Justify your answer! | Calculate the Yellow Square area | (Tutorial) |
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- Опубліковано 3 чер 2024
- Learn how to find the area of the Yellow Square. Important Geometry and Algebra skills are also explained: Quadratic formula; right triangles; area of a square formula; Pythagorean Theorem . Step-by-step tutorial by PreMath.com
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Mr. Premath, pleading you again that you must find other non-geometry problems, challenging or not.
At about 11:40, PreMath decides that x = 82 + 2√(511) will produce a value of a which is too large. The value of a is approximately 11.28, which exceeds both PD (10) and PB (8). Actually, if P is allowed to be inside the square, then x = 82 + 2√(511) is a valid solution.
The requirement that P be outside the square is implied by the diagram, but, probably, should also be explicitly stated, since "This diagram may not be 100% true to the scale" could be interpreted to allow P to be inside the square. In some other problems, presented on other youtube channels, the only valid solution resulted in a significant deviation from the diagram. That would be the case here if a solution with P inside the square was considered valid.
UPDATE: Reviewing the video, I do find that, at the very beginning, PreMath states (verbally) that point P is external to the square. At about 11:40, he states that x = 82 + 2√(511) is "way too much" and "not possible" without further explanation. Offering a suggestion for possibly improving the video, PreMath could have calculated a, the side length of the square, and shown that it will be longer than either PD or PB, so P would have to be internal to the square. Furthermore, he could have shown that x = 82 + 2√(511) produces a valid solution if P is allowed to be internal to the square.
You just proved his point, though. It's pretty basic math, just Pythagorean and Quadratic equations. I don't think he's looking at it in that way.
But I'll store that away as future knowledge. i is even trippier because it turns the whole graph, and alters its dimensions.
In triangle DCP, angle DCP is an obtuse angle, try extending ray CB and it turns out that angle DCP is (90 + a) where variable a is a positive real number, and it's commonly known that the side opposite to the largest angle is the largest side in the triangle, therefore the side of the square or to be more meticulous and pedantic, DC can never be equal to 11.27, therefore side 6 is the only possible solution.
Excellent analysis. I figured the square area had to be 82 minus and not plus 2root511 because I saw that 6 was the minimum side length for the square from the upper triangle. And with a little figuring with the triangle to the right I saw that 7 was going to be a little too big for the square side length. So the area had to be between 36 and 49.
Let's use Heron's formula to calculate the areas of ΔDCP and ΔPCB. S(DPC)=√(7+0,5a)(7-0.5a)(0.5a-3)(05a+3), S(PCB)=√(6+0,5a)(6-0.5a)(0.5a-2)(05a+2). To simplify the expressions, you can use the formula of abbreviated multiplication. On the other hand, S(DPC)=0.5*a*4*sin(∠DCP), S(PCB)=0.5*a*4*sin(270° -∠DCP)=2*a*cos(∠DCP). We equate the areas for each of the triangles, square them and add two equalities. It is easy to see that using the basic trigonometric identity, we get rid of trigonometric functions. We obtain a biquadrate equation with respect to a. -0.125a^4+20.5a^2-585=0. Multiply by -8 and get the equation a^4-164a^2+4680=0. I don't like additional constructions.
Thanks for sharing and information❤
The missing side of the right triangle at the top is radical 84. Which means that each side of the square is radical 84, and therefore the area is 84.
3 parameters a,b,c a is the side of the square, b^2+c^2=16, (a+c)^2+b^2=64, (a+b)^2+c^2=100, 36=c^2-b^2+b^2-c^2+2ab-2ac+2a(b-c), ......😮
Professor, the drawing initially looked like a projection with depth (3D) such that the 2 blue triangles formed a flap at crease CP. under this interpretation, CB would be a leg of right triangle PCB = sqrt(8^2 - 4^2) = a
Sometimes it looks like a quadrilateral pyramid from the base face😂😂😂😂
I did it with an orthonormal as usual, but it is too painful to write the whole solution. I give here onlya summary.
I use an orthonormal center C and first axis (DC)
The equation of the circle center C and radius 4 is x^2 + y^2 = 16. The equation of the circle center D and radius 10 is (x -c)^2 + y^2 = 100 (with c the side length of the square). Then I solve the system to have the coordinates of point P (I choose the one with a positive ordinate)
I find P((c^2 -84)/2.c; sqrt(-c^4 +232.c^2 -7056)/2.c)
Then I write that the distance from P to B(0; -c) is equal to 8 . It is a bit painful but possible to solve.. At the end I fond the eqution c^4 - 164.c^2 +4680 = 0 that gives the same solution than the one proposed by PreMaths.
This solution is not elegant but I like to transform geometric problems into algebraic ones.
I also tried to write P(4.cos(t); 4.sin(t)) and use the fact that DP = 10 and BP =8 but I stopped. As we see the final result of c it is evident that trigonometr is not a good idea here.
Let _∠DCP = θ_
Then _∠BCP = 270° - θ_
Applying the cosine rule in _ΔDCP:_
_a² + 4² - 8acosθ = 10²_
⇒ _8acosθ = a² - 84_ ... ①
Doing the same in _ΔBCP:_
_a² + 4² - 8acos(270° - θ) = 8²_
⇒ _8asinθ = 48 - a²_ ... ②
Squaring both sides of ① and ② then adding them gives:
_64a²(cos²θ + sin²θ) = 2a⁴ - 264a² + 9360_
⇒ _a⁴ - 164a² + 4680 = 0_
⇒ _a² = 82 ± 2√511_
I tried using the law of sines and law of cosines but didn't succeed as you did. Nice work!
Solve the question
Many parameters to determine one unknown ?😢
11:39 You discarded the + value as being too large, but you didn't explain it.
Otherwise, great question!
Actully I solve in the same way, but I cannot believe the true answer is not the whole number, so I suspect that I still makes some mistake 😢😢😢
For yesterdays video, If a shapes length is x and it is a triangle how is x ^2 I thought that would be only true of a square please can you tell me? Or is that only with some triangles?
DCP=α..teorema dei coseno ..10^2=l^2+4^2-2*4*l*cosα...8^2=l^2+4^2-2*4*l*cos(270-α)...calcolo sinα,cosα..e trovo ,dopo i calcoli..l^2=82-√2044...l=6,0654...mah???...toh, è corretto!!!
This was not an Easy Task!!
01) x^2 + y^2 = 16
02) y^2 = 16 - x^2
03) y = sqrt(16 - x^2)
04) Given the Function y = f(x) the Coordinates of Point P are (x ; f(x))
05) As f(x) = sqrt(16 -x^2), with Domain x -> ] sqrt(8) ; 4 [ we have :
06) Coordinates of Point P (x ; sqrt(16 - x^2). The Coordinates of Point C are (0 ; 0).
07) Point P distance from Point D is equal to 10
08) Point P distance from Point B is equal to 8
09) As the Sides of a Square (z) must be equal :
10) Coordinates of Point D (- z ; 0)
11) Coordinates of Point B (0 ; - z)
12) (x + z)^2 + (sqrt(16 - x^2))^2 = 100
13) x^2 + (sqrt(16 - x^2) + z)^2 = 64
14) Solving this System of Two Nonlinear Equations we have the Two Possible Solutions (with "x" inside the defined Domain of the Function):
15) x ~ 3.89178 and z ~ - 6.06543. The Coordinates of Point P are (3,89178 ; 0,92415)
16) So, the Yellow Square Area must be z^2 ~ 36,79 Square Units.
NOTE : What I've done is not Mathematics but Engineer. Approximate Numerical Calculus!!
cumbersome!
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