As a 70 y.o. man who hasn't worked with these types of problems for >45 years, I could visualize this approach in principle, but couldn't remember the quadratic equation anymore and had to look it up. It also took me much longer than the video lasted.
I had the same problem but got the correct answer 🙂buy working out (guessing) the length of the sides then subtracting the total area of the rectangle from that given for the 3 tringles 3x10 =30 /2 =15 so the height is 10, 6x9=54/2=27 so the length is 9 total area for rectangle = 90... Total given triangle area 27+12+15=54 centre triangle 90-54 = 36 not a quadratic in sight ;-)
@Harold L Potts...So that means a*b = 90 is a constant for this problem. It should be possible to fit all 4 triangles with these areas in the rectangle as long as a*b = 90, otherwise not.
@@Jack_Callcott_AU It's important to recognize that the area of the rectangle is a*b, so the quantity a*b is measured in area units. This allows the assertion that a*b = 12+15+27+Blue. So solving for a*b permits calculation of Blue.
There is a forth equation. The blue area can be obtained from the three sides of this inner triangle with Heron’s formula. And the lengths of these three sides can be calculated with the help of Pythagore. That’s a bit complicated but doing so, we can add the areas of the 4 triangles. This sum must be equal to the area of the rectangle. Logically, this constraint should give us a fourth equation. The thing is the blue area always equals 36, whatever the proportion between a and b, as long as a*b equals 90. So we have an infinity of solutions for a and b.
@Alain Peugny: I posted a set of parametric equations to define the length of each of the 6 outer line segments, for a given constant. (Infinite real solutions, but just 4 integer-only, with just 1 with x-dim > y-dim.)
I like how the answer does not constrain the actual values of a and b, individually, but shows an infinite family of triangles in an infinity of rectangles satisfy this set of constraints.
Yeah, at first, I was thinking that's great, but what are the actual values of a and b. But after playing around a bit, I realized that any values for a and b work as long as: 1) ab=90 2) x=(3/5)a 3) y=(1/3)b For example: a=1, b=90 x=(3/5) x 1=0.6 y=(1/3) x 90=30 Basically, it's a really wide short rectangle. Example 2: a=2, b=45 X=(3/5) x 2=1.2 y=(1/3) x 45=15 Etc., etc. As you said, there are an infinite number of solutions.
I watch a fair few vids of this kind of thing, but I must say your explanations are very clear and concise. And yes, there are several ways of doing that, but you chose one and explained it very well.
we have 3 equations: bx=54; ay=30; (b-y)(a-x)=24; add the 3 equations together, we have: ab + xy = 108; multiply the first two equations together, we have (ab) (xy)=54*30. Therefore, both the sum and the product are known for the two numbers ab and xy, and they are the two roots of the quadratic equation: U^2 - 108 U + 54*30 = 0. Solving the quadratic equation, we have ab=90, xy=18 (since ab > xy). Finally, the answer to the original question is ab - 27-15-12 = 36.
It's not a mathematical way of deriving, but if you multiply the area of each triangle by doubling it, that is, assume the length of the side of the quadrangle as the area of the quadrangle, you get the following. 27 = (1/2)*9*6 ... x = 6, b = 9 12 = (1/2)*4*6 ... (a-x)=4, (b-y)=6 15 = (1/2)*3*10 ... y=3, a=10 b = (b-y) + y = 6 + 3 = 9 a = x + (a-x) = 6 + 4 = 10 a * b = 90 Blue Triangle = 90 - 27 - 12 - 15 = 36
Did without pen and paper by assuming the values are integers Starting with the triangle of area 15, the sides are 3* 10 ( of course, 2*15, 6*5 are possible, but 3*10 looks more reasonable). Hence, one side of the rectangle is 10 for now, and part of the other side =3 For the triangle with area 12, L*W = 24. Let's try 6* 4, and add 6 + 3 = 9. Hence 4 goes to the other side, which for now is 10. Hence we need a 6 from the triangle with an area of 2 For the triangle with this area L* W =54. Let's use 6* 9 since we are looking for a 6 and 9 . Hence the sides for the rectangle are 9 and 10. Hence the area of the rectangle = 90. And the area of the triangle: 90-(27-15- 12 ) = 90 - 54 = 36 Answer
As an engineer this is what I hate about math, they intentionally give you the minimum amount of information. In reality, you almost always have enough information to solve by multiple means.
its 330am and this popped up on my recommended for some reason :) Only 6 seconds in (wanted to make sure the numbers shown were the 'area' of the known triangles). But here's how i'd go about solving this. 1) Solve the A2, B2, C2, for all 3 known triangles. 2) add the two 'lengths' (ie: a2 b2) on the left that connect 27 and 12, and add the lengths of where 12 and 15 meet. 3) use that to figure out the total area of the rectangle formed. 4) subtract the total area of the 3 triangles. 5) left with the area for the blue triangle? :) *quick edit* When i say "solve for a2,b2,c2" i also mean to take into account that the two sides that connect 12 and 15 at the bottom combined must also = the x2 that makes up the top side of 27, and that the x2's for 27 and 12 = that of the right side of 15
The important feature of this type of problem is that the blue area is exactly defined but the dimensions of the rectangle are not unique. Therefore one dimension of the rectangle can be conveniently decided by you, such as a width of 9 units, or a height of 10 units. Either of those choices leads to a simple quadratic that gives the other dimension and consequently the area of the rectangle.
Great insight, thanks. Just to test your shortcut in extremis, I chose 54 for the lenght of the rectangle which resulted in the width of the rectangle being 1 2/3, the area of the rectangle equal to 90, and the area of the triangle equal to 36.
Using Pythagoras,I calculated the sides of the blue triangle sqrt(117),sqrt(52) and sqrt(109) and using a Herons Formula calculator and surprise,surprise the area of the blue triangle came out at 36 sq. units. let a=sqrt117,b=sqrt52 and c=sqrt109 Heron's formula; in terms of the sides a,b and c. Ablue=sqrt(4a^2*b^2-(a^2+b^2-c^2)^2)/4 =sqrt(4*117*52-(117+52-109)^2)/4 =sqrt(208*117-60^2)/4 =sqrt(24336-3600)/4 =sqrt(20736)/4 =144/4 =36 sq.units. This checks our answer out using Pythagoras and Heron.
Once I worked out the answer and the fact that we can't determine the actual dimensions, I figured out the relative lengths: The left side is broken into segments in a ratio of 3 (top) to 2 (bottom), and the bottom side is broken into segments in a ratio of 2 (left) to 1 (right).
Yes, but you can create a set of parametric equations to generate a solution set based on a chosen constant. With i > 0, the parametric solution for the 6 outer segments is: x = 3*i, x1 = 2*i, x2 = i, y = 30/i, y1 = 12/i, y2 = 18/i. I posted the general solution separately.
I just guessed at the numbers that make up 54, 30, and 24 (1/2bh of each triangle * 2) and came up with rectangle 10 * 9 = 90. Then just subtracted 90-12-15-27 = 36. It would have been much harder if the sides were fractions instead of whole numbers. I was a math major 50 years ago, but this would have been harder if the triangles were 25, 14 and 17 for example.
A rather convoluted solution. If you assume from the outset that the base and height of each given triangle would be integers, you can quickly find all the factors of 54, 24, and 30. Then by simple inspection you can just sort out that the dimensions of the large triangle must be 18*5. If you were wrong about your guess that these would be integers, you would find that out pretty quickly and do it pretty much the way explained in the video.
There's also no need for "normal" quadratic equation. You can consider all lenghts in terms of x and this lenght (a-x), which should be y. Then you ll get equation which yelds 18y^2 = 8x^2, so 3y=2x, and then y=2/3 x (of course only positive solutions) . Now you got lenghts of all sides in terms of x (another words, we know the proportions of all lenghts) :) Amazing how many approaches can be done here :) im curious if there is pure 'visual' solution.
You should have used letters other than a and b for the length and width (l and w, for example) because the letters are also used for the quadratic equation, which can cause confusion.
I have prepared myself a formula for such a problem,which enables to find area of any one triangle out of these four triangles if areas of remaining three triangles are known.
Let consider rectangles instead of triangles. A1 = bx , A2 = ay, A3 = (b-y)(a-x) and A4 = xy. A1, A2 and A3 are known. Then A4*A4 - (A1+A2+A3)*A4 + A1*A2 = 0. This comes from if a rectangle is subdivised by 4 subretangles R1, R2, R3 and R4 (clockwise), then Area(R1) * Area(R3) = Area(R2) * Area(R4) In this problem, we find A4 = xy = 18 and then ab = 90.
What's so interesting about this particular problem is we never go back and re-substitute; that is, we only arrive at the product (a*b) - - we never solve for either one; or for that matter, for (x) or (y), either.
There is no unique solution for a & b, so you couldn't solve for those anyway. It works for any rectangle where a * b = 90. Also, if you know the area of any 3 of the triangles you can also work out the area of the fourth. All very satisfactory.
In another comment, I created a set of 6 parametric equations to generate any of the infinite real solutions. There are just 4 integer-only solutions, and just 1 with x-dim > y-dim as in the drawing.
I found the area of the rectangle being 45, the sides being clockwise 5, 3+6, 2+3 and 9 on top. All the areas of the side triangles were more than 45 and I could figure out how that was possible until I realized that the area of a triangle is HALF the sides multiplication!!!
Here's a general solution to find the lengths of each of the 6 outer segments. The inner segments can be found with the Pythagorean formula. We place the lower left corner on the (0,0) origin for the X & Y axes, give the segments x & y names, and then solve. The top segment is x; the bottom 2 segments are left x1 & right x2. The right segment is y; the left 2 segments are bottom y1 & top y2. All segments & areas are greater than 0. We have these 5 equations using 6 variables: ε1: x = x1+x2 -> x1 = x-x2 ε2: y = y1+y2 -> y1 = y-y2 ε3: x*y2/2 = 27 -> y2 = 54/x ε4: x2*y/2 = 15 -> x2 = 30/y ε5: x1*y1/2 = 12 -> ... (x-x2)*(y-y2) = 24 (x-30/y)*(y-54/x) = 24 (x*y-30)*(x*y-54) = 24*x*y (x*y)^2 - 108*(x*y) + 1620 = 0 The quadratic equation in x*y is solved as: x*y = [108 ± √(108^2 - 4*1620)]/2 = 54 ± √(54^2 - 1620) = 54 ± √(54*(54 - 30)) = 54 ± √(54*24) = 54 ± √(81*16) = 54 ± 36 ε6: The area of the rectangle is x*y and the area of inscribed triangle, T, is: T = x*y - 27 - 15 - 12 = (54 ± 36) - 54 = ± 36 ∴ T = 36, x*y = 90. We may further solve for each symbol: ε3': y2 = 54/(90/y) = y*3/5 ε2': y1 = y-(y*3/5) = y*2/5 ε4': x2 = 30/(90/x) = x/3 ε1': x1 = x-(x/3) = x*2/3 Thus, we have the relative length of each segment. Let x/3 = i, y/5 = j: x1 = 2*i, x2 = i, x = 3*i, y1 = 2*j, y2 = 3*j, y = 5*j. Parameters i & j are linked by: ε6: (3*i)*(5*j) = 90 -> j = 6/i With i > 0, the parametric solution for the 6 outer segments is: x = 3*i, x1 = 2*i, x2 = i, y = 30/i, y1 = 12/i, y2 = 18/i. Although there are an infinite number of real solutions, there are just 4 solutions having only integers. The previous equations in i & j all have integer coefficients; if i & j are integers, all 6 lengths are integers. We choose i to be an integer and a factor of 6, so j = 6/i is an integer, too. For i = 1, we have: x = 3, x1 = 2, x2 = 1, y = 30, y1 = 12, y2 = 18. For i = 2, we have: x = 6, x1 = 4, x2 = 2, y = 15, y1 = 6, y2 = 9. For i = 3, we have: x = 9, x1 = 6, x2 = 3, y = 10, y1 = 4, y2 = 6. For i = 6, we have: x = 18, x1 = 12, x2 = 6, y = 5, y1 = 2, y2 = 3. The last integer solution is the best fit for the drawing, where x > y.
A more geometric way of understanding this: Annotate the rectangle as ABCD from top-left corner in clockwise order. Let its area be x. Annotate the left tip of the shaded triangle as E (E is located on edge AD), the bottom tip as F (F is on edge CD). Now, connect the opposite apexes of the rectangle as BD. S(BDE)/S(ABD) = S(BDE)/0.5x = (0.5x-27)/0.5x = (x-54)/x. S(BDF)/S(BCD) = S(BDF)/0.5x = (0.5x-15)/0.5x = (x-30)/x. So, DE/AD = (x-54)/x, DF/CD = (x-30)/x. Thus, S(DEF) = 12 = 0.5*[(x-54)/x]*[(x-30)/x]*x. Move terms to the left: 24x = (x-54)(x-30). We get x^2 - 108x + 1620 = 0.
I eyeballed the known-size triangles and realized that their xy proportions appear to correspond to their most obvious denominators. In other words, the area 15 triangle appears to be 3×5, the area 12 triangle appears to be 2×6, and the area 27 triangle appears to be 3×9. Obviously those aren't their actual height and base values, as they would be off by a factor of 2, so I assumed an additional factor of √2 for each dimension, and the math fell into place. The area 15 triangle has height 5√2 and base 3√2, the area 12 triangle has height 2√2 and base 6√2, etc. From that I ended up with a rectangle of dimensions 9√2 (w) by 5√2 (h), and from there it was fairly simple to determine the rectangle area to be 90 and the blue triangle area to be 36.
We denote the vertices of the rectangle with A F D B We denote the vertices of the blue triangle with A E C First: we draw from C a column (plumb) so it cuts AF at point G and cuts AE at point N Second: we draw from point E a straight line parallel to straight line DB then cuts AB at point S and cuts CG at point M Third: We draw from point N ((note the point N is already defined as the point of intersection of the line CG with AE)) We draw a line parallel to both ES and DB that cuts AB at point R and cuts FD at point J CM=MN because ME is an internal bisector of angle E. We note that: EMC=EDC=12 and also NME=NJE=12 Because EMC is half of the rectangle EMCD And because NME is half of the rectangle NMEJ It is clear that CM=MN Because ME is the internal bisector of angle E and EMC=EDC=NME=NJE Now JNFA is a trapezoid whose area is JNFA=AFE-NJE = 27-12 =15 and also the area of the ANSM trapezoid is ANSM= AFES-(AFE+NME) =(2AFS )- (AFE+NME) =2(27) - (27+12) =15 since ARN=ANG and AHMN=JFNA= 15 then NRMH=FJNG hence FG=2AG JN=2AG hence AF=3AG (AF+JN)×FJ/2=15 (3AG+2AG)FJ=30 5(AG)×FJ=30 AG×FJ=6 =ARNG ARN=AGN=3 AGBC=15×2=30 ANC=AGBC-(ABC+AGN) ANC=30-(15+3)=12 ACE=ANC+NEM+EMC ACE=12+12+12=36 .................................... (marginal note): including If NM=JE=ED=2FJ=2NG=2AR, and because DF=DE+EJ+JF, then DF=5JF .......................... ......................
That's one way to do it. Meanwhile we should be able to solve this in our heads in about 15 seconds. This video had way too many unnecessary steps. But, it did work.
Usually we are given the area of the whole rectangle also to calculate the blue shaded area . HOwever, this need not to be including the rectangle’s area and yet there still be such a solution !!!
let's make the multiplication easier, and make the catheti of the bottom-left rectangle as x and y, then we have 1) xy/2=12 2) b(a-x)/2 = 27 3) a(b-y)/2=15. obviously, 12+15=27, let's add them up and get: xy+ab-ay=ab-bx => bx = ay - xy => b/y = (a-x)/x This means the 2 right rectangles with the areas 27 and 12 are similar rectangles. so b/y = (a-x)/x = sqrt(27/12) = sqrt(9/4) = 3/2 => y=2b/3, x = 2a/5 => xy/2 = 2ab/15=12 => ab = 90 => S = ab - 27 - 12 -15 = 36
@The Math District: I continued the work on the problem to create a set of 6 parametric equations to generate a solution set for a chosen parameter. There are an infinite number of solutions. With further analysis, I found 4 integer-only solutions, of which, only 1 has x-dim > y-dim like in the drawing. I hope you review it!
A few quibbles: Writing ay would be more consistent than ya as you are comparing to bx. To be further consistent, why not multiply the third equation by 2 right away, as well? But, I would not have used a and b for the rectangle, given that you are using a, b, c quadratic formula and that could cause confusion with a and b each being used two different ways. On the plus side, sound strategy, so great on that. But you should redo this to overcome these three points... Oh, well...
The algebra looked too tedious so I tried throwing in some values for side lengths that would make the 3 triangles work out and I came up with 6x9 for the 27 triangle, 4x6 for the 12 triangle, and 10x3 for the 15 triangle (a=10, b=9, x=6, y=3 as diagrammed in the video). So, (assuming there would be one unique solution whatever the actual values of a and b), total rectangle area = 90, blue triangle = 90-27-12-15.
These arbitrary values must satisfy bx=54, ay=30, (a-x)(b-y)=24 Without knowing in advance the area of the rectangle, how do you know the values must be integers? Also, do you proceed with trial and error to check that the values satisfy the three equations of the area of a triangle?
Actually, there are an infinite number of real solutions, and just 4 with only integers. Your answer is unlike the drawing which has a longer horizontal dimension than the vertical. In fact, there's only 1 integer solution that has x-dim > y-dim. I posted my general solution separately.
If in the blue triangle we draw a horizontal line and a vertical line through the left and lower vertices respectively, the initial rectangle that contains all the triangles is divided into a grid of 2x2 rectangular cells that we will call, from left to right and from top to bottom , A, B, C and D ⇒ C=2x12=24; A+B=2x27=54; B+D=2x15=30 ⇒ A=54-B; D=30-B ⇒ A/C=B/D ⇒ (54-B)/24=B/(30-B) ⇒ (54-B)x(30-B)=24B ⇒ B²-108B+1620= 0 ⇒ B=18 ⇒ Initial rectangle area = (A+B)+(B+D)-B+C = 54+30-18+24=90 ⇒ Blue triangle area = 90-27-12-15=36
Wow so compilicated. The top area = 54/2 so you end up with 9x6 for the size of the rectangle, the right triangle fits in a rectangle of 3x10, the bottom left fits in a rectangle of 6x4. So the outer rectangle has a size of 9x10. And then 90-27-15-12 = 36. No need for quadratics at all.
What you showed was that if the area of the triangle is the same for all possible rectangles and triangles with the dimensions shown, then it's 36. That's not the same thing as showing it's always 36.
Чтобы найти синюю площадь наверное можно было найти общую площадь прямоугольника a*b, и от этой площади отнять площади остальных фигур, останется только площадь синей фигуры
Area(Blue Δ) = √{(A+B+C)² - 4.A.B} = √{(27+15+12)² - 4.(27).(15)} = √{4.(27)² - 4.(27).(15)} = √{4(27)(3)}.√{9-5} = √{4(9)(9)}.√4 = 36. Here 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. ------------ Let Δ= area of blue triangle, A = area of 27, B = area of 15 and C= area of 12. Let u=horizontal side of C, v= vertical side of C, a=width, b=height of rectangle. Then A+B+C+Δ=ab. Also uv=2C=2(12), b(a-u)=2B=2(15), and a(b-v)=2A=2(27). So bu=ab-2B & av=ba-2A. ∴ 0= (ab-2B)(ba-2A) - ab(uv)= (ab-2B)(ab-2A) - ab(2C). So (ab)² -2(ab)(A+B+C)+ 4AB=0. ∴ (A+B+C+Δ)² -2(A+B+C+Δ).(A+B+C)+ 4AB=0. Let E=A+B+C. Then (E+Δ)² -2(E+Δ).E +4AB=0. ∴ E²+2EΔ+Δ² -2E² -2ΔE +4AB=0. So Δ² = E² - 4AB = (A+B+C)² - 4AB. ∴ Δ= √{(A+B+C)² - 4AB}, since Δ>0. .
In the second part of your comment, you have written 0 = (ab - 2A)(ab - 2B).. How the product can be zero.. could you kindly explain pl.. though I have got the answer using the quadratic method.. but I am wondering how the direct formula has been derived and I saw ur derivation.. but that product step equals 0 am not getting that.. could u pl help @Ramkabharosa
with guessing 15x2=30=3x10 12x2=24=6x4 27x2=54=9x(10-4) so we can easily see one lane is10, the other is 9, area of rectangle is 9x10=90 blue are is 90-27-12-15=36 this looks like a simple way.
There is no need for such complicated calculations, just use the staring method to get the side lengths of the three triangles together. 9*6, 4*6, 3*10, you can easily calculate the area of the rectangle as 9*10=90. Then the area of the middle triangle is 90-27-12-15=36.
Not efficient, you need to utilize the geometry properties reassign (a-x) as a, and (b-y) as b, let x be (w*a) instead and let y be (k*b) We know that 27+12 does not equal to half the rectangle, However, 27+12+(k*12) is the same size as half the rectangle (if you r wondering why, this is pretty much the same as 27+a*(b+k*b)/2 = 27+a*b/2+k*a*b/2, if a*b/2=12, then k*a*b/2=k*12) By analogy, 15+12 does not equal half of the rectangle, but 15+12+(w*12) does Thus 27+12+(k*12)= 15+12+(w*12) k+1=w (a*b)/2=12 and (w*a)*(k*b+b)/2=27 since w=k+1, (w*a)*(k*b+b)/2=27 is the same as [(k+1)^2]*a*b/2=27 divide a*b from both sides you’ll get (k+1)^2=9/4, thus k+1=3/2, k=1/2 With this, you should be able to solve the area of the blue triangle pretty easily
and to simplify, determine the area of the rectangle and minus the existing areas. did it in my head in about 3 minutes, no need for those calculations for this problem. and Im a carpenter
Identical to the way that I did it except for the way I labelled the side of the squares and the triangles. I also called the product of the sides of the rectangle (it's area) "z" in the last few steps to keep the equations simpler. In any event, it's a very satisfying problem.
Not at all, she made the rest of us feel like 5 year old children, the way she had to explain the OBVIOUS. And Math is BEAUTIFUL EVEN MORE if your tiny brain cannot understand it, but you would not know.
a+b=108 and ab=1620. we know a or b must be less the square root of 1620. Then it gets the greatest integer 40. From the following pairs, you could find the clue. (a, b): (40,68), (20,88), (18,90) --- for ab = 1620, one of the pair should be I*10
As a 70 y.o. man who hasn't worked with these types of problems for >45 years, I could visualize this approach in principle, but couldn't remember the quadratic equation anymore and had to look it up. It also took me much longer than the video lasted.
I had the same problem but got the correct answer 🙂buy working out (guessing) the length of the sides then subtracting the total area of the rectangle from that given for the 3 tringles 3x10 =30 /2 =15 so the height is 10, 6x9=54/2=27 so the length is 9 total area for rectangle = 90... Total given triangle area 27+12+15=54 centre triangle 90-54 = 36 not a quadratic in sight ;-)
Él lado de un triángulo siempre es mayor que la suma de los otros dos. En ese triángulo azul eso no se cumple
Excelente. Bastante didático. Muito útil.
When you have 3 equations and 4 unknowns, it's important to recognize that a solution for a*b is sufficient.
@Harold L Potts...So that means a*b = 90 is a constant for this problem. It should be possible to fit all 4 triangles with these areas in the rectangle as long as a*b = 90, otherwise not.
@@Jack_Callcott_AU It's important to recognize that the area of the rectangle is a*b, so the quantity a*b is measured in area units. This allows the assertion that a*b = 12+15+27+Blue. So solving for a*b permits calculation of Blue.
There is a forth equation. The blue area can be obtained from the three sides of this inner triangle with Heron’s formula. And the lengths of these three sides can be calculated with the help of Pythagore. That’s a bit complicated but doing so, we can add the areas of the 4 triangles. This sum must be equal to the area of the rectangle. Logically, this constraint should give us a fourth equation.
The thing is the blue area always equals 36, whatever the proportion between a and b, as long as a*b equals 90.
So we have an infinity of solutions for a and b.
@@alainpeugny1146 Well, yes! That's what I thought. Also , I like Heron's formula, but using it in this case would be complicated. Thanks!
@Alain Peugny: I posted a set of parametric equations to define the length of each of the 6 outer line segments, for a given constant. (Infinite real solutions, but just 4 integer-only, with just 1 with x-dim > y-dim.)
تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين
I like how the answer does not constrain the actual values of a and b, individually, but shows an infinite family of triangles in an infinity of rectangles satisfy this set of constraints.
Yeah, at first, I was thinking that's great, but what are the actual values of a and b. But after playing around a bit, I realized that any values for a and b work as long as:
1) ab=90
2) x=(3/5)a
3) y=(1/3)b
For example:
a=1, b=90
x=(3/5) x 1=0.6
y=(1/3) x 90=30
Basically, it's a really wide short rectangle.
Example 2:
a=2, b=45
X=(3/5) x 2=1.2
y=(1/3) x 45=15
Etc., etc. As you said, there are an infinite number of solutions.
I watch a fair few vids of this kind of thing, but I must say your explanations are very clear and concise. And yes, there are several ways of doing that, but you chose one and explained it very well.
we have 3 equations: bx=54; ay=30; (b-y)(a-x)=24; add the 3 equations together, we have: ab + xy = 108; multiply the first two equations together, we have (ab) (xy)=54*30. Therefore, both the sum and the product are known for the two numbers ab and xy, and they are the two roots of the quadratic equation: U^2 - 108 U + 54*30 = 0. Solving the quadratic equation, we have ab=90, xy=18 (since ab > xy). Finally, the answer to the original question is ab - 27-15-12 = 36.
Sum of two sides equal to third side, hence area equal to zero... as all vertices are collinear....
It's not a mathematical way of deriving, but if you multiply the area of each triangle by doubling it, that is, assume the length of the side of the quadrangle as the area of the quadrangle, you get the following.
27 = (1/2)*9*6 ... x = 6, b = 9
12 = (1/2)*4*6 ... (a-x)=4, (b-y)=6
15 = (1/2)*3*10 ... y=3, a=10
b = (b-y) + y = 6 + 3 = 9
a = x + (a-x) = 6 + 4 = 10
a * b = 90
Blue Triangle = 90 - 27 - 12 - 15 = 36
So a>b? ;-)
I like the way you explain everything. Nice video.
Solution is simple to understand, but needs a creative mind to produce. Neatly done.
I’ve seen a lot of geometry problems that have triangles everywhere, usually your best bet is to define everything
Did without pen and paper by assuming the values are integers
Starting with the triangle of area 15, the sides are 3* 10 ( of course, 2*15, 6*5 are possible, but 3*10 looks more reasonable). Hence, one side of the rectangle is 10 for now, and part of the
other side =3
For the triangle with area 12, L*W = 24. Let's try 6* 4, and add 6 + 3 = 9. Hence 4 goes
to the other side, which for now is 10. Hence we need a 6 from the triangle with an area of 2
For the triangle with this area L* W =54. Let's use 6* 9 since we are looking for a 6 and 9 .
Hence the sides for the rectangle are 9 and 10. Hence the area of the rectangle = 90. And the area
of the triangle:
90-(27-15- 12 ) = 90 - 54 = 36 Answer
As an engineer this is what I hate about math, they intentionally give you the minimum amount of information. In reality, you almost always have enough information to solve by multiple means.
I like your approach as much as I like your accent. The former is appealing, the latter comforting. Thanks.
Reminded me of my good old school days!! Thanks, Maa'm, for the nice explanation!! 🙏😊
its 330am and this popped up on my recommended for some reason :)
Only 6 seconds in (wanted to make sure the numbers shown were the 'area' of the known triangles). But here's how i'd go about solving this.
1) Solve the A2, B2, C2, for all 3 known triangles.
2) add the two 'lengths' (ie: a2 b2) on the left that connect 27 and 12, and add the lengths of where 12 and 15 meet.
3) use that to figure out the total area of the rectangle formed.
4) subtract the total area of the 3 triangles.
5) left with the area for the blue triangle?
:)
*quick edit* When i say "solve for a2,b2,c2" i also mean to take into account that the two sides that connect 12 and 15 at the bottom combined must also = the x2 that makes up the top side of 27, and that the x2's for 27 and 12 = that of the right side of 15
Muy bueno, como los otros que he visto en este canal . 👍
Really nice explanation - and a nice problem too! Thank you
The important feature of this type of problem is that the blue area is exactly defined but the dimensions of the rectangle are not unique. Therefore one dimension of the rectangle can be conveniently decided by you, such as a width of 9 units, or a height of 10 units. Either of those choices leads to a simple quadratic that gives the other dimension and consequently the area of the rectangle.
..
There's just 1 integer-only solution with x-dim > y-dim like the drawing. You found 1 of 3 integer-only solutions that's unlike the drawing.
Great insight, thanks. Just to test your shortcut in extremis, I chose 54 for the lenght of the rectangle which resulted in the width of the rectangle being 1 2/3, the area of the rectangle equal to 90, and the area of the triangle equal to 36.
Using Pythagoras,I calculated the sides of the blue triangle sqrt(117),sqrt(52) and sqrt(109) and using a Herons Formula calculator and surprise,surprise the area of the blue triangle came out at 36 sq. units.
let a=sqrt117,b=sqrt52 and c=sqrt109
Heron's formula; in terms of the sides a,b and c.
Ablue=sqrt(4a^2*b^2-(a^2+b^2-c^2)^2)/4
=sqrt(4*117*52-(117+52-109)^2)/4
=sqrt(208*117-60^2)/4
=sqrt(24336-3600)/4
=sqrt(20736)/4
=144/4
=36 sq.units.
This checks our answer out using Pythagoras and Heron.
Great solution. Explained it very well :)
Once I worked out the answer and the fact that we can't determine the actual dimensions, I figured out the relative lengths: The left side is broken into segments in a ratio of 3 (top) to 2 (bottom), and the bottom side is broken into segments in a ratio of 2 (left) to 1 (right).
Yes, but you can create a set of parametric equations to generate a solution set based on a chosen constant. With i > 0, the parametric solution for the 6 outer segments is:
x = 3*i, x1 = 2*i, x2 = i,
y = 30/i, y1 = 12/i, y2 = 18/i. I posted the general solution separately.
It’s not what you do,it’s the way you do it.Mathematics has Theorems,Equations,,Etc.Simplification is of the Essence.❤️.
Muito bom!!!
I enjoyed this video, thanks a lot!
very neat method!
Which text editor are you using?
I just guessed at the numbers that make up 54, 30, and 24 (1/2bh of each triangle * 2) and came up with rectangle 10 * 9 = 90. Then just subtracted 90-12-15-27 = 36. It would have been much harder if the sides were fractions instead of whole numbers. I was a math major 50 years ago, but this would have been harder if the triangles were 25, 14 and 17 for example.
Thank u mam. So good explanation🙏
A rather convoluted solution.
If you assume from the outset that the base and height of each given triangle would be integers, you can quickly find all the factors of 54, 24, and 30. Then by simple inspection you can just sort out that the dimensions of the large triangle must be 18*5.
If you were wrong about your guess that these would be integers, you would find that out pretty quickly and do it pretty much the way explained in the video.
THIS WAS NOT THE PROBLEM GIVEN. YOU CANNOT (REPEAT CANNOT) assume that the continuous variables are INTEGERS!!!!
@@anastassiosperakis2869 Of course you can. In this case it worked just fine.
Outstanding explantion without delays.
Funny EO2 don't look paralel to CB in your draw, take a bit more care with your draw 😊😊
Velmi pěkný příklad.
Cool ! I've just come back to school 😊. Everything is clear
Explained very well
I just guessed the sides of the triangles. It worked after a bit of fiddling around.
Thanks for this ❤
Good Job!
There's also no need for "normal" quadratic equation. You can consider all lenghts in terms of x and this lenght (a-x), which should be y. Then you ll get equation which yelds 18y^2 = 8x^2, so 3y=2x, and then y=2/3 x (of course only positive solutions) . Now you got lenghts of all sides in terms of x (another words, we know the proportions of all lenghts) :) Amazing how many approaches can be done here :) im curious if there is pure 'visual' solution.
You should have used letters other than a and b for the length and width (l and w, for example) because the letters are also used for the quadratic equation, which can cause confusion.
The confusion of the lesson, with the symbols, can be lessened. But the vessel with the pestle has the brew that is true.
Tuka method- It can be also done using hit and trial method using co-factors of the given three triangle
awesome, many thanks!
so cool, so good. And your voice is very well :-)
I wonder what that second solution is all about.
little bit hard for me to understand but I do believe that constant practice makes a man's perfect
I have prepared myself a formula for such a problem,which enables to find area of any one triangle out of these four triangles if areas of remaining three triangles are known.
Pls share those formulas....
Okay, I got it! Thanks.
Thank you for the daily brain breakfast.
Let consider rectangles instead of triangles. A1 = bx , A2 = ay, A3 = (b-y)(a-x) and A4 = xy. A1, A2 and A3 are known. Then A4*A4 - (A1+A2+A3)*A4 + A1*A2 = 0. This comes from if a rectangle is subdivised by 4 subretangles R1, R2, R3 and R4 (clockwise), then Area(R1) * Area(R3) = Area(R2) * Area(R4)
In this problem, we find A4 = xy = 18 and then ab = 90.
What's so interesting about this particular problem is we never go back and re-substitute; that is, we only arrive at the product (a*b) - - we never solve for either one; or for that matter, for (x) or (y), either.
There is no unique solution for a & b, so you couldn't solve for those anyway. It works for any rectangle where a * b = 90. Also, if you know the area of any 3 of the triangles you can also work out the area of the fourth. All very satisfactory.
In another comment, I created a set of 6 parametric equations to generate any of the infinite real solutions. There are just 4 integer-only solutions, and just 1 with x-dim > y-dim as in the drawing.
very interesting. I'm very out of practice with this kind of question.
I like your method.
Is there a way of getting the values of a and b?
If a and b represent the rectangle dimensions, you can't find them because the are infinite solutions !!! Strange but true...
I found the area of the rectangle being 45, the sides being clockwise 5, 3+6, 2+3 and 9 on top. All the areas of the side triangles were more than 45 and I could figure out how that was possible until I realized that the area of a triangle is HALF the sides multiplication!!!
(ab)^2-108ab+1620まで出てしまえば解の公式を使わなくても
たすき掛けの因数分解で(ab-18)(ab-90)=0が出せますね。
Nice solution.
Glad you think so!
Here's a general solution to find the lengths of each of the 6 outer segments. The inner segments can be found with the Pythagorean formula.
We place the lower left corner on the (0,0) origin for the X & Y axes, give the segments x & y names, and then solve.
The top segment is x; the bottom 2 segments are left x1 & right x2. The right segment is y; the left 2 segments are bottom y1 & top y2. All segments & areas are greater than 0.
We have these 5 equations using 6 variables:
ε1: x = x1+x2 -> x1 = x-x2
ε2: y = y1+y2 -> y1 = y-y2
ε3: x*y2/2 = 27 -> y2 = 54/x
ε4: x2*y/2 = 15 -> x2 = 30/y
ε5: x1*y1/2 = 12 -> ...
(x-x2)*(y-y2) = 24
(x-30/y)*(y-54/x) = 24
(x*y-30)*(x*y-54) = 24*x*y
(x*y)^2 - 108*(x*y) + 1620 = 0
The quadratic equation in x*y is solved as:
x*y = [108 ± √(108^2 - 4*1620)]/2
= 54 ± √(54^2 - 1620)
= 54 ± √(54*(54 - 30))
= 54 ± √(54*24)
= 54 ± √(81*16)
= 54 ± 36
ε6: The area of the rectangle is x*y and the area of inscribed triangle, T, is:
T = x*y - 27 - 15 - 12
= (54 ± 36) - 54
= ± 36
∴ T = 36, x*y = 90.
We may further solve for each symbol:
ε3': y2 = 54/(90/y) = y*3/5
ε2': y1 = y-(y*3/5) = y*2/5
ε4': x2 = 30/(90/x) = x/3
ε1': x1 = x-(x/3) = x*2/3
Thus, we have the relative length of each segment. Let x/3 = i, y/5 = j:
x1 = 2*i, x2 = i, x = 3*i,
y1 = 2*j, y2 = 3*j, y = 5*j.
Parameters i & j are linked by:
ε6: (3*i)*(5*j) = 90 -> j = 6/i
With i > 0, the parametric solution for the 6 outer segments is:
x = 3*i, x1 = 2*i, x2 = i,
y = 30/i, y1 = 12/i, y2 = 18/i.
Although there are an infinite number of real solutions, there are just 4 solutions having only integers. The previous equations in i & j all have integer coefficients; if i & j are integers, all 6 lengths are integers. We choose i to be an integer and a factor of 6, so j = 6/i is an integer, too.
For i = 1, we have:
x = 3, x1 = 2, x2 = 1,
y = 30, y1 = 12, y2 = 18.
For i = 2, we have:
x = 6, x1 = 4, x2 = 2,
y = 15, y1 = 6, y2 = 9.
For i = 3, we have:
x = 9, x1 = 6, x2 = 3,
y = 10, y1 = 4, y2 = 6.
For i = 6, we have:
x = 18, x1 = 12, x2 = 6,
y = 5, y1 = 2, y2 = 3.
The last integer solution is the best fit for the drawing, where x > y.
Good job 💕❤️👌🏿
A more geometric way of understanding this:
Annotate the rectangle as ABCD from top-left corner in clockwise order. Let its area be x.
Annotate the left tip of the shaded triangle as E (E is located on edge AD), the bottom tip as F (F is on edge CD).
Now, connect the opposite apexes of the rectangle as BD.
S(BDE)/S(ABD) = S(BDE)/0.5x = (0.5x-27)/0.5x = (x-54)/x.
S(BDF)/S(BCD) = S(BDF)/0.5x = (0.5x-15)/0.5x = (x-30)/x.
So, DE/AD = (x-54)/x, DF/CD = (x-30)/x.
Thus, S(DEF) = 12 = 0.5*[(x-54)/x]*[(x-30)/x]*x.
Move terms to the left: 24x = (x-54)(x-30).
We get x^2 - 108x + 1620 = 0.
I eyeballed the known-size triangles and realized that their xy proportions appear to correspond to their most obvious denominators. In other words, the area 15 triangle appears to be 3×5, the area 12 triangle appears to be 2×6, and the area 27 triangle appears to be 3×9. Obviously those aren't their actual height and base values, as they would be off by a factor of 2, so I assumed an additional factor of √2 for each dimension, and the math fell into place. The area 15 triangle has height 5√2 and base 3√2, the area 12 triangle has height 2√2 and base 6√2, etc.
From that I ended up with a rectangle of dimensions 9√2 (w) by 5√2 (h), and from there it was fairly simple to determine the rectangle area to be 90 and the blue triangle area to be 36.
We denote the vertices of the rectangle with A F D B
We denote the vertices of the blue triangle with A E C
First:
we draw from C a column (plumb) so it cuts AF at point G and cuts AE at point N
Second:
we draw from point E a straight line parallel to straight line DB then cuts AB at point S and cuts CG at point M
Third:
We draw from point N ((note the point N is already defined as the point of intersection of the line CG with AE))
We draw a line parallel to both ES and DB that cuts AB at point R and cuts FD at point J
CM=MN
because ME is an internal bisector of angle E.
We note that: EMC=EDC=12
and also
NME=NJE=12
Because EMC is half of the rectangle EMCD
And
because NME is half of the rectangle NMEJ It is
clear that
CM=MN
Because ME is the internal bisector of angle E and EMC=EDC=NME=NJE Now
JNFA is a trapezoid whose area is JNFA=AFE-NJE
= 27-12 =15
and
also the area of the ANSM trapezoid is ANSM= AFES-(AFE+NME)
=(2AFS )- (AFE+NME) =2(27) - (27+12) =15 since
ARN=ANG
and
AHMN=JFNA= 15
then
NRMH=FJNG
hence
FG=2AG
JN=2AG
hence
AF=3AG
(AF+JN)×FJ/2=15
(3AG+2AG)FJ=30 5(AG)×FJ=30
AG×FJ=6 =ARNG ARN=AGN=3 AGBC=15×2=30 ANC=AGBC-(ABC+AGN) ANC=30-(15+3)=12
ACE=ANC+NEM+EMC ACE=12+12+12=36 .................................... (marginal note): including If NM=JE=ED=2FJ=2NG=2AR, and because DF=DE+EJ+JF, then DF=5JF .......................... ......................
U’re powerfull
That's one way to do it.
Meanwhile we should be able to solve this in our heads in about 15 seconds.
This video had way too many unnecessary steps.
But, it did work.
Can you make the sizes of the print bigger so it become easier to see the letters
What about the triangle inequality theorem doesnt this violate that?
THANK FOR GOOD EXPLANATION,
Usually we are given the area of the whole rectangle also to calculate the blue shaded area . HOwever, this need not to be including the rectangle’s area and yet there still be such a solution !!!
let's make the multiplication easier, and make the catheti of the bottom-left rectangle as x and y, then we have
1) xy/2=12 2) b(a-x)/2 = 27 3) a(b-y)/2=15.
obviously, 12+15=27, let's add them up and get:
xy+ab-ay=ab-bx
=> bx = ay - xy
=> b/y = (a-x)/x
This means the 2 right rectangles with the areas 27 and 12 are similar rectangles.
so b/y = (a-x)/x = sqrt(27/12) = sqrt(9/4) = 3/2
=> y=2b/3, x = 2a/5
=> xy/2 = 2ab/15=12
=> ab = 90
=> S = ab - 27 - 12 -15 = 36
nice, i kinda did an uneffective way, found factors of the areas - got to 90, subtracted to 36.
Thank you so much for the comment. It is always a good idea to try new methods for solving problems.
@The Math District: I continued the work on the problem to create a set of 6 parametric equations to generate a solution set for a chosen parameter. There are an infinite number of solutions. With further analysis, I found 4 integer-only solutions, of which, only 1 has x-dim > y-dim like in the drawing. I hope you review it!
Beatiful
So will all possible rectangles that inscribe the blue triangle have the same area?
Maybe yes...or not ! Find out why...
Great little problem.
A few quibbles: Writing ay would be more consistent than ya as you are comparing to bx. To be further consistent, why not multiply the third equation by 2 right away, as well? But, I would not have used a and b for the rectangle, given that you are using a, b, c quadratic formula and that could cause confusion with a and b each being used two different ways. On the plus side, sound strategy, so great on that. But you should redo this to overcome these three points... Oh, well...
These types of questions are fun but damn are they lenghty... It felt like i was playing hide and seek with this until i finally got 36
ok. very good. but what happens when you have to solve it within 45 seconds in the Turkish university field exam AYT?
Well explained
Thanks ⚘
The algebra looked too tedious so I tried throwing in some values for side lengths that would make the 3 triangles work out and I came up with 6x9 for the 27 triangle, 4x6 for the 12 triangle, and 10x3 for the 15 triangle (a=10, b=9, x=6, y=3 as diagrammed in the video). So, (assuming there would be one unique solution whatever the actual values of a and b), total rectangle area = 90, blue triangle = 90-27-12-15.
🤣🤣🤣
I also did this method through visualization, and I was so relieved I was right.
This is Math, not Physics or Engineering to make 'assumptions' and 'approximations' as you like 🤦🙄
These arbitrary values must satisfy bx=54, ay=30, (a-x)(b-y)=24
Without knowing in advance the area of the rectangle, how do you know the values must be integers?
Also, do you proceed with trial and error to check that the values satisfy the three equations of the area of a triangle?
Actually, there are an infinite number of real solutions, and just 4 with only integers. Your answer is unlike the drawing which has a longer horizontal dimension than the vertical. In fact, there's only 1 integer solution that has x-dim > y-dim. I posted my general solution separately.
If in the blue triangle we draw a horizontal line and a vertical line through the left and lower vertices respectively, the initial rectangle that contains all the triangles is divided into a grid of 2x2 rectangular cells that we will call, from left to right and from top to bottom , A, B, C and D ⇒ C=2x12=24; A+B=2x27=54; B+D=2x15=30 ⇒ A=54-B; D=30-B ⇒ A/C=B/D ⇒ (54-B)/24=B/(30-B) ⇒ (54-B)x(30-B)=24B ⇒ B²-108B+1620= 0 ⇒ B=18 ⇒ Initial rectangle area = (A+B)+(B+D)-B+C = 54+30-18+24=90 ⇒ Blue triangle area = 90-27-12-15=36
So simple! Cheers.
Wow so compilicated. The top area = 54/2 so you end up with 9x6 for the size of the rectangle, the right triangle fits in a rectangle of 3x10, the bottom left fits in a rectangle of 6x4. So the outer rectangle has a size of 9x10. And then 90-27-15-12 = 36. No need for quadratics at all.
What you showed was that if the area of the triangle is the same for all possible rectangles and triangles with the dimensions shown, then it's 36.
That's not the same thing as showing it's always 36.
I like the way you say a and also like the way you say b
Thank you :)
Чтобы найти синюю площадь наверное можно было найти общую площадь прямоугольника a*b, и от этой площади отнять площади остальных фигур, останется только площадь синей фигуры
i don't see a final answer above. It must be a NUMBER.
@@anastassiosperakis2869 ок
Any Euclidean solution?
Area(Blue Δ) = √{(A+B+C)² - 4.A.B} = √{(27+15+12)² - 4.(27).(15)}
= √{4.(27)² - 4.(27).(15)} = √{4(27)(3)}.√{9-5} = √{4(9)(9)}.√4 = 36.
Here 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.
------------
Let Δ= area of blue triangle, A = area of 27, B = area of 15 and C= area of 12.
Let u=horizontal side of C, v= vertical side of C, a=width, b=height of rectangle.
Then A+B+C+Δ=ab. Also uv=2C=2(12), b(a-u)=2B=2(15), and a(b-v)=2A=2(27).
So bu=ab-2B & av=ba-2A. ∴ 0= (ab-2B)(ba-2A) - ab(uv)= (ab-2B)(ab-2A) - ab(2C).
So (ab)² -2(ab)(A+B+C)+ 4AB=0. ∴ (A+B+C+Δ)² -2(A+B+C+Δ).(A+B+C)+ 4AB=0.
Let E=A+B+C. Then (E+Δ)² -2(E+Δ).E +4AB=0. ∴ E²+2EΔ+Δ² -2E² -2ΔE +4AB=0.
So Δ² = E² - 4AB = (A+B+C)² - 4AB. ∴ Δ= √{(A+B+C)² - 4AB}, since Δ>0.
.
I like this method, but I do not know where you got the first formula for the area of the blue triangle.
In the second part of your comment, you have written 0 = (ab - 2A)(ab - 2B).. How the product can be zero.. could you kindly explain pl.. though I have got the answer using the quadratic method.. but I am wondering how the direct formula has been derived and I saw ur derivation.. but that product step equals 0 am not getting that.. could u pl help @Ramkabharosa
What was written was: ∴ 0 = (ab-2B)(ba-2A) - ab(uv) = (ab-2B)(ab-2A) -ab(2C).
You forgot about the " - ab(uv) " . The rest is algebraic manipulations.
@@Ramkabharosa Thanks a lot I understood ur derivation
15를 보고 3*5를 생각했고 27을 보고 3*9를 떠올렸고 12를 보고 3*4나 2*6을 떠올렸는데 대충 15랑 27에서 떠올린 3,5,3,9를 통해 12=2*6을 떠올렸고 삼각형이니까 모든 숫자에 루트2씩을 곱해주면 가로길이는 9루트2, 세로길이는 5루트2라는걸 알게됐습니다
찍은 거에요?
@@LK_99. 직관이죠 뭐 15보고 3*5라는건 바로 떠올릴수있잖아요
@@broomaster 아하.. 중2인데 이해될듯 말듯 하네요
with guessing 15x2=30=3x10 12x2=24=6x4 27x2=54=9x(10-4) so we can easily see one lane is10, the other is 9, area of rectangle is 9x10=90 blue are is 90-27-12-15=36
this looks like a simple way.
There is no need for such complicated calculations, just use the staring method to get the side lengths of the three triangles together. 9*6, 4*6, 3*10, you can easily calculate the area of the rectangle as 9*10=90. Then the area of the middle triangle is 90-27-12-15=36.
Is the triangle with given sides possible?
great !
Not efficient, you need to utilize the geometry properties
reassign (a-x) as a, and (b-y) as b, let x be (w*a) instead and let y be (k*b)
We know that 27+12 does not equal to half the rectangle, However, 27+12+(k*12) is the same size as half the rectangle
(if you r wondering why, this is pretty much the same as 27+a*(b+k*b)/2 = 27+a*b/2+k*a*b/2, if a*b/2=12, then k*a*b/2=k*12)
By analogy, 15+12 does not equal half of the rectangle, but 15+12+(w*12) does
Thus 27+12+(k*12)= 15+12+(w*12)
k+1=w
(a*b)/2=12 and (w*a)*(k*b+b)/2=27
since w=k+1, (w*a)*(k*b+b)/2=27 is the same as [(k+1)^2]*a*b/2=27
divide a*b from both sides you’ll get (k+1)^2=9/4, thus k+1=3/2, k=1/2
With this, you should be able to solve the area of the blue triangle pretty easily
Can you make the size of the writing a little bigger?
and to simplify, determine the area of the rectangle and minus the existing areas. did it in my head in about 3 minutes, no need for those calculations for this problem. and Im a carpenter
Identical to the way that I did it except for the way I labelled the side of the squares and the triangles. I also called the product of the sides of the rectangle (it's area) "z" in the last few steps to keep the equations simpler.
In any event, it's a very satisfying problem.
Решил в уме за 2 минуты без всех этих расчетов... 27=3x9 12=2x6 15=3x5 3+2=5 5x9=45 45x2=90 27+12+15=54 90-54=36 :)
Or else we could try the integers 😅 yes, Find out the solution. However, we need to assume and set the unknown by method.
You made math more beaultifull than it is..thanks...very creative
Not at all, she made the rest of us feel like 5 year old children, the way she had to explain the OBVIOUS. And Math is BEAUTIFUL EVEN MORE if your tiny brain cannot understand it, but you would not know.
Hey! Maths is stunningly beautiful.
What does ab = 18 mean? Does it have significance?
ab = total area = (27＋15＋12＋blue area) ≠ 18
Done!
a+b=108 and ab=1620. we know a or b must be less the square root of 1620. Then it gets the greatest integer 40. From the following pairs, you could find the clue. (a, b): (40,68), (20,88), (18,90) --- for ab = 1620, one of the pair should be I*10
We could make the solving a bit easier, if, at the end of the solving, we assign: m=a*b.
By this the process is more clear.
Thank you for sharing, well explained
excellent
i solved it in the same way u did
Super! Thank you for watching.
If 27=A, 12=B and 15=C, then the blue area is sqrt((A+B+C)²-4AC).