Can you solve for the area?
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- čas přidán 18. 05. 2024
- Problem 1: What is the area of the inscribed square to the right triangle? Problem 2: there are 4 triangles whose side lengths are given, and the areas of 3 triangles are known. What is the area of the 4th triangle? Thanks to Nifty who received the problem from Aditya!
0:00 problems
1:24 solution 1
4:00 solution 2
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@noahtaul found a counter-example to problem 2 with a=9.73712, b=3.88159, c=3, d=5.11613 with final area equals 7.67141. These triangles are obtuse which makes those opposite sides NOT parallel. So question: will the answer be the same for acute triangles? While the question was incomplete and wrong as stated, any student in math knows that mistakes have inspired many interesting discoveries. So what's the proper condition for this question?
dont care lmao
Bro it is easy br s similar triangle because in similar triangle if we noticed the sides are multiplying and the equal will be the area of the square so it would have the area 100. Made by kushagra 10 year
if we assume that we can build all 4 triangles into 1 figure (sum of angles at common point are 360 degrees), then we get a quadrilateral with equal sides. And that's a rhombus! And its sides are parallel. And so this particular geometric case (when we assume that triangles are folded into one figure) - will have only one solution, and it is a square.
The case where the triangles cannot be folded into a single figure - is the main problem.
I think the original problem has an incomplete description, otherwise it cannot be solved geometrically and need a hard math
I believe this topic could be interesting for a follow-up video. I am currently working on a construction that I believe can produce infinite possible solution sets of acute triangles, though I haven't quite proven it yet. Should I send an email to your business inquiries email with a more detailed description of this construction than a youtube comment can give if my conjecture is correct?
update: been a fun 5 hours slowly realizing just how much more complicated it was to prove that my construction exists. That said, I believe I now am extremely close to proving that there are infinite solution sets with triangles and even closer to producing a specific acute counter example which would serve as either the upper or lower bound on what the area of the fourth triangle can be using only acute triangles; I have lengths for a, b, and d, I just need to solve one system of equations to get c. The problem is, I don't know how to solve this system of equations, as it's in a form I'm unfamiliar with, with one of the variables appearing exclusively inside trig functions and the other being the coefficient of some of the terms containing said trig functions.
All that having been said, I am confident that there exists some angle θ (probably somewhere between 2π/3 and π) such that the following values produce a counterexample:
a = 5/3
b = sqrt(25/9+28+6sqrt(3))
c = sqrt((5/3+6sin(θ))^2+((1+3sqrt(3))+6cos(θ))^2)
d = sqrt(241)/3
As I don't know the value of theta yet, I also don't know what the area comes out to, but I am confident that it is not 14
In the second solution, it is not obvious that we have parallel sides at the top and bottom. The angles between the sides "6" of the triangles can be any, not necessarily 90 degrees.
I have the same remark
The logic in the video holds to show the side length in the construction should be 6, and that gets us a kite, so the sides are definitely parallel but we are very much missing showing it’s a square
@@centryll1313you are assuming the four triangles form a closed quadrilateral at all, which is not clearly demonstrated
I think it's already established. When he created the heights of the triangles, he used perpendicular heights. If the two heights aren't arranged properly so as to make a single segment, just ignore the triangle on the right entirely and arrange the top and bottom triangles so that these heights are collinear. By construction, the top and bottom sides are perpendicular to the segment formed by the two heights. Therefore they are parallel. Now, draw the sides in on both the left and right. By equidistance of parallel lines, the perpendicular distances we just drew are both 6, which he showed to be that equidistance value with his calculation. The triangle on the right has an established area (SSS congruence for rigor), and the rest of the proof holds.
Also at the common point of all triangles the sum of angles can be different from 360 degrees.
In general, the second solution is based more on visual coincidence, but not on geometry, math, or logic
I did the first problem with pythagorean theorem, definitely one of the longest ways to tackle it
Same, well I tried and didnt finish. The solution was so much simpler😂
@@jeremy6384 simpler, but wrong.
@@samueldeandrade8535 how is the solution wrong? by solution I am talking about the solution provided in the video.
@@jeremy6384 badly justified. It is assumed that the triangles together make a square.
@@samueldeandrade8535 No assumptions were made. The directions said to "construct an inscribed square" so... from that we knew it was a square.
the two green triangles are similar so:
x/20 = 5/x
x² = 5 × 20
x² = 100
I solved it the same way. The solution in the video is just too complicated and this solution is a lot better.
I also did this took me 10 seconds
That was the video solution.
Am I the only one to look at problem and said tangent?
THIS! (took me less than 10 seconds to solve this way).
There is flaw in second q.s as many have indicated: There is infinite solution, and that's why heron didn't help you: You place the first triangle in some position, we only know one side length, and area, so basically the third point of first triangle can vary anywhere in line joining first two points that are 6cm apart to give area as 13,this variation allows to change dimensions/orientation of other triangles, you have constructed a square, but it need not be only solution, there can be rhombus, or anything.. 4th side is not 6cm,proof is flaw
100. Similar triangles. tan@ = X/20 ; tan@ = 5/X ;
tan@ drops out, and X/20 = 5/X ; ==> X^2 = 100
The square could even be a rectangle.
@@robertvaneersel3741wdym? All squarea are already a special kind of rectangles
@@robertvaneersel3741 No , it is mentioned in the problem that it is a square
that's what i did too
You can solve the first problem in one step by using similar triangles and setting up an equation 5/x=x/20.
Yup 👍
That's what I did. Took me 20 seconds in my head.
That’s literally the first solution the video offered…
@@kylen6430These problems are posed as challenges. A lot of people try to post an answer first if they can, and then watch the video.
@@Sam_on_CZcams ok. So pause the video, solve it, check your answer and move on. Posting the answer contributes nothing and Nobody is patting this person on the back for commenting. Maybe this person is just that insecure and needs the validation 🤷🏼♂️
While I also used the „textbook“ approach for solving a^2=100 in 5 sec by using proportionate triangles, I find the alternative „out of the box“-solution quite impressive! I admit I wasn‘t able to solve the second question, because I didn‘t realize that you can group the triangles into a square. Though I had the idea to calculate the heights of the triangles, and this should have let me realize that two of the heights as up to six, which could have let me to the solution. So I give myself at least some credit for the height-idea…!😅
Never give yourself credit - you will end up being in debt to yourself!
I solved the first problem by constructing a diagonal of the square, observing that it bisected the right angle of the right triangle, and then using the theorem that states that a line segment that bisects one angle of a triangle divides the opposite side into segments whose lengths have the same proportion as the lengths of the other two sides. The algebra was messier, but I got the same solution as you did.
Two lines, both perpendicular to the same line segment, are parallel.
And, if another line segment between the two parallel lines has the same length, then the two line segments are also parallel because they together make a parallelogram.
For the first problem I did something a little different. I put the lower left vertex of the triangle on the origin of a graph. Then the hypotenuse can be described by some linear function. The upper right vertex of the shaded square must have coordinates where the x-coordinate equals the y-coordinate. I plugged everything into y=mx+b and solved the resulting quadratic.
Definitely took the long way around there.
Outside the box thinking is amazing 👏 🙌 😍
but it's not as simple as the textbook solution so that makes the textbook solution the better solution.
@@solidpixel but we all should appreciate this method too because it involves imagination
@@Nothingx303 ... but takes more time - I always appreciate the shorter solution.
I actually solved the first problem in a way more complicated way.
I considered that there were 2 hypothenuses, the one of the upper triangle, and the one of the lower, I called the upper triangle's one D, and the lower triangle's one F, and the hypothenuse of the big triangle I labelled it as C, so that means that C=D+F, and at the same time, I also calculated both hypothenuses using the Pythagorean theorem, so I calculated that D=√(25+x^2) and F=√(400+x^2), and then I also calculated the hypothenuse of C doing this (5+x)^2+(20+x)^2=C^2, and after simplifying it, it became C=√(2x^2+50x+425), after that I realized that I had 2 equations both equal to C, one being C=D+F (it's valid because I know that D=√(25+x^2) and F=√(400+x^2) so C=√(25+x^2)+√(400+x^2)) and C=√(2x^2+50x+425), I didn't known if they were redundant at the time, however, I still took the risk and decided to do √(2x^2+50x+425)=√(25+x^2)+√(400+x^2) and solve for X, and what do you know, turns out they weren't redundant and I got X=10 as solution, which squared, gives 100, the solution for the area.
I also did that!
@@hypersonicrush Lol I thought I was the only one who did that since it took a lot of time out of me, good to know I wasn't the only one
Shortest solution
5/x = x/20 because green and black triangles are similar (have same angles)
100/x = x
100 = x^2
x=10
The second problem. The triangles form a square when arranged together. Can’t see how you prove this. I must/be missing something obvious!
In the first soln , let the side of the square be x
Then the area of small upper triangle = 1/2 × x × 5
Area of square = x²
Area of lower big triangle = 1/2 × x × 20
Now if we add all these areas so we know the area is equal toh the whole big triangle whose area would be 1/2 base × height i.e 1/2 × (5+x) × (20+x)
Now,
(1/2 × 5x) + s² + 10x = 1/2 × (5+x) × (20+x)
On solving the equation we get x=10
x² =100 = area of the square
its much easier with trignometry tan(theta) = 5/x and tan(theta)=x/20. where x is the side of the square, then x^2=100. (area of the square=100)
Problem 2 : I can't see where you demonstrated that the "h" and the "k" heights are colinear : without it, you couldn't say that the two opposite sides are parallel, and you couldn't say either that you may correctly add their lengths. Did I miss anything ?
I don't think you've missed anything; I have the same problem with this proposed solution.
I think the drawing is wrong: b and d cannot be on the same line. If they are, then a and c must be equal and the two areas cannot be 4 and 5.
By stacking the two triangles so that their bases are parallel, their heights, which are perpendicular to their bases, must be colinear.
@@dr.johnslab7502 What allows us to think that their bases are parallel ? We should have (or get) information about some of the triangles angles, and we have none, so nothing can make us assume that the bases are parallel once the equal-length sides are "merged".
@@R_V_ I'm saying that if you stack any two triangles so that their bases are parallel, then their heights would have to be colinear.
I did in a way without involving similar triangles and out-of-box thinking (because I forgot about the similar triangle theorem).
1. Found the area of the square alone = x²
2. Found the area of the triangle with side 5 = 5x/2=2.5x
3. Found the area of the triangle with side 20 = 20x/2=10x
4. Added all that up for the area of square = x² + 12.5x
5. Added the sides of the triangle and multiplied them and divided by half to find area = [(x+5)(x+20)]/2
=(x²+25x+100)/2
We know the area of the two triangles and the square is equal to the big triangle, so;
(x²+25x+100)/2=x²+12.5x
=>x²+25x+100=2x²+25x
(Subtracting similar terms from both sides)
=>100=x²(2-1)+(25-25)x
=>100=x²
=>x²=100(Ans)
I solved the first problem sort of the long way: by setting it up as an algebra problem and setting the combined areas of the triangles and square to the area of the triangle as a whole. I assigned x to be the side of the square. The problem came out as x^2 + 0.5*5*x + 0.5*20*x = (5+x)(20+x)*0.5. The left side is the combined areas of the triangles and square and the right side is the triangle as a whole. Just work out of the problem, and you get x^2 = 100, which is the answer.
That's the exact approach I used. It wasn't a very long way to do it, but the solutions in the video are certainly shorter.
I am not sure of the reliability of the solution to the second problem…
Considering the triangles with given areas, I suggest a little more labelling:
Triangle 1 (with area=4) - label angles A1 (opposite side a) and B1 (opposite side b)
Triangle 2 (with area=5) - label angle C1 (opposite side c)
Triangle 3 (with area=13) - label angle D1 (opposite side d)
The core assumption in setting out the solution seems to be that the A1, B1, C1 and D1 sum to at least 180 degrees - the solution seems to rely entirely on this assumption with the evaluation of the heights of Triangles 2 and 3. That is, if those angles sum to more than 180 degrees, the sum of the two triangle heights would be greater than 6 and, assuming a four-sided shape with equal sides, a triangle height sum of 6 would only be possible for a square. However, can the core assumption also be proven (namely that the sum of those angles is not less than 180 degrees)?
I wonder if this is resolved by specifying that the lengths a, b, c and d must all be less than (or equal to?) 6 - it occurs to me that a smaller total for the above-mentioned angles would result in a shorter available outer edge length for the fourth triangle, which would likely require lengths c and d to be somewhat longer…
Really very interesting solution
100
The difference between A and B on both wedges need to be the same;
1a = 5
2b = 20
1b = 2a
I’m arbitrarily picking a number that’s between the two known values;
If
1b = 2a = 10
Then
Wedge 1;
a = 5, 5 * 2 = 10 = b
Wedge 2;
b = 20, 20 / 2 = 10 = a
So;
1b = 10 = 2a
Thus;
For square
x = 10
y = 10
10^2 = 100
for the first one because the triangles are similar i set up the equation 5k = 20/k, and then solved that k = 2, and got each side length as 2
I don't wanna try problem 2, rather it may be obscure and too much time-consuming so I'll try only problem 1.
Wow, what an easy question this !!! 😊
Solved it orally now, by equating the sum of areas of each portions of triangle, with full area of triangle using (0.5 × base × height)
x^2 + 12.5 x
= 0.5 x^2 + 12.5 x + 50
Solving further,
Area of square, x^2
= 100 sq. units ☺️
One objection to the second problem is that the solution is based on the assumption that these four triangles will make a square. It's good that you did the check to prove that a square was formed by putting three triangles together. But there is no reason to assume, in advance, that this would be true.
A way to counter this objection would be to include some key angles in the original question -- such that complimentary angles may be noticed and give justification to the assumption that they might be able to form a square.
Let the unknown side to be x.
The area of the large triangle is (5+x)(20+x)/2
The area is also the sum of areas of the square and two triangles: x^2+5x/2+20x/2
Equate both expressions and solve x^2=100.
I did the same thing! Much easier
Good
in problem 1 we can use the thales theorem.
Let the horizontal side be x and vertical side be y.
now we have
(5/5+y) = (x/20+x)
100+5x=5x+xy
xy=100
Thanks
For me, the solution for the 2nd problem, it's not right. I intuitively verified that the heights indicated as adding up to 6 (in first 2 triangles assembled), even though they have this property, do not need to be aligned. Furthermore, 2nd. triangle positioned through side A doesn't need to have side 6 in vertical.
To show this, I made a drawing with the 1st. triangle with sides D and A with area 13, with base 6 below, and height (13/3) coinciding with side D (right triangle). So the second triangle (side 'A' in common) can be drawed rotating side 6 and height 4/ 3 so that the height was very close to the upper vertex. As for the 3rd. triangle (side 'B' in common), so that side 6 would be a little to the right of side 6 of the 2nd. triangle, forming an angle to the right slightly less than 180o.
This all stems from the fact that the problem is not deterministic. The 1st. triangle has great drawing freedom, because we only know the base 6 and the height 13/3, although the 2nd. and 3rd. triangle are deterministic (2 known sides and 1 known height), many different triangles with different areas are generated, composing a totally irregular quadrilateral.
I did the exact calculation with these assumptions, just using Pythagoras and the law of cosines, in the order of the triangles added to the proposed solution:
The side A of the 1st. triangle is =ROOT((13/3)^2 +6^2)~ 7.401. One starts from side 'A' relative to the 2nd. triangle is ROOT(6^2-(4/3)^2) ~ 5.85, so the remainder of side 'A' is 1.55. So side 'B' is ROOT(1.551^2+ (4/3)^2) ~ 2.046. In the 3rd. triangle, a part of side 6 is ROOT(2.046^2-(5/3)^2) ~ 1.186, so the rest of side 6 is 4.814. Therefore, side 'C' will be ROOT(4.814^2+(5/3)^2) = 5.094. Now the angles: the sine of the angle of sides 'D' and 'A' of the first triangle is 6/7.401 ~ 0.8107 which gives ~ 54.16 degrees. By the law of cosines, cos of the angle between 'B' and 'A' in the 2nd. triangle = (6^2-7.401^2-2.045^2)/(-2*7.401*2.046) = 0.7584, which gives an angle of 40.68 degrees. Finally the angle between sides 'C' and 'B' of the 3rd. triangle =(6^2-2.046^2-5.094^2)/(-2*2.046*5.094) = -0.281, so the angle is 106.34 degrees. Adding the 3 angles and taking what is missing for 360 degrees, we have the angle of sides 'C' and 'D' of the 4th. triangle is 158.82 degrees, which corresponds to 2.772 radians. Using the law of cosines for the 4th. triangle we have that the unknown side is the ROOT((13/3)^2+5.094^2-2*(13/3)*5.094*COS(2.772) = 9.27. Applying Heron's law, the area of the 4th triangle gives only 3.988 and not 13.
If I applied a different assumption to draw the first triangle, I would definitely arrive at a different area for the 4th. triangle.
when you draw 4 lines from 1 points( these lines have lenght a b c d) the only figures you can make are any quadrilateral with 4 triangles with different 3th sides, But the only way to this 3th side has the same lenght( in this problem 6) is for Square or Rhombus. Maybe you must consider the Rhombus case also
For 1st problem: 5 / x = 5+x / x+20 Crossmultiply to get x=10.
Yes by using trigonometry tan`A = 5/x for smaller triangle and also tan`A = 5+x/20+x equating both equations we get x²=100
Use trig - since triangles are similar we know that 5/x = x/20 (tangents of the small angles). Simple to solve x = 10 and area = 100
Problem 1:
The two small triangles are clearly similar to each other and to the larger triangle. They each share an angle with the larger triangle, and the opposite angles on the larger triangle hypotenuse are separated by a 90° corner of the square.
20/s = s/5
s² = 100
s = 10
Confirm:
(20+10)/(5+10) = 20/10 = 2
30/15 = 2 ✓
Area = 100
Problem 2:
A = bh/2
4 = 6h/2 = 3h
h = 4/3 (blue - ab)
A = bh/2
5 = 6h/2 = 3h
h = 5/3 (green - bc)
A = bh/2
13 = 6h/2 = 3h
h = 13/3 (yellow - da)
A = bh/2
x = 6h/2 = 3h (orange - cd)
Each lettered side is present exactly twice, and the bases are all the same (6). Is conceivable that they fit together in a square, as the orientations match. Leaving the blue ab triangle as the base, we can rotate the green bc triangle 90° ccw and the b's match, and we can rotate the yellow da triangle 90° cw and the a's match. If we then rotate the orange cd triangle 180°, both the c's and d's match, but we're not confirmed yet. We need to make sure the heights match up as well, and the only two given opposing heights are for the yellow and green triangles.
5/3 + 13/3 = 18/3 = 6 ✓
This means that the height for the orange triangle can be determined from the blue height, and the orange area x can be determined either from that or from the total area of a side 6 square (6² = 36). We'll do the former since we already have the equation.
h + 4/3 = 6
h = 6 - 4/3 = 14/3
x = 3h = 3(14/3) = 14
1. tan(a/20) = tan(5/a) ; a/20 = 5/a ; a^2 = 100 ; ? = 100 area of square
2. Align triangles into square ; 6^2 = 36 area of a square ; 36-13-4-5=14 area of triangle
For the second one, I thought the difference in the first two triangles must be the same for the last two. So I just randomly said 14, and I was shocked to see that I was right.
Another way for the first one:
Area of the triangle= area of the small triangle with 5 as height and x as base + area of the larger teiangle with 20 as base and x as height + area of square. You get 25x+2x²=100+25x+x², so x²=100
in the first one you could just add the areas of the two smaller triangles and the square and put it equal to the area of the big triangle. x would come 10 easily
Don't like answer 2. Don't see why the corners should be right angles.
Doesn't matter if it's a right angle or not. It's easier to demonstrate with a perfect square, but the math works out the same if it's a parallelogram, because the area is still 36, just shaped differently.
@@alanlafond9705 I can see no reason why h and k form a straight line. Maybe they do, but PT is just assuming it, not proving it.
@@stevebailey805 They don't need to form a straight line. If it's square instead of parallelogram, they do form a straight line, due to all the abcd segments meeting at the same spot. However, in a parallelogram, even if the h and k segments are offset from each other, they only need to have a combined height of 6 for it to work.
The line segments share a point (the common vertex in the image). If the shape is a parallelogram, then h and k would be colinear, since they share a point and are both perpendicular to parallel lines. However, we haven't shown that the shape in question is a parallelogram, much less that the sides are parallel, so we don't know if h and k are colinear or are 2 sides of a new smaller triangle.
Nice solution to prob1
I solved the first problem the same way you did using similar triangles.
I found a much simpler solution to the second problem.
All the triangles have one side that is 6 inches.
I put the triangles with areas 5 and 13 together at the 6 inch sides to make a four sided polygon with sides a, b, c, and d.
This four sided polygon has an area of 5 + 13 = 18.
I put the triangles with areas 4 and unknown area (I'll call it x) together at the 6 inch sides to make a four sided polygon with sides a, b, c, and d. This four sided polygon has an area of 4 + x. This 4 sided polygon is the same as the first 4 sided polygon, because they both have the same sides a, b, c, and d, in the same orientation, so they have the same area.
Therefore, 4 sided polygon abcd area = 4 + x = 18
and x = 14.
the triangles are similar, so you can set the ratios of the sides equal to each other, solve for x^2
First triangle area is 5x/2 , 2nd triangle area is 20x/2. Square area is x². The area of the big triangle is
(5+x).(20+x)/2=(x²+25x+100)/2. This area is equal to the sum of the 3 areas 20x/2+5x/2+x² so there you have it x²=100
Μόνο εμείς οι Έλληνες βρίσκουμε το πιο εύκολο τρόπο ;-)
@@giovannicole5561 😝
100.
If the side of the square is x.
5/x=x/20 because the 2 small triangles are analog.
x²=100. One positive solution x=10
Area x²=100.
Little check: 5/10=10/20=½
Also (5+x)/(x+20)=15/30=½
The big triangle has the same ratio.
You can solve it by creating a qudratic equation. That is how I first solved it. (x^2= {[(5+x)(20+x)]/2}-((5x/2)+10x)). Therefore, x = 10.
Solved first one with 5x * 0.5 + 20x*0.5 + x^2 =0.5(10+x)(5+x) which is summ of two areas of triangles and a square equales to area of big triangel that has sides x+20 and x+5
Nice
A more intuitive way is to say that the area of the big trangle is equal to the sum of the area of the square and of the two smaller triangles.
So you have
(x+20)(x+5)/2 = x^2 + 5x/2 + 20x/2 =>
(x+20)(x+5) = 2x^2 +5x + 20x =>
x^2 + 5x + 20x + 100 = 2x^2 + 5x + 20x =>
x^2 + 100 = 2x^2 =>
x^2 = 100
This problem is solved by similarity of triangles
& by this theory the area of shaded part should be 100 sq. unit for the 1st figure
20/x = x/5
x^2 = 100
Got the same answer. Fwiw, the slope is. 26.56 degrees. Figured it would be a round number for the box size. A little trial and error. You could cheat with the goal seek function in a spreadsheet and come up with the right answer the first time.
I did x/20 = (5+x)/(20+x) to find x^2=100. Obviously, using the two smaller triangles to solve the problem is more elegant and easier to calculate the answer. Anyway, I found the solution to the problem quickly and that is what counts, isn't it? 😃
The second task isn't complete. The area of the last triangle actually varies depending on side lengths from 14.0 to ~14.734.
How can u please elaborate
The area can't be more than 14. those four triangles creates rhombus, so maximum combined ares is 36 (for square) and less than 36 for any not square.
You can also solve the second problem with:
(A + B) / 6 = 4
A + B = 24
B = 24 - A
And then
(B + C) / 6 = 5
B + C = 30
C = 30 - (24 - A) = 6 + A
And then
(D + A) / 6 = 13
D + A = 78
D = 78 - A
And then
(D + C) / 6 = ? lets make ? = x
D + C = 6x
(78 - A) + (6 + A) = 6x
84 = 6x
14 = x
So ? = 14
For the first problem I got so fixated on solving the triangles that I derived x=100/y (from TOA on both triangles) and didn't put together that I solved the problem for like 5 more minutes. Needless to say solving the other box didn't occur to me haha.
Concerning problem 2, Presh didn't mind his decisions.
There is an easier solution for the first one.
We get two triangles that have the same slope. So their height/width has to be equal. If we say the squares side is x, that öeans 5/x=x/20 solve for x you get 10.
Excuse me, but please check a=4.9286, b=1.83119, c=5.5, d=5.66697, with area 14.1217. I think the second question is impossible to answer
You are correct that the second problem has no unique answer. There are a range of values for b that work, resulting in a range of values for a, c, d, and also a range for the area.
I wrote the equations for Heron's formula for the 3 triangles with known areas, and I ended up with 3 equations and 4 unknowns. I took b as a parameter and solved for a, c, d and then used c and d to compute the area for the 4th triangle. The algebraic expression is very long and messy, so I will just give the numerical range:
the range of areas for the 4th triangle is from 14 at b=2.1343748 to 14.734 at b = 5/3
This is assuming acute triangles. If obtuse angles are allowed, then there may be a greater range (I did not check that).
By the way, the numbers given in your comment for b=1.83119 agree exactly with what I computed when choosing the negative signs in the formulas below (corresponding to acute triangles with no obtuse angles)
a^2 = b^2 + 36 +/- 4*sqrt(9*b^2 - 16)
c^2 = b^2 + 36 +/- 4*sqrt(9*b^2 - 25)
d^2 = b^2 + 72 +/- 4*sqrt(9*b^2 - 16) +/- 4*sqrt( 9*b^2 + 155 +/- 36*sqrt(9*b^2 - 16) )
You can use algebra to solve the 1st problem. 1/2*x*5+x*x+1/2*20*x=1/2*(x+20)*(5+x) by solving the equation to get the area of square is 100.
Lol I did it the same way😂
Piece of cake. The general case: let the square have side x (the square's area is x^2) and the two leftover parts of the legs of the triangle are a and b in any order WLOG. Since the two small triangles are similar, a/x = x/y. Cross-multiply to get x^2 = ab.
Tresh Palwalker
I don’t get the second problem
Is there a geometrical clean method to solve it?
I have actually proved the bigger triangle similar to the triangle which is present on the right hand side 😊 and my answer is 100.0000000000000000 sq units
with the second problem youre assuming a lot about the triangles (mainly the size of their angles). how can you be sure connecting the three triangles will give you 3 right angles? isnt it possible they might have combined into a trapezoid?
I found the second petty fast. We already know all outer sides are 6. This means the outer sides form a square or parallelogram. And both the upper and lower sides with length 6 and the left and right sides with length 6 are parallel to each other. The heights of the left and right triangle are in the same direction and sum up to 6. With that we know the total area is 36.
problem no 2: how do you assume that composition of triangles gives right angles and form square?
edit: im pretty sure this can not be solved unless given one of a, b, c or d, or answer will be function of one of those
here's examplary drawing. area of triangle DC6 is aproximetly 14,11
easyimg.io/i/x05nba1nl/myd.jpg
In the case of the 1st problem i was like "think about outside the box solution... wait a sec, its similar triangles and solution is extremely easy, what's outside the box about it? "
After watching the video " ok, it works and makes sense, but the textbook solution is way easier and faster and you do exactly the same calculations, 5*20 "
Letme provide my way to answer... a little diffrent from both ways
Area of triangle=1/2×(X+5)×(X+20)= 1/2×20×X +1/2×5×X+X^2
SOLVE THIS EQUATION everything will cancel out and we directly get value of X^2 as 100 hence area=100
Niiiiice
I made it by similar triangle which is the easiest method if you notice the side are miltiplying
It is not obvious that h and k are parallel, because it is not obvious that the top and bottom lines of length 6 are parallel.
If we make h and k parallel, it is not obvious that we don't have a trapezium instead of a square and the "vertical" sides are not necessarily the 6's from teh original triangles.
I was seriously expecting to see some quadratic equations with the Pythagorean theorem for that first problem...
Only solved 1 as that was the only one in the thumbnail:
Information Given (Pythagoras):
[1]: 25 + LL = AA
[2]: 400 + LL = BB
[3]: (20+L)(20+L) + (5+L)(5+L) = (A+B)(A+B)
Adding [1] and [2]:
[4]: AA+BB = 425 + 2LL
Expanding [3] and Substituting [4] into it:
[5]: 25L = AB
Multiplying [1] and [2] and Substituting [5] into it:
[6]: 10000 - 200LL + LLLL = 0
Solving [6] via the quadratic formula:
2LL = 200 +|- sqrt(40000 - 40000)
Yields:
LL = 100
In the second problem I found that all sides were the same indicating it's a square so I found the area which was 36 then I did 36 - (13+4+5) which was equal to 14 but idk.
Frack, for first problem I (brute forcely) use Pythagoras on 3 rect triangle and solve resulting quadratic equation after several substitution. Right answer but slower path.
10×10， simply.😊
Basic similar triangles, solved in 15 seconds without pen and paper.
You can put the 4 triangles together to a rhombus by connecting equal sides b, c, and d of the triangles. In a rhombus both heights are equal. Also since both heights are each the sum of the heights of two opposing triangles you get 4/3 + h = 5/3 + 13/3, so h = 14/3. So the area of the unknown triangle is 14.
ok so the second problem is leaving me with a question. why would we have to ignore the knowledge that the C/D/6 triangle's length is equal to 6? it's stated in the picture.
3:23 I really don't get why when you take out all the triangle the square part area equals the rectangle area.
And why don't this work in the other way ? Why the two little triangle area don't equals the 2 big triangles area ?
I know im bad at maths but i would love an explanation please
مرحبا اخي قم بترجمة هذا للحصول علي الشرح
عندما قام هو بإنشاء مثلث مساوي للمثلث الاخر الذي حدث ان كل قسم هو قام بتقسيمه ساوي الاخر حيث عندما قام بتقسيم المثلثات التي علي الأعلي من جهه اليسار قام بحذفها ثم قام بحذف المثلثان الكبيران
والمقصود هو أنه في البدايه قام برسم مثلث مساوي للآخر ثم حذف منهما المثلثان الذان اعلي يسار الشاشة ثم المثلثان الاخران فتبقي المستطيل والمربع ونحن نعلم المثلثان الذان رسمهما في البدايه مستاويين
About 2nd solution: How do you know arranging 6-lenght-sides in a square makes triangles allign into square perfectly, and not overlapping themselves or leaving gaps in the area?
There is no unique solution for the 2nd problem. The solution in the video is flawed, since it makes some assumptions that are not warranted. I solved for a, b, c, d using Heron's formula, but since there are 3 equations and 4 unknowns, the problem is underspecified. I chose to use b as a parameter, and solved for a, c, d in terms of b:
a^2 = b^2 + 36 +/- 4*sqrt(9*b^2 - 16)
c^2 = b^2 + 36 +/- 4*sqrt(9*b^2 - 25)
d^2 = a^2 + 36 +/- 4*sqrt(9*a^2 - 169)
= b^2 + 72 +/- 4*sqrt(9*b^2 - 16) +/- 4*sqrt( 9*b^2 + 155 +/- 36*sqrt(9*b^2 - 16) )
Since side lengths must be real, 9*b^2 - 16 > 0 and 9*b^2 - 25 > 0 so b > 5/3
The equation for d^2 also gives a maximum for b if we take the - of +/- (which means acute angles in the triangles, the + corresponds to an obtuse triangle), so the range for b is
5/3 < b < 1/3 * sqrt(493 - 108*sqrt(17))
1.667 < b < 2.302
These values for a, c, d can be plugged into Heron's formula to find a range for the area of the 4th triangle. I used the - sign for each possibility, corresponding to acute angles, and the result is long and messy:
(1/4)*sqrt(-(32*(sqrt(-(36*sqrt(9*b^2-16))+9*b^2+155)*(sqrt(9*b^2-16)-sqrt(9*b^2-25))+18*sqrt(9*b^2-25)+sqrt(9*b^2-16)*(-sqrt(9*b^2-25)-18)+9*b^2-105)))
I found the range of areas for the 4th triangle is from area of 14 at b=sqrt(41)/3 to area of 2*sqrt(67-9*sqrt(2)) at b = 5/3
Or numerical approximations,
A = 14 for b = 2.134375
A = 14.73392 for b = 1.666667
and of course the area can take any value between 14 and 14.73392 with b between 1.666667 and 2.134375 (and corresponding values for a, c, d).
1/2.x.5 +1/2.20.x + x^2=1/2(20+x).(5+x)...simplify to get x^2 =100....clunky & cumbersome--- but it does the job.
In the second question, why mist it be a square? Can't it be a rhombus?
Hi Presh, great problems! But please, if you're going to do multiple problems, either give the second problem after the first solution or at least refresh us on the second problem. If I have to skip around to watch your video properly, there is something wrong with your order!
Or just minus the two small triangle from bigger one and equate it with a squared
Btw love from India
similar triangle, 5/x=x/20, so the area is 100
Area will be 100 cm2
Problem 1: Tangent of the small angle is (5+x)/(20+x) = x/20. Algebra then gets you the answer. No trig needed.
Sin( θ)= 5/x = x/20 => x=10.
I am amazed that I got both of these solutions the same way you did! Also, CONGRATS ON 3 MILLION SUBSCIBERS!!!
Problem 2. Can a triangle be obtuse?
Rubber goose
rearrange the triangles so that they form a irregular quadrilateral with sides A, B, C, and D. there should be some overlap. from there you can find the area of the whole thing (5+13) and then subtract 4.
Yes! This solution is very straightforward. I prefer it. And no need to prove there is square… is other terms, if you add the triangles two by two, you have two quadrilaterals that have the same area. Then 5 + 13 = 4 + 14
Because of similar triangles the square side lengths are 10
That first one is easy-peasy. That second one, well it would be a lot easier for me to solve with scissors and paper. The first thing to see, if those four triangles will make up a square or rhomboid (all sides with length 6). If it's a square we get 14, but that wouldn't fly on a math contest, because I would need to prove those 4 triangles can make up a square. Not sure how to do that.
For problem 2, I instantly thought about putting the triangles together which result in a parallelogram with side lengths 6. This means the area of the parallelogram will be 36. Now to get the area of the triangle we need, we can just do 36-13-4-5=14.
Ah!
for the second problem i just put the triangle AB just below the triangle we are trying to find the area. You will see that this immediately re create triangle cb and da which we already have the area. you just add the area of cd + da, then substracte ba. took me 10 seconds to solve using this
Doesn't this solution presuppose that the distance between the vertices cd and ab in the proposed figure is 6?
This is where fun begins! The verteces you mentioned in proposed arrangement are at the same time the end points for the base of triangle cb, which is 6!
For me this approach is even "cooler", than the used one, which is fun too❤😊!
They used hedge clippers?
First one took a few seconds spotting the similar triangles. The second one I missed the trick that showed that the square based wonky pyramid was flat. Very neat!
No way I just solvef the first question with a "test side" of the square (10) and tried using the proportionality formula and got the correct answer
5/x = x/20
x² = 100
Area is 100