Great question professor!!

@PreMath

3 ай бұрын

Glad you liked it! Thanks ❤️

Me and my love of trigonometric identities. A relaxing algebraic solution! [0.1]  tan 2θ = 2 tan θ / (1 - tan² θ) It'll be useful in a minute. First though, that π circle. [1.1]  Area = π𝒓² [1.2]  π = π𝒓² [1.3]  1 = 𝒓² [1.4]  1 = 𝒓 While NOT knowing θ, we do know the value of tan θ, if we label the side of the whole square 𝒔 [2.1]  tan θ = 1 / (𝒔 - 1) … and therefore [3.1]  tan 2θ = [ 2 / (𝒔 - 1) ] / [ 1 - (1 / (𝒔 - 1)²) ] Which through a bit o algebraic shuffling is also [3.2]  tan 2θ = [ 2(𝒔 - 1) ] / [ (𝒔 - 1)² - 1 ] Here's the fun part: (𝒔 times tan 2θ) must be the length of the top of the square, minus ONE, that chunk taken out at the top right: [4.1]  𝒔 tan 2θ = 𝒔 - 1 … subbing in [3.2] [4.2]  2𝒔(𝒔 - 1) / [(𝒔 - 1)² - 1] = (𝒔 - 1) … divide by (𝒔 - 1) [4.3]  2𝒔 / [(𝒔 - 1)² - 1] = 1 … then cross multiply [4.4]  2𝒔 = (𝒔 - 1)² - 1 … expand the square [4.5]  2𝒔 = 𝒔² - 2𝒔 ⊕ 1 - 1 … cancel and rearrange [4.6]  4𝒔 = 𝒔² … divide by 𝒔 [4.7]  4 = 𝒔 Yay! Now it all is simple for the area of a trapezoid [5.1]  area = (square part) + (triangle part) [5.2]  area = (1 × 4) + (½ 3 × 4) [5.3]  area = 4 ⊕ 6 [5.4]  area = 10 Which answers the question by a rather different route. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

Very nice video sir

Solved the equation (a+b-c)/2=r, where: “a” and “b” are the legs, and “c” is the hypotenuse ∆ADE; r is the radius of the circle. Trapezoid area 10 cm^2

@PreMath

3 ай бұрын

Thanks ❤️

1/ Let r and a be the radius of the circle and the side of the square. Just label EP=FE= m and DP=DG =n and A = the area of the triangle DAE We have A=Area of the square AGOF + 2 area of triangle FOE + 2 area of GOD = sq r + mr +nr (1) and A= 1/2 AExAD= =1/2 (r+m)) (r+n)) = 1/2(sqr + r n+rm+mn)= 1/2(A+mn) (2) (1) =(2) ----->A= 1/2(A+mn) ----> 2A= A+mn-----> mn = A 2/ r=1 ---> m= a-2 and n= a-1 we have 1/2 AExAD= mn----> 1/2 a. (a-1)= (a-2)(a-1)---->a/2= a-2--->a= 4 3/ Area of the trapezoid= 1/2 (1+4). 4= 10 sq cm

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

Area ABCD=16 ; AreaADE=6 ; ฉะนั้นAreaBCDE=16-6=10

@PreMath

3 ай бұрын

Thanks ❤️

(l-1+l-2)^2=l^2+(l-1)^2..(2l-3)^2=2l^2-2l+1…..2^2-10l+8=0...l^2-5l+4=0...l=4...

@PreMath

3 ай бұрын

Thanks ❤️

Nice, I used x+1 and x+2 as sides of ADE, which led to x² -x -2. X=2, so side of sqr is 2+1+1=4

@PreMath

3 ай бұрын

Thanks ❤️

There are two formulas for the radius of a circle inscribed in a right triangle: _r = (a + b − c) / 2_ and _r = ab / (a + b + c)._ In our case, _r_ = 1, _a_ is the side of the square and _b_ = _a_ − 1. So we have 1 = (a + a − 1 − c) / 2 and 1 = a (a − 1) / (a + a − 1 + c). Let's isolate _c_ in both equations so we get c = 2a − 3 and c = a² − 3a + 1. Then we can write 2a − 3 = a² − 3a + 1. This also leads to the quadratic equation a² − 5a + 4 = 0 and, of course, the same result as in the video.

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

Veri nice sharing❤❤❤

@PreMath

3 ай бұрын

Thanks ❤️

Thank you!

It's well known that a circle of radius 1 (as this one is given that it's area is pi) inscribed in a triangle like that implies that the triangle's side lengths are 3, 4, and 5. Therefore, the square's side length L is 4. The square's area is 16, the triangle's area is 6, so the shaded area is the difference, 10 cm^2.

Since the radius of the blue circle is 1, the famous (3, 4, 5) Pythagorean triangle is a good suspect 🧐 Innocent until proven guilty? Judge PreMat and his quadratic equation gave the guy zero chance! 👨‍⚖

@Se-La-Vi

3 ай бұрын

Я крокував тим же шляхом і далі від площини квадрата відняв площину трикутника...

@PreMath

3 ай бұрын

Thanks ❤️

Si el lado del cuadrado es BC=c=CD ; EB=1 y el área del círculo azul =π→ r=1 → OF=OP=OG=AF=AG=r=1 ; FE=c-2=PE ; GD=c-1=PD → c² =1*(c-1)+1*(c-2)+(1*1)+c*[(1+c)/2] → c²-5c+4=0→ c=4 → Área del trapezoide verde EBCD =4*[(1+4)/2] =10 cm². Gracias por el vídeo. Un saludo cordial.

It suffices to find the side s of the square, clearly 2s^2-2s+1=s^2+(s-1)^2=(s-1+s-2)^2=(2s-3)^2=4s^2-12s+9, s^2-5s+4=0, s=1, too small, rejected or s=4， therefore the area is 1/2×(1+4)×4=10.😊 ,

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

πr^2=π r=1 Let EF=x AB=AD=BC=CD=2+x EF=EP=x AF=AG=1 DG=DP=2+x-1=1+x BD=1+x+x=2x+1 (x+1)^2+(x+2)^2=(2x+1)^2 So: x=2 CD=2+2=4 Area of the green shaded region trapezoid=1/2(1+4)(4)=10 cm^2.❤❤❤ Thanks sir.

@PreMath

3 ай бұрын

Excellent! You are very welcome! Thanks ❤️

@SkinnerRobot

3 ай бұрын

Excellent writeup, but don't you mean ED=1+x+x=2x+1?

@prossvay8744

3 ай бұрын

@@SkinnerRobot BD is not ED

@SkinnerRobot

3 ай бұрын

Check PreMath's diagram.

Let's use an adapted orthonormal, center D, first axis '(DC), second axis (DA). WE have D(0;0) A(0;c) E(c -1;c) and O(1;c-1) where c is the length of the square and 1 the radius of the circle (evident). Then VectorDE(c -1;c) and the equation of (DE) is: (x).(c) - (y).(c -1) = 0 or c.x -(c -1).y = 0. The distance from O to ((DE) is abs(c - (c -1)^2)/sqrt(c^2 + (c -1)^2) and this distance is 1 (the raduis of the circle), so c - (c -1)^2 = sqrt( c^2 + c-1)^2) We put into square to eliminate the radical. Then: (c - (c -1)^2)^2 = c^2 + (c -1)^2, or (-c^2 +3.c -1)^2 = 2.c^2 -2.c +1 We develop and obtain c^4 -6.c^3 +11.c^2 -6.c +1 = 2.c^2 -2.c +1, or c^4 -6.c^3 +9.c^2 -4.c = 0. As c0 we also have c^3 -6.c^2 +9.c -4 = 0 C = 1 is an evident solution, we factorize the third degree polynom by (x -1) by division and obtain (c -1). (c^2 -5.c +4) = 0 So c =1 or c^2 -5.c +4 = 0 and finally c =1 or c = 4 (the solutions of c^2 -5.c +4 = 0 beeing evident). As c must be superior than 1 o the diagram, we have that c = 4 is the only solution. So the length of the square is 4. Now the area of the trapezoid is ((EB + DC)/2). DA = ((1 + 4)/2).4 = 10.

@robertlynch7520

3 ай бұрын

While I appreciate your HARD work, take a look at my (trigonometric identities) solution. Never exceeds the quadratic. Thanks though ... you've given me a headache! LOL

@marcgriselhubert3915

3 ай бұрын

@@robertlynch7520 I look at the given solution and try to find an alternative one.

@PreMath

3 ай бұрын

Thanks ❤️

In this problem I came to the geometrical conclusion that, the given Square, holds 4 Equal Circles (tangent to each other and tangent to the sides of the square) with Radius 1 lin un and Area Pi sq cm. So the Side Length of the Square must be equal to 2 diameters; 4 cm. Ao, AB = AD = BC CD = 4 cm Now the Area of the Green Shaded Region is equal to (B + b) * (B/2) = (4 + 1) * (4/2) = 5 * 2 = 10 sq cm My answer is: The Green Shaded Region Area (Trapezoid) is equal to 10 Square Cm.

Si l'aire du cercle est pi alors le triangle rectangle est 3-4-5. Donc les côtés du trapèze sont 1,4,4. L'aire du trapèze est (1+4)*4/2=10

A = πr² π = πr² 1 = r² r = 1 The radius of the blue circle is 1 cm. Draw radii FO, GO, & PO. By the Circle Theorem, ∠AFO, ∠AGO, & ∠DPO are right angles. And since ∠A is a right angle by definition of squares & FO = GO, quadrilateral AFOG is a square. AF = AG = 1. Label EF = x & DG = y. By definition of squares, y = x + 1. (s = x + 2, s = y + 1) By the Two-Tangent Theorem, EF = EP & DG = DP. Therefore, EP = x, DP = x + 1, & DE = 2x + 1. Drop a line perpendicular to sides AB & CD thru point E. Label the intersection of the line and side CD as point H. EH = x + 2. Because CH = 1 by the Parallelogram Opposite Angles Theorem (rectangle BCHE), DH = x + 1. Apply the Pythagorean Theorem on △DHE. a² + b² = c² (DH)² + (EH)² = (DE)² (x + 1)² + (x + 2)² = (2x + 1)² x² + 2x + 1 + x² + 4x + 4 = 4x² + 4x + 1 2x² + 4x + 1 = 6x + 5 2x² - 2x + 1 = 5 2x² - 2x - 4 = 0 x² - x - 2 = 0 (Time to factor!) (x - 2)(x + 1) = 0 x - 2 = 0 or x + 1 = 0 x = 2 or x = -1 But x represents the length of a segment and it can't be negative, so x ≠ -1 & x = 2. Thus, y = 3 & s = 4. Now, the green shaded region is a trapezoid. A = h[(a + b)/2] = 4[(1 + 4)/2] = 4(5/2) = 10 So, the area of the green region is 10 square centimeters.

@ChuzzleFriends

3 ай бұрын

Don't be confused; in the video, x is the side length of the square. in this solution, x is the length of segment EF.

@PreMath

3 ай бұрын

Thanks ❤️

Easy brother

@PreMath

3 ай бұрын

Excellent! Thanks dear❤️

2x(45°)=90° 2x(45°)°90°x 3x(15°)°45°x 3x(15°)=45°x (90°x+90°x+45°x+45°x)=270°x^423x(15°)=45°x 3x(15°)=45°x (90°+45°x+45°x)=180°x^2 (180°x^2-270°x^4) =√90°x^2 3^√30x^2 3^√5^√6 x^2 √31^3^√2x^2 √1^√13^√1x^2 3x^2 (x+2x-3)