I spent far too long trying to solve for a*b like you did, but didn't hit on the obvious way you did it. I think the problem was that I was labeling the sides of the "4" triangle as x and y, which does not produce nice expressions. I gave up for a while, and as I was trying to sleep last night, I came up with an entirely different solution. Let's see if I can intelligibly describe it in pure text... While I was flailing around with your solution, I had drawn some lines breaking the rectangle into quadrants at the corners of the "4" triangle. I'll label the quadrants A B C D I had this random thought: If I know the areas of A, C, and D, can I find B? So I derived what is probably a well-known equality that A*D = B*C. So, "4" fills up half of C so C = 8. "9" fills up half of the top two, so A + B = 18, or A = 18 - B. "5" fills up half of the right two, so B + D = 10, or D = 10 - B Plugging these into the equality, we get (18 - B) * (10 - B) = B * 8 => 180 - 28*B + B^2 = 8*B => B^2 - 36*B + 180 = 0 which is the same quadratic equation you ended up with. In this case, the appropriate solution is "6", so A = 18 - 6 = 12, and D = 10 - 6 = 4, and the total area of the rectangle is A+B+C+D = 30. At this point, the last step is identical, and I got the answer of 12.
@ThePhantomoftheMath I agree! Half the reason why i go to math channels is to see how people solve these kind of problems. Its like comparing tool kits.
Let A = area of Δ with one side being the Width of the rectangle, and B = area of Δ with one side being the Height of the rectangle, & C = Area (3rd Outer Δ). Then area (Inner Δ) = √{(A+B+C)² - 4AB} = √{(9+5+4)² - 4(9)(5)} = √{4(9)² - 4(45)} = √{4(36)} = √4.√36 = 2(6) =12.
Ok, let the rectangle have height h and width w. Let the area 4 triangle have base x and height y. Let the sought after area be R. Then we can write down all of the following: 1) x*y = 8 2) h*w = R+18 3) h*(w-x) = 10 --> h*w-h*x = 10 --> R+18-h*x = 10 --> R-h*x = -8 4) w*(h-y) = 18 --> h*w - w*y = 18 --> R+18-w*y = 18 --> R-w*y = 0 Now multiply together the last versions of (3) and (4): (R-h*x)*(R-w*y) = 0 R^2 - h*x*R - w*y*R + h*w*x*y = 0 R^2 - (h*x+w*y)*R + 8*(R+18) = 0 R^2 - (h*x+w*y)*R + 8*R + 144 = 0 R^2 - (h*x+w*y-8)*R + 144 = 0 Now add the last versions of (3) and (4): R-h*x + R-w*y = -8 2*R - (h*x+w*y) = -8 h*x+w*y-8 = 2*R But see that this last is exactly what appears as the R coefficient in the quadratic above - substitute it in and all of the "working variables" disappear and we are left with only R as our unknown: R^2 - 2*R*R + 144 = 0 -R^2 + 144 = 0 R^2 = 144 R = 12 Q.E.D. Later on I'm tempted to try to solve this with variables for the given areas, say u, v, and w, and see if I can get a quadratic with coefficients being simple functions of u, v, w. That would be an interesting solution to have in one's back pocket. PS: I always work these out before watching the video, and then use the video to check my answer. This time I let a sign error creep into my algebra so I initially got the wrong numerical answer, but it was pretty easy to find and once I fixed that it was fine.
As @DangRenBo noted, we have 3 equations and 4 variables, so we don't have enough information to solve for the 4 variables. However, we find that we only need to solve for the product of a and b, not the individual values, so we have 3 variables: x, y, and ab. When we attempt to solve the system of equations, we find that it can be reduced to one equation with the product ab as the only variable. Then, we replace ab with A and solve for A. We find that A = 30 is a valid solution. Any pair of positive real values for a and b whose product is 30 is valid. The orange triangle will always have the same area for any valid pair a, b.
If that's no bother, next time, could you draw something that bears some kind of resemblance with reality? Not that it made any difference in this specific case. Actually, the incoherence between shape and numbers was so outrageous and obvious that I immediately redraw the shape according to the numbers. It took me less than 3 minutes from there on. You could say that your obvious attempt at making this very easy problem harder than it was made it even easier. Or maybe that was your plan all along.
Hi. Thank you for your feedback. I appreciate you taking the time to watch the video and share your thoughts. I apologize if the visual representation didn’t match the numbers as clearly as it should have. My intention is always to make the content helpful and accurate, and I’m glad to hear you were able to work through it quickly on your own. Just a quick note: the sketches in math videos are never to scale; they’re meant to illustrate concepts rather than provide precise measurements. However, I’ll make sure to be more careful in future videos to ensure the shapes and numbers align better. Thanks again for pointing this out, and I hope you continue to find value in the channel. All the best!
@ThePhantomoftheMath Actually, thinking about it, maybe the best formula for learning AND remembering would be to describe the shape clearly without actually showing it, and listing the available data, then let the viewer draw a shape accordingly and which is coherent with the data provided. That would prevent the viewer to be tempted to find a quick shortcut looking at the shape for clues (if this one has the right proportions) and, at the same time, prevent any misleading proportions to lead to huge waste of time. It's just an idea. The visual in the thumbnail would be lacking, but the essence of the problem can be described briefly in a colorful snapshot, with attractive words and symbols.
@@kextrz I'll definitely take your idea into consideration for future videos. Balancing clarity, engagement, and educational value is always a priority, so exploring different presentation styles could be beneficial.
Thats a fun one. Love geometry problems that dont solely rely on the Pythagoras' a^2+b^2=c^3
Impressive solution presented using colorful animation. Wel-done
Thanks! Glad you liked it!
Love the animation.
Thank you friend!
I am in touch of exceptionally good lessons. Thanks a lot sir🙏
@@bkp_s Thank you sir!
I spent far too long trying to solve for a*b like you did, but didn't hit on the obvious way you did it. I think the problem was that I was labeling the sides of the "4" triangle as x and y, which does not produce nice expressions. I gave up for a while, and as I was trying to sleep last night, I came up with an entirely different solution. Let's see if I can intelligibly describe it in pure text...
While I was flailing around with your solution, I had drawn some lines breaking the rectangle into quadrants at the corners of the "4" triangle. I'll label the quadrants
A B
C D
I had this random thought: If I know the areas of A, C, and D, can I find B? So I derived what is probably a well-known equality that A*D = B*C.
So, "4" fills up half of C so C = 8. "9" fills up half of the top two, so A + B = 18, or A = 18 - B. "5" fills up half of the right two, so B + D = 10, or D = 10 - B
Plugging these into the equality, we get (18 - B) * (10 - B) = B * 8 => 180 - 28*B + B^2 = 8*B => B^2 - 36*B + 180 = 0 which is the same quadratic equation you ended up with. In this case, the appropriate solution is "6", so A = 18 - 6 = 12, and D = 10 - 6 = 4, and the total area of the rectangle is A+B+C+D = 30. At this point, the last step is identical, and I got the answer of 12.
Great job, man! Really nice! I love it when people come up with their own solutions and approaches to the problem! Nice! 👍
@ThePhantomoftheMath
I agree! Half the reason why i go to math channels is to see how people solve these kind of problems.
Its like comparing tool kits.
Excellent problem
Let A = area of Δ with one side being the Width of the rectangle,
and B = area of Δ with one side being the Height of the rectangle,
& C = Area (3rd Outer Δ). Then area (Inner Δ) = √{(A+B+C)² - 4AB} =
√{(9+5+4)² - 4(9)(5)} = √{4(9)² - 4(45)} = √{4(36)} = √4.√36 = 2(6) =12.
How do you get the formula for inner triangle
Bakıdan salamlar.Əla həll etdiniz.Təşəkkürlər.
Needed Google Translate for this but: Çox sağ ol dostum!
Ok, let the rectangle have height h and width w. Let the area 4 triangle have base x and height y. Let the sought after area be R. Then we can write down all of the following:
1) x*y = 8
2) h*w = R+18
3) h*(w-x) = 10 --> h*w-h*x = 10 --> R+18-h*x = 10 --> R-h*x = -8
4) w*(h-y) = 18 --> h*w - w*y = 18 --> R+18-w*y = 18 --> R-w*y = 0
Now multiply together the last versions of (3) and (4):
(R-h*x)*(R-w*y) = 0
R^2 - h*x*R - w*y*R + h*w*x*y = 0
R^2 - (h*x+w*y)*R + 8*(R+18) = 0
R^2 - (h*x+w*y)*R + 8*R + 144 = 0
R^2 - (h*x+w*y-8)*R + 144 = 0
Now add the last versions of (3) and (4):
R-h*x + R-w*y = -8
2*R - (h*x+w*y) = -8
h*x+w*y-8 = 2*R
But see that this last is exactly what appears as the R coefficient in the quadratic above - substitute it in and all of the "working variables" disappear and we are left with only R as our unknown:
R^2 - 2*R*R + 144 = 0
-R^2 + 144 = 0
R^2 = 144
R = 12
Q.E.D.
Later on I'm tempted to try to solve this with variables for the given areas, say u, v, and w, and see if I can get a quadratic with coefficients being simple functions of u, v, w. That would be an interesting solution to have in one's back pocket.
PS: I always work these out before watching the video, and then use the video to check my answer. This time I let a sign error creep into my algebra so I initially got the wrong numerical answer, but it was pretty easy to find and once I fixed that it was fine.
@@KipIngram Very well done!!!
Fine . Can we calculate the lengths a , b , x and y ?
Three equations and four variables. A=a*b, so we didn't need to solve for the variable. Rescued from infinite solutions!
As @DangRenBo noted, we have 3 equations and 4 variables, so we don't have enough information to solve for the 4 variables. However, we find that we only need to solve for the product of a and b, not the individual values, so we have 3 variables: x, y, and ab. When we attempt to solve the system of equations, we find that it can be reduced to one equation with the product ab as the only variable. Then, we replace ab with A and solve for A. We find that A = 30 is a valid solution. Any pair of positive real values for a and b whose product is 30 is valid. The orange triangle will always have the same area for any valid pair a, b.
Area of the orange rectangle=30-(9+5+4)=12
I came up with 10
Phantom😊
If that's no bother, next time, could you draw something that bears some kind of resemblance with reality?
Not that it made any difference in this specific case.
Actually, the incoherence between shape and numbers was so outrageous and obvious that I immediately redraw the shape according to the numbers. It took me less than 3 minutes from there on.
You could say that your obvious attempt at making this very easy problem harder than it was made it even easier. Or maybe that was your plan all along.
Hi. Thank you for your feedback. I appreciate you taking the time to watch the video and share your thoughts.
I apologize if the visual representation didn’t match the numbers as clearly as it should have. My intention is always to make the content helpful and accurate, and I’m glad to hear you were able to work through it quickly on your own.
Just a quick note: the sketches in math videos are never to scale; they’re meant to illustrate concepts rather than provide precise measurements. However, I’ll make sure to be more careful in future videos to ensure the shapes and numbers align better.
Thanks again for pointing this out, and I hope you continue to find value in the channel. All the best!
@ThePhantomoftheMath Actually, thinking about it, maybe the best formula for learning AND remembering would be to describe the shape clearly without actually showing it, and listing the available data, then let the viewer draw a shape accordingly and which is coherent with the data provided.
That would prevent the viewer to be tempted to find a quick shortcut looking at the shape for clues (if this one has the right proportions) and, at the same time, prevent any misleading proportions to lead to huge waste of time.
It's just an idea. The visual in the thumbnail would be lacking, but the essence of the problem can be described briefly in a colorful snapshot, with attractive words and symbols.
@@kextrz I'll definitely take your idea into consideration for future videos. Balancing clarity, engagement, and educational value is always a priority, so exploring different presentation styles could be beneficial.