Very Very useful video sir 👍

@PreMath

3 ай бұрын

Thanks and welcome🌹❤️

2:44 No. You cannot assume that the two triangles are congruent because you cannot assume that AF = EB unless OF = OE. And that was not stated as a starting condition of the problem. You have to prove it and show why it is the case. The appearance of visual symmetry isn't good enough, especially since you've already warned us that the diagram is not to scale.

@mikeparfitt8897

3 ай бұрын

The thumbnail states the "green trapezoid" which has to have two parallel sides by definition. The diagram might not be to scale, but it would be madness to claim that the two parallel sides were EC and DA. The congruence is proved when you accept that DC is parallel to the diameter AB.

@johnbrennan3372

3 ай бұрын

Triangles dfo and coe are congruent so |fo|= |oe|=7/2.Then in triangle dfo fo=7/2, do=(2x/7)/2 and df = sqroot (225- x^2)? Using Pythagorous you end up with x=9 etc.

@71littlerabbit

3 ай бұрын

​@@johnbrennan3372you can't say that, unless O is located exactly in the middle of EF...

@71littlerabbit

3 ай бұрын

CB can't be equal to AB!!!! The only case when this happens would be if O is at the same distance from F and from E. Otherwise, the assumption of CB=AD is wrong!

@md.abdurrahmantarafder2256

3 ай бұрын

Right.

I like your solution, yet I approached it with a different 'knowns' point of view. Basically, I noted that the DC segment, if divided by a not-shown [O→top] vertical line, is always symmetrically divided in half. Always. Each half, I called (𝒂) [0.1]  2𝒂 = 7 [0.2]  𝒂 = 3.5 Therefore, can write [1.1]  15² - (𝒓 - 𝒂)² = 𝒉² = 𝒓² - 𝒂² … with numbers [1.2]  225 - (𝒓 - 3.5)² = 𝒓² - 3.5² … expanding [1.3]  225 - 𝒓² + 7𝒓 - 3.5² = 𝒓² - 3.5² … cancelling and rearranging [1.4]  225 - 2𝒓² + 7𝒓 = 0 … is quadratic so [1.5]  𝒓 = [12.5 or -9] With the positive 𝒓, 𝒉 is [2.1]  𝒉 = √(12.5² - 3.5²) [2.2]  𝒉 = √144 [2.3]  𝒉 = 12 So, the area is the sum of the square, and the triangle [3.1]  area = (□ = 7 × 12) + (△ = ½ (𝒓 - 𝒂 = 9) × 12) [3.2]  area = 84 + 54 [3.3]  area = 138 And that's the solution! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

@PreMath

3 ай бұрын

Super GoatGuy! Thanks ❤️

i used the proportions of the sides of the semi circle. That give y^2 = (r+3.5)*(r-3.5). That yields y=(r^2-12.25)^1/2. Plug that into Pythagoras you get (r-3.5)^2 + r^2 -12.25 =225. Simplify that and solve for r you get r equals 12.5 of -9. Of course the -9 is invalid. If r = 12.5, y = 12. Plug that into the trapezoid area equation and you get 138 (of course). Good problem Premath. You keep my 61 year old brain in shape.

@PreMath

3 ай бұрын

Excellent! Glad to hear that! Thanks for the feedback❤️

Thank you Thale!

@PreMath

3 ай бұрын

😀 Excellent! Thanks ❤️

Great!

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

Alternate solution: Draw vertical line through point O that intersects chord CD at point M. Any line that passes through center of circle and is perpendicular to a chord, bisects that chord. Therefore, CM = DM = 7/2 and since CDFE is a rectangle and OM is parallel to CE and DF, then OM also bisects EF Therefore, OE = OF = 7/2 Let AF = x, DF = h Since OA and OD are both radii, then both are equal and OD = OA = x + 7/2 Using Pythagorean Theorem in △ODF, we get h² + (7/2)2 = (x+7/2)² h² + 49/4 = x² + 7x + 49/4 h² = x² + 7x Using Pythagorean Theorem in △ADF, we get x² + h² = 15² x² + (x² + 7x) = 225 2x² + 7x − 225 = 0 2x² −18x + 25x − 225 = 0 2x (x − 9) + 25 (x − 9) = 0 (x − 9) (2x + 25) = 0 Since x is a length, then x > 0 x = 9 h² = x² + 7x = 81 + 63 = 144 h = 12 Now we can calculate area top base = 7 bottom base = 7+x = 7+9 = 16 height = h = 12 A = 1/2 (7 + 16) * 12 = 138

Area of the green trapezoid=1/2(7+9+7)(12)=138 square units. ❤ Thanks dir

@PreMath

3 ай бұрын

Excellent! You are very welcome! Thanks ❤️🌹

I propose something different (I don't say that it is better). Let'a name t = angleAOD and t' = angleDOC, we have t + t' + t = Pi radians, so t' = Pi - 2.t, and R is the radius of the circle. In triangle AOD: AD^2 = AO^2 + DO^2 - 2.AO.DO.cos(t), or 225 = 2.R^2 - 2.R^2.cos(t) or 225 = 2.R^2. (1 -cos(t)) In triangle DOC, in the same way we have 49 = 2.R^2.(1 -cos(t')). From these two equations we get by division that 225/49 = (1 -cos(t))/(1 -cos(t')) or 1 - cos(t) = 2.sin(t/2)^2 and 1 -cos(t') = 2.sin(t'/2)^2, so we simplify and obtain that 15/7 = sin(t)/sin(t'). Now let's note sin(t) = x As t' = Pi -2.t, we have t'/2 = Pi/2 -t and sin(t') = cos(t) = 1 - sin(t/2)^2 = 1 -2.x^2. So we have the equation 15/7 = x/(1 -2.x^2) That gives: 30.x^2 +7.x -15 = 0. Delta = 49 +4.30.15 = 1849 = 43^2, so x = (-7 -43)/60 (rejected as beeing negative) or x = (-7 +43)/60 = 36/60 = 3/5 Now we have that sin(t/2) = 3/5. Then 1 -cos(t) = 2.sin(t/2)^2 = 2.(9/5) = 18/25. We replace in the first equation we had: 225 = 2.R^2.(18/25) That gives R^2 = (225.25)/36 and R = (15.5)/6 = 25/2. Now in triangle ODF: DF = OD.sin(t) = R.2.sin(t/2).cos(t/2) = (25/2).2.(3/5).(4/5) = 12 The area of the trapezoid in now: ((AE + DC)/2). DF ,with AE = R + 7/2 = 11, so this area is ((16 +7)/2). 12 = 138

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

I solved it almost the same as you. 1/ ADCB is an isosceles trapezoid. We have: sq h = (7+x) .x ( by using the right triangle altitude theorem) (1) and sq h= sq AD- sq x = sq15 - sq x (Pythagorean theorem) (2) From (1) and (2) we have: 2sq x + 7x -225= 0----> x= 9 and h= 12 and AE=16 Area= 1/2 (7+16) .12 =138 sq units

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

My solution with 2x Pythagoras (r=Radius, h=CE): OCE: h² = r² - 3.5² AFD: h² = 15² - (r - 3.5)² Solve r² - 3.5² = 15² - (r - 3.5)² and get r=12.5 and h=12 A=0.5*(r+3.5 + 7)*h = 138

@PreMath

3 ай бұрын

138 Thanks ❤️

I agree with the solution, but i dont agree with skipping to the assumption that triangle CEB is necessarily congruent with ADF based on "symmetry"... For younger learners, i would've "proved" that by reminding that the radius is the same from the center to any point along the perimeter of the circle THEN moving to prove symmetry from there... Just an observation rather than assume symmetry.....ESPECIALLY, when you go out of the way to say the drawing isn't to scale. Love all these videos though.

AD = BC = 15 → AC = BD = 20 → AB = 25 → r = 25/2 → cos⁡(δ) = 4/5 = AE/20 → AE = 16 → CE = h = 12 → green area = (1/2)(16+7)h = 138; ∆ ABD = ∆ ABC = pyth. triple = 5(3 - 4 - 5) → ∆ AEC = pyth. triple = 4(3 - 4 -5)

@phungpham1725

3 ай бұрын

Could you explain how AC=20?

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

@murdock5537

3 ай бұрын

@@phungpham1725 AO = BO = CO = DO = r sin⁡(BDA) = 1 ↔ AD = 15 → ABD = pyth. triple = 5(3 - 4 - 5) → 2r = 25 → BD = 20

@ybodoN

3 ай бұрын

@@murdock5537 with sin⁡(BDA) = 1 ↔ AD = 15, DB could have any length and we would still have AO = BO = CO = DO = r 🧐

@murdock5537

3 ай бұрын

@@ybodoN or try this: φ = 30°; DF = h = NO → DN = CN = CD/2 = 7/2 → ∆ AOD → AO = AF + FO = (r - 7/2) + 7/2 = r = DO AOD = α; ODF = β → α + β = 3φ → sin⁡(α) = cos⁡(β) → cos⁡(α) = sin⁡(β); AD = 15 DAO = δ → AO = DO = r → DAO = ODA = δ → AD = AT + DT ↔ AT = DT = AD/2 → AOT = TOD = α/2 → sin⁡(α/2) = cos⁡(δ) = (r - 7/2)/15 = (2r - 7)/30 → cos⁡(α/2) = sin⁡(δ) = h/15 → sin⁡(α) = h/r = 2sin⁡(α/2)cos⁡(α/2) → 1/r = 2(2r - 7)/450 → r1 = 25/2; r2 = -9 h = 12 → r - 7/2 = 9 → green area = (1/2)(7 + 16)12 = 6(23) = 138

Me be like😂- If i use pythagoras theory,, so the length and hight of the triangle 🔺DBA will be 9 and 12 So,the area of the triangle is 54 And the area of the rectangle is 12×7 = 84 So, GREEN TRAPEZOID AREA IS 54+84= 138

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

There is no need to construct triangle BCE (Its congruence to triangle ADF by symmetry is just by intuition and not a proper congruence test.) to get BE = AF. Perpendicular line from centre to chord is a line bisector. Hence chord DC can be bisected with perpendicular line from centre O into two equal segments. In the 2 equally divided rectangles formed from the rectangle CDFE, OF = OE (opposite sides equal). AF = radius - OF and BE = radius - OE. Hence AF = BE.

@PreMath

3 ай бұрын

Thanks ❤️

I could make a lucky guess knowing that 15^2 = 225 and x^2 + y^2 = 225 can be equal to 9^2 + 12^2 = 225 ; 81 + 144 = 225 Then I could say that the Area of Trapezoid is equal to 138 Square Units, because A = ((9 + 7) + 7) * 12/2 = 23 * 6 = 138 Sq un. But I didn't!! What have I done? Defining Point F: Perpendicular Line passing point D and intersecting Line AB. Let's call DF = h (height) and Radius = R A) h^2 = 15^2 - (R - 3,5)^2 B) h^2 = R^2 - 3,5^2 225 - (R^2 - 7R + 12,25) = R^2 - 12,25 225 - R^2 + 7R - 12,25 = R^2 - 12,25 225 +7R -2R^2 = 0 R = - 9 or R = 12,5 R = 12,5 lin un AE = (12,5 + 3,5) li un = 16 lin un AF = (12,5 - 3,5) lin un = 9 lin un Now I know that h^2 = 225 - 81 = 144 ; h = sqrt(144) = 12 A = [(16 + 7) * (12 /2)] sq un = 23 * 6 sq un = 138 sq un Answer: The Trapezoid Area is Equal to 138 Square Units. QED.

😎

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

Let AE be x, then AF=x-7, so the diameter is x+(x-7)=2x-7, the radius is x-7/2, thus the square of the height h^2 will be (x-7/2)^2-(x-(x-7/2))^2=(x-7/2)^2-(7/2)^2, it is also equal to 15^2-(x-7)^2, thus x^2-7x=15^2-x^2+14x-49, 2x^2-21x-176=(x-16)(2x+11), so x=16, reject negative root, and h=sqrt(225-81)=sqrt(144)=12, therefore the area is 1/2×(7+16)×12=23×6=138.😊

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

S=138

@PreMath

3 ай бұрын

Thanks ❤️

Intanto calcolo r...15^2=r^2+r^2-2r^2cos(90-arcsin3,5/r)...(Teorema dei coseno)...r=12,5..h=√(12,5^2-3,5^2)=12...A=(7+√(15^2-12^2)+7)*12/2=(7+9+7)*6=23*6=138

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

Bakıdan salamlar.Azərbaycan.Əla həll etdiniz.Təşəkkürlər.

@PreMath

3 ай бұрын

Bunu eşitməyə şadam! Çox xoş gəlmisiniz! Təşəkkür canım ❤️🌹 ABŞ-dan sevgi və dualar! 😀

this is much more difficult compared with yesterday's one.😅😅😅

@PreMath

3 ай бұрын

Yes! Thanks ❤️

(15)^2=225 (7)=49 2x(45°)=90°x. 2x(45°)=90°x. 3x(15°)=45°x 3x(15°)=45°x (90°x+90°x+45°x+45°x)!=,270°x^4 (49+225)=474 (270°x^4-474)=√196x^4 √10^√108^√12 x^4√ 5^√2√5√^2 √8^3^√4 x^√4 √1^√1√ 1^√1√ 2^√3 3^√2^√2 x^√2^2. √1^√1 3^√1^√1x√1^2 32 (x+2x-3)

@PreMath

3 ай бұрын

Thanks ❤️

I'll say 138

Let F be the point on AB where DF is perpendicular to AB. As CD is a chord, and is parallel to diameter AB, ∆DFO and ∆OEC are congruent and OF = OE = 7/2. Let DF = CE = h. Triangle ∆DFO: OF² + DF² = OD² (7/2)² + h² = r² h² = r² - 49/4 h = √[r²-(49/4)] Triangle ∆AFD: AF² + FD² = DA² (r-(7/2))² + [h²] = 15² r² - 7r + 49/4 + [r² - 49/4] = 225 2r² - 7r - 225 = 0 2r² + 18r - 25r - 225 = 0 2r(r+9) - 25(r+9) = 0 (2r-25)(r+9) = 0 2r - 25 = 0 | r + 9 = 0 r = 25/2 ✓ | r = -9 ❌ r > 0 h² = (25/2)² - 49/4 = 625/4 - 49/4 h² = 576/4 = 144 h = √144 = 12 Trapezoid AECD: A = h(a+b)/2 A = 12(7+(7+9))/2 A = 6(7+16) = 6(23) A = 138 Edit: Interesting how with the quadratic equation in this problem, if you solve for r, the discarded solution is -x, and if you solve for x, the discarded solution is -r. They're effectively the same quadratic equation, just with opposite signs on the x coefficient and constant.

@PreMath

3 ай бұрын

Thanks ❤️