 and let that be your hypotnuse and then solve for x since it is a 45-45-90 triangle
Andy is cool.
sqrt(144) also equals -12 though so x could also = 1
Me at a glance = OK i think i can solve this
Me at a minute later = WTH 😭😭
32.06CM?
Edit: oops, i timesed a few measurements by 2 when i shouldnt have... nerts
Radiuses has only become accepted due to the overwhelming ignorance of people who don't know that the plural is, and should only have ever been, radii.