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.
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.
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
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
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
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.
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.
An easier way: Use a straight line split the "∆15" as 2 pieces: small right∆ and small obtuse∆. Let the height of the small right ∆ is as same as the the "∆12". we call it "∆x", and then we call another obtuse piece "∆(15-x)". And then you can consider that: ∆12 and ∆x could combine to a new triangle as long as ∆27. We have this formula: 27/(12+x)=(15-x)/x 180-24x-x^2=0 x=6 Continue, if we combine more∆: we can found that : 27+12+x = half rectangle. Then (27+12+6)x2 = 27+12+15+∆blue= whole rectangle. ∆blue=36
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.
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
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
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...
At 0:04, Label rectangle ABCDEF in a clockwise direction starting at top LHS of rectangle. Compare areas of triangles. 54 =9*6..................(1) 30=10*3.................(2) 24=4*6...................(3) AB=9 from (1) BC=10 from (2) CD=3 from (2) DE=6 from (3) EF=4 from (3) FA=6 from (1) Width of rectangle AB=9=CD+DE=3+6=9 units Height of rectangleBC=10=AF+FR=6+4=10 units. Area of rectangle ABCE=AB*BC=10*9=90 sq. units. Iblue triangle BFDI=IABCEI-IABFI-IBCDI-IDEFI Iblue triangle BFDI=I90I-I27I-I15I-I12I=I90I-I54I=36 sq.units. and that is the answer.
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.
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 .......................... ......................
Central triangle area is always equal to Square Root of [(X+Y+Z)^2 - 4XY] ie Square root of [(27+15+12)^2-(4*27*15)] = √(2916-1620) = √1296 = 36 Note - Two Triangles have their one common vertex coinciding with the vertex of the rectangle. That's your XY.
I worked out the answer in my head, just by looking at the thumbnail. Here's how: the top triangle has an area of 27 units. Assume that the area of a triangle is length x height instead of length x height / 2. This is just a scale factor and we'll correct for that factor of 2 later. Ignoring this scale factor, the sides of the top triangle are probably 3 x 9 if they're whole numbers, since this equals 27. Similarly, the sides of the other two triangles are 2 x 6 and 3 x 5 respectively. Looking at how these could fit together, it all works if the top of the rectangle has length 9 and the bottom 3 + 6 = 9 as well. The right side of the rectangle will have length 5 and the left side 2 + 3 = 5 as expected. So the area of the rectangle is 9 x 5 = 45. Because of this scale factor of 2, the real linear dimensions will all be √2 times as much as the above figures and the area 2 times as much. So now we need to correct for that missing factor of 2. Therefore the area of the rectangle after correction is actually 2 x 45 = 90 square units. The area of the blue shaded section is therefore 90 - 15 - 27 - 12 = 36 square units.
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.
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.
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.
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.
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.
My solution: Let's call a and b the length of the rectangle and x and y the lengths from the left bottom corner to the points where the little triangles meet each other. So we have, from the formula of the area of the right triangles: a.(b-x)=54 ; xy=24 ; (a-y).b=30. And the area of the blue triangle is just A=ab-54. From the first equation we know that a=54/(b-x) and from the second one we know that x=24/y then a=54/(b-24/y)=54y/(yb-24) Reintroduce that into the third equation: yb.(54/(yb-24)-1)=30 yb.(78-yb)/(yb-24)=30 yb(78-yb)=30(yb-24) (yb)²-48yb-720=0 (yb-24)²=(yb)²-48yb+576 then (yb-24)²=720+576=1296=36². We know that yb-24>0 since 24 is the area of the bottom part of the rectangle we can draw in the bottom left corner from the small triangle, and yb is the area of a bigger rectangle with the little rectangle inside it. Then yb-24=36 then yb=60. But from the third equation we know that ab=yb+30=90 Then the area is 90-54=36
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.
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.
@@Khusbuhasrat I have shown my method by factoring the three rectangles 54,30 and 24. 54=9*6,30=10*3 and 24=4*6. I then manipulated these numbers to calculate the sizes required to to calculate the lengths of the sides of the triangle.I then used the Heron Formula for the sides of the triangle to calculate its area.I have presented two methods of solving this prob.
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.
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
I could have set up some equations and solve it easily, but I found it so easy that I solved it mentally: * Triangle com 15 m² = 3 * 10 / 2 = 15 * Triangle com 12 m² = 4 * 6 = 24 /2 = 12 * Triangle com 27 m² = 6 * 9 = 54 / 2 = 27 Area blue triangle, 90 - 54 = 36 m² Bingo! Saludo from Brazil!!!!!
نرمز رؤوس المستطيل ب A F D B نرمز رؤوس المثلث الأزرق ب A E C أولا: نرسم من C عمود ( شاقول) فيقطعAF في النقطة G ويقطع AE في النقطة N ثانيا: نرسم من النقطة E مستقيم موازي للمستقم DBفيقطع AB في النقطة S ويقطع CG في النقطة M ثالثا: نرسم من النقطة N ((ملاحظة النقطة N معرفة مسبقا بانها نقطة تقاطع المستقيم CG مع AE )) نرسم مستقيم موازي لكلا منESو DBيقطع ABفي النقطة Rويقطع FD في النقطة J CM=MN لان MEمنصف داخلي للزاوية E نلاحظ ان : EMC=EDC=12 وكذلك NME=NJE=12 لان EMC هو نصف المستطيل EMCD ولان NME هو نصف المستطيل NMEJ واضح بان CM=MN لان MEمنصف داخلي للزاوية E وكذلك EMC=EDC=NME=NJE الآن JNFAهو شبه منحرف مساحته هي JNFA=AFE-NJE =27-12=15 وايضا شبه المنحرف ANSM مساحته هي ANSM= AFES-(AFE+NME) =(2AFS )- (AFE+NME) =2(27) - (27+12) =15 بما ان ARN=ANG و AHMN=JFNA=15 فإن NRMH=FJNGومنه بالتالي FG=2AG JN=2AG ومنه 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 .............................................. ( ملاحظة هامشية ): بما ان NM=JE=ED=2FJ=2NG=2AR ولان DF=DE+EJ+JF فان DF=5JF ................................................
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: #1: x = x1+x2 -> x1 = x-x2 #2: y = y1+y2 -> y1 = y-y2 #3: x*y2/2 = 27 -> y2 = 54/x #4: x2*y/2 = 15 -> x2 = 30/y #5: 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 ± sqrt(108^2 - 4*1620)]/2 = 54 ± sqrt(54^2 - 1620) = 54 ± sqrt(54*(54 - 30)) = 54 ± sqrt(54*24) = 54 ± sqrt(81*16) = 54 ± 36 #6: 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: #3': y2 = 54/(90/y) = y*3/5 #2': y1 = y-(y*3/5) = y*2/5 #4': x2 = 30/(90/x) = x/3 #1': 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: #6: (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.
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=NMR=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 .......................... ......................
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.
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
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.
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.
Why go for Quadratic Formula? Basic factoring is an easier way (ab)²-108ab+1620=0 (a-90b)(a-18b) a= 90b, 18b Then of course, we follow the larger value
yep, and 90 = 18*3, which are the only integers that will fit. Of course if they were not integer values, you'd need t do it the way shown on the video.
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.
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.
a+b x c+d, let the rectangle 54=a(c+d), bc=24, d(a+b)=30, (a+b)(c+d)=54/a 30/d=1620/ad, ad-54=-ac, ad-30=-bd, (ad-54)(ad-30)=adbc=24ad, let x be ad, x^2-108x+1620=0, x=18 or 90, the area of the rectangle is 90 or 18, rejected, therefore the answer is 90-27-15-12=36.😊
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!!!
Let's solve this in an entirely general way. We can describe the problem as follows. (1) Impose a standard x/y coordinate system on the problem (2) Rectangle dimensions: width w, height h (3) Rectangle corners: (0, 0), (w, 0), (w, h), (0, h) (4) Target triangle corners: (u, 0), (w, h), (0, v) (5) Area of triangle (0, 0), (u, 0), (0, v): X (6) Area of triangle (0, v), (w, h), (0, v): Y (7) Area of triangle (u, 0), (w, 0), (w, h): Z (8) Area of triangle (u, 0), (w, h), (0, v): a (this is the variable we seek) (9) Define S = X + Y + Z Note that we have used uppercase variables for quantities that are "given" and lower case for quantities that we do not know. So our ultimate goal is to get an expression for a that is in terms of uppercase variables only. We can immediately write the following four relationships: (10) w*h = a + S (11) u*v = 2*X (12) (w-u)*h = 2*Y (13) w*(h-v) = 2*Z This is simple, but as a linear system it has five unknowns: w*h, a, u*v, u*h, and w*v, but there are only four equations. This tells us immediately that no amount of linear algebra will get us to the solution we seek. We have to do something nonlinear. First, expand equations (12) and (13): (14) w*h - u*h = 2*Y (15) w*h - w*v = 2*Z Now form the sum and product (14)*(15) and substitute (10). The product is our nonlinear action: (16) 2*w*h - (w*v + u*h) = 2*Y + 2*Z (17) -(w*v + u*h) = 2*Y + 2*Z - 2*(S + a) (18) (w*h)*(w*h) - (w*h)*(w*v) - (u*h)*(w*h) + (u*h)*(w*v) = (2*Y)*(2*Z) Substitute in (10), (11), and (17): (19) (a + S)^2 - (a+S)*(w*v) - (a+S)*(u*h) + (a+S)*(2*X) = (2*Y)*(2*Z) (20) (a + S)^2 + (2*X - (w*v + u*h))*(a+S) = 4*Y*Z (21) (a+S)^2 + (2*X + 2*Y + 2*Z - 2*S - 2*a)*(a+S) = 4*Y*Z (22) (a + S)^2 - 2*a*(a + S) = 4*Y*Z Finally just expand and simplify: (23) a^2 + 2*S*a + S^2 - 2*a^2 - 2*S*a = 4*Y*Z (24) -a^2 + S^2 = 4*Y*Z (25) a^2 = S^2 - 4*Y*Z (26) a = sqrt(S^2 - 4*Y*Z) This is our final answer. One might ask why Y and Z appear explicitly while X does not, but if you look at the original diagram you see that X is "special" in the sense that it is the only triangle with a corner at the origin. Note that we can always rotate any problem of this type to bring it into this form - to ensure that only one of the four triangles has a corner at the origin. I'm sure it could be solved in any orientation, but as I've played with this a few times I've felt like the algebra is simplified by putting it into this "standard orientation." Finally, we note that this does give the right answer. For our given problem we have S = 54, Y = 15, Z = 27. If you plug those in you do get 36.
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.
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 solved this by inspection and conjecture. 27 = 9 x 3. Assume, therefore, the long side is 9 sqrt(2). 15 = 3 x 5. Assume, therefore, the short side is 5 sqrt (2). Area of the rectangle is thus 9 x 5 x sqrt(2) x sqrt(2) = 90. Area of blue = 90 - 27 -15 - 12 = 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.
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
تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين
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.)
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.
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....
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.
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
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 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.
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
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.
An easier way:
Use a straight line split the "∆15" as 2 pieces: small right∆ and small obtuse∆. Let the height of the small right ∆ is as same as the the "∆12". we call it "∆x", and then we call another obtuse piece "∆(15-x)".
And then you can consider that: ∆12 and ∆x could combine to a new triangle as long as ∆27. We have this formula:
27/(12+x)=(15-x)/x
180-24x-x^2=0
x=6
Continue, if we combine more∆: we can found that : 27+12+x = half rectangle.
Then (27+12+6)x2 = 27+12+15+∆blue= whole rectangle.
∆blue=36
1/2 base X perpendicular height =6X15=90!!
Easy peasy
How can you determine that the height of the smaller half of the bisected triangle is the same as the that of triangle=12?
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.
Excelente. Bastante didático. Muito útil.
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
I like the way you explain everything. Nice video.
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
I like your approach as much as I like your accent. The former is appealing, the latter comforting. Thanks.
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...
At 0:04,
Label rectangle ABCDEF in a clockwise direction starting at top LHS of rectangle.
Compare areas of triangles.
54 =9*6..................(1)
30=10*3.................(2)
24=4*6...................(3)
AB=9 from (1)
BC=10 from (2)
CD=3 from (2)
DE=6 from (3)
EF=4 from (3)
FA=6 from (1)
Width of rectangle AB=9=CD+DE=3+6=9 units
Height of rectangleBC=10=AF+FR=6+4=10 units.
Area of rectangle ABCE=AB*BC=10*9=90 sq. units.
Iblue triangle BFDI=IABCEI-IABFI-IBCDI-IDEFI
Iblue triangle BFDI=I90I-I27I-I15I-I12I=I90I-I54I=36 sq.units. and that is the answer.
The method used by the Math District is probably the best method for this type of problem.
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.
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...
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
Reminded me of my good old school days!! Thanks, Maa'm, for the nice explanation!! 🙏😊
It’s not what you do,it’s the way you do it.Mathematics has Theorems,Equations,,Etc.Simplification is of the Essence.❤️.
Central triangle area is always equal to
Square Root of
[(X+Y+Z)^2 - 4XY]
ie Square root of [(27+15+12)^2-(4*27*15)]
= √(2916-1620)
= √1296 = 36
Note -
Two Triangles have their one common vertex coinciding with the vertex of the rectangle. That's your XY.
Why?
@@SergeyKykov
I have derived it. It is exactly this way. Long procedure to derive
Excellent performance of resolving the problem
Which text editor are you using?
Cool ! I've just come back to school 😊. Everything is clear
I wonder what that second solution is all about.
I worked out the answer in my head, just by looking at the thumbnail. Here's how: the top triangle has an area of 27 units. Assume that the area of a triangle is length x height instead of length x height / 2. This is just a scale factor and we'll correct for that factor of 2 later. Ignoring this scale factor, the sides of the top triangle are probably 3 x 9 if they're whole numbers, since this equals 27. Similarly, the sides of the other two triangles are 2 x 6 and 3 x 5 respectively. Looking at how these could fit together, it all works if the top of the rectangle has length 9 and the bottom 3 + 6 = 9 as well. The right side of the rectangle will have length 5 and the left side 2 + 3 = 5 as expected. So the area of the rectangle is 9 x 5 = 45.
Because of this scale factor of 2, the real linear dimensions will all be √2 times as much as the above figures and the area 2 times as much. So now we need to correct for that missing factor of 2. Therefore the area of the rectangle after correction is actually 2 x 45 = 90 square units.
The area of the blue shaded section is therefore 90 - 15 - 27 - 12 = 36 square units.
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.
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.
What about the triangle inequality theorem doesnt this violate that?
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.
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 💕❤️👌🏿
So will all possible rectangles that inscribe the blue triangle have the same area?
Maybe yes...or not ! Find out why...
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.
little bit hard for me to understand but I do believe that constant practice makes a man's perfect
My solution:
Let's call a and b the length of the rectangle and x and y the lengths from the left bottom corner to the points where the little triangles meet each other.
So we have, from the formula of the area of the right triangles:
a.(b-x)=54 ; xy=24 ; (a-y).b=30.
And the area of the blue triangle is just A=ab-54.
From the first equation we know that a=54/(b-x) and from the second one we know that x=24/y then a=54/(b-24/y)=54y/(yb-24)
Reintroduce that into the third equation: yb.(54/(yb-24)-1)=30 yb.(78-yb)/(yb-24)=30 yb(78-yb)=30(yb-24) (yb)²-48yb-720=0
(yb-24)²=(yb)²-48yb+576 then (yb-24)²=720+576=1296=36². We know that yb-24>0 since 24 is the area of the bottom part of the rectangle we can draw in the bottom left corner from the small triangle, and yb is the area of a bigger rectangle with the little rectangle inside it. Then yb-24=36 then yb=60.
But from the third equation we know that ab=yb+30=90
Then the area is 90-54=36
I did something similar.
Outstanding explantion without delays.
Funny EO2 don't look paralel to CB in your draw, take a bit more care with your draw 😊😊
If 27=A, 12=B and 15=C, then the blue area is sqrt((A+B+C)²-4AC).
(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。
15=3*5; 27=3*9; 12=3*4=2*6
2+3=5; 6+3=9; a=5*2^0,5; b=9*2^0,5
аb=2*5*9=90
90-27-12-15=36
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.
Muy bueno, como los otros que he visto en este canal . 👍
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.
Ok how u solved sides of triangle
@@Khusbuhasrat I have shown my method by factoring the three rectangles 54,30 and 24.
54=9*6,30=10*3 and 24=4*6. I then manipulated these numbers to calculate the sizes required to to calculate the lengths of the sides of the triangle.I then used the Heron Formula for the sides of the triangle to calculate its area.I have presented two methods of solving this prob.
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.
Thank u mam. So good explanation🙏
Is the triangle with given sides possible?
I enjoyed this video, thanks a lot!
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.
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
I could have set up some equations and solve it easily, but I found it so easy that I solved it mentally:
* Triangle com 15 m² = 3 * 10 / 2 = 15
* Triangle com 12 m² = 4 * 6 = 24 /2 = 12
* Triangle com 27 m² = 6 * 9 = 54 / 2 = 27
Area blue triangle, 90 - 54 = 36 m²
Bingo!
Saludo from Brazil!!!!!
Really nice explanation - and a nice problem too! Thank you
THANK FOR GOOD EXPLANATION,
very neat method!
نرمز رؤوس المستطيل ب A F D B
نرمز رؤوس المثلث الأزرق ب A E C
أولا:
نرسم من C عمود ( شاقول)
فيقطعAF في النقطة G
ويقطع AE في النقطة N
ثانيا:
نرسم من النقطة E مستقيم موازي
للمستقم DBفيقطع AB في النقطة S
ويقطع CG في النقطة M
ثالثا:
نرسم من النقطة N
((ملاحظة النقطة N معرفة مسبقا
بانها نقطة تقاطع المستقيم CG مع AE ))
نرسم مستقيم موازي لكلا منESو DBيقطع ABفي النقطة Rويقطع FD في النقطة J
CM=MN
لان MEمنصف داخلي للزاوية E
نلاحظ ان :
EMC=EDC=12
وكذلك
NME=NJE=12
لان EMC هو نصف المستطيل
EMCD
ولان NME هو نصف المستطيل
NMEJ
واضح بان
CM=MN
لان MEمنصف داخلي للزاوية E
وكذلك
EMC=EDC=NME=NJE
الآن JNFAهو شبه منحرف مساحته هي
JNFA=AFE-NJE
=27-12=15
وايضا
شبه المنحرف ANSM مساحته هي
ANSM= AFES-(AFE+NME)
=(2AFS )- (AFE+NME)
=2(27) - (27+12)
=15
بما ان
ARN=ANG
و
AHMN=JFNA=15
فإن NRMH=FJNGومنه
بالتالي
FG=2AG
JN=2AG
ومنه
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
..............................................
( ملاحظة هامشية ):
بما ان
NM=JE=ED=2FJ=2NG=2AR
ولان
DF=DE+EJ+JF
فان
DF=5JF
................................................
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:
#1: x = x1+x2 -> x1 = x-x2
#2: y = y1+y2 -> y1 = y-y2
#3: x*y2/2 = 27 -> y2 = 54/x
#4: x2*y/2 = 15 -> x2 = 30/y
#5: 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 ± sqrt(108^2 - 4*1620)]/2
= 54 ± sqrt(54^2 - 1620)
= 54 ± sqrt(54*(54 - 30))
= 54 ± sqrt(54*24)
= 54 ± sqrt(81*16)
= 54 ± 36
#6: 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:
#3': y2 = 54/(90/y) = y*3/5
#2': y1 = y-(y*3/5) = y*2/5
#4': x2 = 30/(90/x) = x/3
#1': 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:
#6: (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.
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=NMR=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 .......................... ......................
This is the most complete and exhaustive solution
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....
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
Tuka method- It can be also done using hit and trial method using co-factors of the given three triangle
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.
Can you make the sizes of the print bigger so it become easier to see the letters
I just guessed the sides of the triangles. It worked after a bit of fiddling around.
Can you make the size of the writing a little bigger?
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.
Why go for Quadratic Formula?
Basic factoring is an easier way
(ab)²-108ab+1620=0
(a-90b)(a-18b)
a= 90b, 18b
Then of course, we follow the larger value
very interesting. I'm very out of practice with this kind of question.
Thanks for this ❤
The area of the rectangle is 90 minus the areas of the three triangles given as 27, 15, and 12 so 90-54 =36
yep, and 90 = 18*3, which are the only integers that will fit.
Of course if they were not integer values, you'd need t do it the way shown on the video.
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.
What does ab = 18 mean? Does it have significance?
ab = total area = (27＋15＋12＋blue area) ≠ 18
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.
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 :)
you should have swapped x and y... horizontal is usually x, and vertical is usually y...
Any Euclidean solution?
It's weird that you can just guess x or y and it all works. if y=3, x=6. if y=5, x=3.6
محاضره جدا راقيه اشكرك ياأستاذه الورده 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
ok. very good. but what happens when you have to solve it within 45 seconds in the Turkish university field exam AYT?
so cool, so good. And your voice is very well :-)
Hi, may I know is for which level? For primary or secondary? Please let me know. Thanks
Good Job!
Thank you for the daily brain breakfast.
a+b x c+d, let the rectangle 54=a(c+d), bc=24, d(a+b)=30, (a+b)(c+d)=54/a 30/d=1620/ad, ad-54=-ac, ad-30=-bd, (ad-54)(ad-30)=adbc=24ad, let x be ad, x^2-108x+1620=0, x=18 or 90, the area of the rectangle is 90 or 18, rejected, therefore the answer is 90-27-15-12=36.😊
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!!!
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.
Nice simple solution.
At first glance it looks like there are too many variables and too few equations.
Explained very well
Let's solve this in an entirely general way. We can describe the problem as follows.
(1) Impose a standard x/y coordinate system on the problem
(2) Rectangle dimensions: width w, height h
(3) Rectangle corners: (0, 0), (w, 0), (w, h), (0, h)
(4) Target triangle corners: (u, 0), (w, h), (0, v)
(5) Area of triangle (0, 0), (u, 0), (0, v): X
(6) Area of triangle (0, v), (w, h), (0, v): Y
(7) Area of triangle (u, 0), (w, 0), (w, h): Z
(8) Area of triangle (u, 0), (w, h), (0, v): a (this is the variable we seek)
(9) Define S = X + Y + Z
Note that we have used uppercase variables for quantities that are "given" and lower case for quantities that we do not know. So our ultimate goal is to get an expression for a that is in terms of uppercase variables only. We can immediately write the following four relationships:
(10) w*h = a + S
(11) u*v = 2*X
(12) (w-u)*h = 2*Y
(13) w*(h-v) = 2*Z
This is simple, but as a linear system it has five unknowns: w*h, a, u*v, u*h, and w*v, but there are only four equations. This tells us immediately that no amount of linear algebra will get us to the solution we seek. We have to do something nonlinear. First, expand equations (12) and (13):
(14) w*h - u*h = 2*Y
(15) w*h - w*v = 2*Z
Now form the sum and product (14)*(15) and substitute (10). The product is our nonlinear action:
(16) 2*w*h - (w*v + u*h) = 2*Y + 2*Z
(17) -(w*v + u*h) = 2*Y + 2*Z - 2*(S + a)
(18) (w*h)*(w*h) - (w*h)*(w*v) - (u*h)*(w*h) + (u*h)*(w*v) = (2*Y)*(2*Z)
Substitute in (10), (11), and (17):
(19) (a + S)^2 - (a+S)*(w*v) - (a+S)*(u*h) + (a+S)*(2*X) = (2*Y)*(2*Z)
(20) (a + S)^2 + (2*X - (w*v + u*h))*(a+S) = 4*Y*Z
(21) (a+S)^2 + (2*X + 2*Y + 2*Z - 2*S - 2*a)*(a+S) = 4*Y*Z
(22) (a + S)^2 - 2*a*(a + S) = 4*Y*Z
Finally just expand and simplify:
(23) a^2 + 2*S*a + S^2 - 2*a^2 - 2*S*a = 4*Y*Z
(24) -a^2 + S^2 = 4*Y*Z
(25) a^2 = S^2 - 4*Y*Z
(26) a = sqrt(S^2 - 4*Y*Z)
This is our final answer. One might ask why Y and Z appear explicitly while X does not, but if you look at the original diagram you see that X is "special" in the sense that it is the only triangle with a corner at the origin. Note that we can always rotate any problem of this type to bring it into this form - to ensure that only one of the four triangles has a corner at the origin. I'm sure it could be solved in any orientation, but as I've played with this a few times I've felt like the algebra is simplified by putting it into this "standard orientation."
Finally, we note that this does give the right answer. For our given problem we have S = 54, Y = 15, Z = 27. If you plug those in you do get 36.
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.
К1, К2, К3, К4, К5, К6 - катеты треугольников 15, 12, 27 по часовой стрелке. Треугольник 15. К1 × К2 / 2 = 15. К1 × К2 = 2 × 3 × 5. К1 = 2 × 5 = 10; К2 = 3. Треугольник 12. К3 × К4 / 2 = 12. К3 × К4 = 2 × 2 × 2 × 3. К3 = 2 × 3 = 6; К4 = 2 × 2 = 4. Треугольник 27. К5 × К6 / 2 = 27. К5 × К6 = 2 × 3 × 3 × 3. К5 = 2 × 3 = 6; К6 = 3 × 3 = 9. Равенство противоположных сторон прямоугольника: К1 = К4 + К5 = 10 = 4 + 6; К6 = К2 + К3 = 9 = 3 + 6. Площадь прямоугольника равна 9 × 10 = 90. Площадь синего треугольника равна 90 -- (15 + 12 + 27) = 36
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 solved this by inspection and conjecture. 27 = 9 x 3. Assume, therefore, the long side is 9 sqrt(2). 15 = 3 x 5. Assume, therefore, the short side is 5 sqrt (2). Area of the rectangle is thus 9 x 5 x sqrt(2) x sqrt(2) = 90. Area of blue = 90 - 27 -15 - 12 = 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.
I would have measured the sides with a ruler.
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
I like the way you say a and also like the way you say b
Thank you :)