I love how this man improves my math culture

@PreMath

14 күн бұрын

Excellent! Glad to hear that! Thanks ❤️

Not difficult. But need lengthy calculation😮, FC=30/sqrt(5), EC=15xsqrt(5), FC/EC=2/5, therefore the answers is 450×3/5=270.😅

@PreMath

14 күн бұрын

Thanks for the feedback ❤️

First find the side which is 30 Then focus on ABE and EDC Both have sides 30 and 15 and a 90 degree angle so they are congruent Then their hypothenuses are equal so BEC is an isoceles triangle The hypothenuse value is 15root5 (by the Pythagorean theorem) Let's focus on BEF It's a right triangle in

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

1)get EC and BE from pythagorean theorem 2)get BF from identity BF*EC=AB*BC 3)get EF from pythagorean theorem 4)get BF*EF/2 as wanted area

CD is esual of 30, DE at 15, we have CE is equal of 15*sqrt5 With the area of BCE we can know BF which is equal of 900/CE it means 12*sqrt5 With the triplet 3,4,5 we can know EF which is equal of 9*sqrt5 For finishing, we calculate BF*EF/2 which is equal at 270

Sorry, if I'm repeating somebody. I didn't see anybody use the method I'll propose. Start at the point in the video (about 5:00) where you decide to find the area of triangle BFC. Instead of drawing an extra line, simply compare triangle EDC with BFC and notice they are similar. Find the length of EC by pythagorean theorem: EC^2 = ED^2 + DC^2. EC^2 = 1125. EC = 15 * sqr(5). BC corresponds to EC. Let's find the ratio: BC/EC = 30 / (15*sqr(5)). This simplifies to 2/(sqr(5)). All sides of BFC are proportional to the sides of EDC by the same ratio. Based on this, the area of BFC is proportional to the area of EDC by that ratio squared. The square of 2/(sqr(5)) is 4/5. The area of EDC is 225. 225 * 4 / 5 = 180 which is the area of BFC. Thus the area of BFE is 450 - 180 = 270.

Finding area ΔEFB almost directly: From Pythagoras, length CE = √(DE² +CD²) = √(15² + 30²) = 15√5. ΔBCF and ΔCDE are similar. CF/BC = DE/CE, CF = (BC)(DE)/CF = (30)(15)/(15√5) = 30/√5 = 6√5. BF/BC = CD/CE, BF = (BC)(CD)/CF = (30)(30)/(15√5) = 60/√5 = 12√5. EF = EC - CF = 15√5 - 6√5 = 9√5. Let b = EF and h = BF for ΔEFB. Then, area = (1/2)(9√5)(12√5) = (54)(5) = 270 square units, as PreMath also found.

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

Square ABCD: A = s² 900 = s² s = √900 = 30 Triangle ∆EDC: ED² + DC² = CE² 15² + 30² = CE² 225 + 900 = CE² CE = √1125 = 15√5 A = bh/2 = 30(15)/2 = 225 BA = DC and AE = ED, so by symmetry, ∆BAE and ∆EDC are congruent. If ∠CED = α and ∠DCE = β, where β = 90°- α, then ∠FCB = α and ∠CBF = β. Thus ∆EDC and ∆BFC are similar. Triangle ∆BFC: FC/CB =ED/CE FC/30 = 15/15√5 = 1/√5 FC = 30/√5 = 6√5 BF/FC = DC/ED BF/6√5 = 30/15 = 2 BF = 2(6√5) = 12√5 EF = EC - FC = 15√5 - 6√5 = 9√5 Pink Triangle ∆EFB: A = bh/2 = 9√5(12√5)/2 A = 9(6)5 = 270 sq units

s² = 900 Due to similarity of triangles CDE and BFC: h² + (h/2)² = s² (h = BF, h/2 = CF) h² + h²/4 = 900 5/4 h² = 900 h² = 900 * 4/5 = 180 * 4 = 720 h = 4 * 3 √5 = 12 * √5 h/2 = 6 * √5 A = square - triangle CDE - triangle ABE - triangle BCF A = 900 - 2 * 225 - 1/2 * h/2 * h A = 450 - 1/2 * 6 * √5 * 12 * √5 A = 450 - 36 * 5 A = 450 - 180 = 270 square units

Thank you, this was a good way to start the day!

@PreMath

14 күн бұрын

Glad you enjoyed it! You are very welcome! Thanks for the feedback ❤️

In this problem using an orthonormal is very simple. We choose center D and first axis (DC). We then have D(0;0) A(0;30) b(30;30) C(30;0) E(0;15) VectorCE(30;-15) is colinear to VectorU(-2;1). The equation of (CE) is: (x-30).(1) -(y).(-2) = 0 or x +2.y -30 = 0 VectorU is orthogonal to (BF), the equation of (BF) is (x-30).(-2) + (y -30).(1) = 0 or 2.x -y -30 = 0 By solving the system x +2.y -30 = 0 and 2.x -y -30 = 0 we find easily the coordinates of F: F(18;6) Now VectorEF(18; -9) and EF^2 = 18^2 + (-9)^2 = 405 and EF = sqrt(405) = 9.sqrt(5); VectorBF(-12;-24) and BF^2 = (-12)^2 + (-24)^2 = 720 and BF = sqrt(720) = 12.sqrt(5). Finally the area of triangle EFB is (1/2).EF.BT = (1/2).(9.sqrt(5)).(12.sqrt(5)) = 270.

1/ Side of the square= 30 2/ All the white right triangles are similar so FC/FB= 1/2 Let FC = x so FB = 2x --> sqr + 4sqx= 900-> sqx= 900/5 -> x=6sqrt5 and FB=12 sqrt5 Area of the smallest white right triangle = 36x 5=180 Area of the red triangle= 450-180= 270 sq units

Let's find the area: . .. ... .... ..... First of all we calculate the side length s of the square: A(square) = s² = 900 ⇒ s = √900 = 30 The area of the triangle BCE is obviously half of the area of the square. Now we need to know the area of the right triangle BCF. Since ∠CDE=∠BFC=90°, ∠DCE=∠CBF=α and ∠CED=∠BCF=β, the triangles CDE and BCF are similar. So we can conclude: BF/CF = CD/DE = s/(s/2) = 2 Since BCF is a right triangle, we can apply the Pythagorean theorem: BF² + CF² = BC² (2*CF)² + CF² = s² 4*CF² + CF² = s² 5*CF² = s² ⇒ CF² = s²/5 Now we can calculate the size of the pink area: A(BCF) = (1/2)*BF*CF = (1/2)*(2*CF)*CF = CF² = s²/5 A(pink) = A(BEF) = A(BCE) − A(BCF) = s²/2 − s²/5 = 900/2 − 900/5 = 450 − 180 = 270 Best regards from Germany

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

@kristypoh9226

12 күн бұрын

bro ai

φ = 30°; ∎ABCD → AB = BC = CD = AD = AE + DE = a/2 + a/2 = 30 → CE = BE = 15√5 = n CE = CF + EF → sin⁡(EFB) = sin⁡(3φ) = 1 → DCE = CBF = δ; CF = k → EF = n - k BF = m → sin⁡(δ) = √5/5 = k/a → k = 6√5 → EF = 9√5 → m = 12√5 → FBE = θ → sin⁡(θ) = 3/5 → area ∆ BEF = (1/2)sin⁡(θ)nm = 270

1) Total Area of Square [ABCD] = 900 2) Side of Square [ABCD] = 30 = sqrt(900) 3) Half Side of Square = 15 4) Area of Triangle [ABE] + Area of Triangle [CDE] = 450, because : 15 * 30 / 2 = 450 / 2 = 225 5) So, the Area of Triangle [BCE] = 450 6) Base of Triangle [EBC] = EC = sqrt(30^2 + 15^2) = sqrt(900 + 225) = sqrt(1.125) = 15*sqrt(5) ~ 33,54 Now, 7) Base * h = 2*Area 8) 15*sqrt(5) * h = 900 ; h = 900 / 15*sqrt(5) ; h = 60 / sqrt(5) ; h = 60*sqrt(5) / 5 ; h = 12*sqrt(5) ; h ~ 26,83 9) sqrt(FC) = sqrt(900 - 720) = sqrt(180) = 6*sqrt(5) ~ 13,42 10) Area of Triangle [BCF] = 12*sqrt(5) * 6*sqrt(5) / 2 = 72 * 5 / 2 = 180 11) Area of Triangle [BEF] = 450 - 180 = 270 12 Answer : The Pink Shaded Region Area is equal to 270 Square Units.

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

The coolest part of this problem is @ 3:05 and that the areas of triangles EAB & EDC are equivalent to area of EBC. A very cool property! One can do a similar problem with a rectangle inscribed in a triangle where one edge of the rectangle lies on any line of the triangle, then the final result of equivalent triangles is the same but It's journey to determine equivalence is different depending on choice of variables to use. 😉

@PreMath

14 күн бұрын

Excellent! Thanks for the feedback ❤️

No need to use trigonometry, angles or costants. Just look at EBF and BFC triangles, they have the longest leg in common.

AB=30..EF=a,FB=b...a^+b^2=900+225=1125...((√1125-a)^2+b^2=900..1350=2a√1125..a=9√5...b^2=1125-81*5=720..b=12√5...Apink=9√5*12√5/2=54*5=270

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

Great Video again. I used your solution finding propotions over simular triangles with success. Now I want to propose another solution : I calculated EC with the theorem of Pythagoras and got EC = 33,45 LE . I know the area of the triangle EDC ( = ABE ) 225 squareunits. The area of triangle EBC must be 450 squareunits. I know EC and now I can find BF the hight of triangle EBC . BF is 900 / EC = 26,83 LE. Using Pythagoras theorem you can find the lenght of EF . EF = 20,13 LE. so I get for the pink area 20,13 * 26,83 /2 = 270,02 squareunits.

By pythag |EB|= sqroot 1125. So also is |EC|=sqroot 1125. The triangles bfc and edc are similar so |FB|\ 30= 30/ sqroot 1125. Therefore|FB|= 900/ sqroot 1125. In triangle ebf 1125= |EF|^2 plus ( 900/sqroot 1125)^2. So 1125= |EF|^2+ (810000/ 1125) that is |EF|^2 = 405. So |EF| = sqroot 405. Pink area = half |FB| by |EF| = 1/2 ( 900/ sqroot1125)( sqroot 405)= 1/2 by (900 by 9 sqroot5)/ 15 sqroot 5= 270.

3 triangles are similar and the shaded triangle is not so it is a simple matter to get the lengths of all the line segments. 1/2*9*root5*12*root5=270

the moment with the rectangular is not clear. We can prove that the area of an inscribed triangle is half of that of a square by considering two white triangles which both have the equal sides: one is the length of the square and another is a half of that length. Then we do not need to do any additional constructions. All can be solved, as usually, using Pythagorean theorem and finding the hight of the pink triangle.

@PreMath

14 күн бұрын

Thanks for the feedback ❤️

S=270

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

Seems a little complicated when BFC is a similar to ABE and DCE. BFC sidelengths are 2/sqrt5 of DCE. So area is 4/5 of DCE. 900 - 225 - 225 - 180 = 270

@allanflippin2453

13 күн бұрын

Yes, I solved it in this way as well. I didn't see your solution at first when I wrote my comment.

Pink Area = 270 cm^2

If the triangle area is given and the square area is unknown, this problem will be more difficult.

@PreMath

14 күн бұрын

Thanks for the feedback ❤️

I solve it in my way and got 270.2788 square units. Is it right?

@allanflippin2453

13 күн бұрын

It seems like a rounding error somewhere. The answer is definitely exactly 270. Give me a little background and maybe I can find where the error crept in.