Nice colorful complex task. Immediately refreshes the memory of several topics in planimetry and trigonometry. Thank you so much for your math and spoken English lesson, Mr PreMath Have a good working week and may God give you all the best in your life and your pedagogical activity! Sincerely Anatoliy from Ukraine

@PreMath

2 жыл бұрын

So nice of you, Anatoliy. You are too generous with your kind words. Thank you! Cheers! You are the best. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed and safe during these tough times. You are always in our thoughts and prayers!

Once figured out the angle of CEB, I got the EC = 2* EB = 30, as DCE = CDE , DE = EC = 30, as BC = 15 sqrt3, the green area = 1/2 * 30 * sqrt3 = 15 sqrt3.

@giuseppemalaguti435

Жыл бұрын

Sbagliato!!

hello i am 12 years old but i can understand you thank you for explaining and explaining the problems with logic 😊❤️

@PreMath

2 жыл бұрын

Happy to help! Excellent! Thanks for sharing! Cheers! You are awesome, Utku. Keep it up 👍 We are all proud of you! Love and prayers from the USA! 😀

@utkugocmen5886

2 жыл бұрын

Thank you

At first I didn't even think of the tangents of the angles, so I did it through the outer corner: ADC= DCE+CED. So CED is 120°, hence it is clear that CDE = 30°, which means the triangle CDE is isosceles. Now we use the outer angle CED, and find the angle BCE equal to 30°, and since in a right triangle BCE the catheter lying against the angle 30 ° is equal to half of the hypotenuse, the hypotenuse CE will be equal to 30. According to the Pythagorean theorem, we find the cathet CB. CB = 15√3. And finally: using the triangle area formula, we find the area of the triangle CDE. S = 1/2 * DE*CB (DE=CE) means S = 1/2*30*15√3 = 225√3. But your method is more beautiful and accurate.

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Thanks for your amazing feedback! Cheers! You are awesome, Nikolai. Keep it up 👍 Stay blessed 😀

@user-tb4jl7bi6k

2 жыл бұрын

Да! Способ через равнобедр.треугольник СDЕ, а Х=СЕ через косин. и h син. 60 прямуг. треугольника CEB куда очевидней и проще.

That was an easy one for you! Pretty straight forward, but the nice thing about your channel is that you explain every step even if it’s rudimentary so the viewer never gets lost in the process. 👍

Much easier to recognize that triangle ECB is a 30/60/90 triangle, making h (CB) = 15sqrt3 and EC = 30. Because triangle DCE is isosceles, DE (the base) is also 30, and area = (1/2)*30*(15sqrt3) = 225sqrt3

@GillAgainsIsland12

2 жыл бұрын

Agree. Solved it in 1 minute that way.

@PreMath

2 жыл бұрын

Excellent! Many approaches are possible to find the solution of this problem! Thanks for your feedback! Cheers! You are awesome, Tim. Keep it up 👍 Stay blessed 😀

@philippeganty

Жыл бұрын

Same approach. I'm always trying to resolve geometry without using trigonometry!

@shaunkruger

Жыл бұрын

This was also the first approach I saw once I realized that the diagram wasn’t exactly to scale.

@JatinderSingh-xh9hg

11 ай бұрын

That’s what I did… much more straightforward than using trig….but always good to see alternative approaches….

I did it by recognizing that both triangle ECB and DCE are 30-60-90 triangles and applying the ratios accordingly, first to ECB, then once I had h, to DCE....led straight to 225sqrt3. Made me happy because I only get about one in five of his problems.

Nice! Many thanks! DCE = α = 30° ECB = φ ADC = θ = 150°→ CDE = β = 180° - θ = α → α + φ = 60°→ α = β = φ → tan⁡(φ) = √3/3 = 15/BC =15/x → x = 15√3 → CE = DE = 30 → (3/2)225√3 - (1/2)225√3 = 225√3 ≈ 389,711

very well done, thanks for sharing

@PreMath

2 жыл бұрын

Thanks for your feedback! Cheers! You are awesome, my dear friend. Keep it up 👍 Stay blessed 😀

Thank you for sharing!

@PreMath

2 жыл бұрын

Thanks for your feedback! Cheers! You are awesome, Sana dear. Keep it up 👍 Stay blessed 😀

Great example, and using trig too... Thanks again 👍🏻

External angle ADC = 150 deg, which makes angle CED 120 deg. and angle CDE = 30 deg. External angle DEC = 120 makes angle ECB = 30 deg & angle BEC = 60 deg. BEC is a 30-60-90 rt triangle with short leg 15, which makes the hyp. CE = 2(BE) = 30 & long leg BC = 15V3 (15 sqrt(3)). CDE is an isosceles triangle (base angles 30 deg) makes DE = CE = 30. Area of green triangle = (1/2) b h = (1/2) (DE) (BC) = (1/2) (30) (15V3) = 225V3. Done!

@keithwood6459

Жыл бұрын

Your observation that CDE is isosceles sure speeds things up. Nice.

Great 😊😊 Thanks for sharing👌👌🎉🎉

@PreMath

2 жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Stay blessed 😀

One of the neat things about mathematics is that sometimes there is more than just one way to figure out the answer. What really helps confirm the solution to a problem is when multiple methods give the same answer. Sometimes geometry problems can be solved using trigonometry and analytic geometry, and sometimes you can use Euclidean techniques and save yourself a lot of calculations. I have a tendency to use the former method rather than the later, failing to see the short cuts that may exist within Euclidean geometry. After solving the problem along the lines shown by Pre-Math, I saw that there were a couple of things from Euclidean geometry that greatly reduce calculations. Since

@PreMath

2 жыл бұрын

Shape is not true to the scale. I should have mentioned it at the beginning. Many approaches are possible to find the solution of this problem! Thanks for your candid feedback! Cheers! You are awesome, David. Keep it up 👍 Stay blessed 😀

Thanks for video. Good luck sir!!!!!!!!

@PreMath

2 жыл бұрын

Thanks, you too! So nice of you. You are awesome. Keep it up 👍 Stay blessed 😀

225rot3 be the answer. Nice question

@PreMath

2 жыл бұрын

Excellent! Glad you think so! Thanks for your feedback! Cheers! You are awesome, Mustafiz. Keep it up 👍 Love and prayers from the USA! 😀

Angle CEB,= 60 . Hence CE = 2x15=30 .Since angles CDE is equal to DCE(=30°) . Hence DE = CE = 30. Hence area of triangle CDE = 1/2 xDE x CE x Sin 120 = 1/2 x 30 x 30 √3/2 = 225√3

Vnice stepwise explanation

@PreMath

2 жыл бұрын

Glad you think so! You are awesome, Niru. Keep it up 👍 Stay blessed 😀

Alternate method in finding h. Focusing on smaller rt triangle EBC Since we see that triangle EBC is a 30-60-90 rt triangle, and h=EB*sqrt(3) Therefore, h= 15 sqrt(3) Then we can use the tan 30 = h/(x+15) in triangle DBC sqrt(3)/3 = 15*sqrt(3)/(x+15) 1/3=15/(x+15) x+15=45 x=30

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Great Thanks for your feedback! Cheers! You are awesome, Kevin. Keep it up 👍

Interesting

@PreMath

2 жыл бұрын

Glad you think so! Thanks for your feedback! Cheers! You are awesome, Mahalakshmi. Keep it up 👍 Stay blessed 😀

I started off the same way by figuring out the base angles: ∠CDE = 30°, ∠CED = 120°, ∠CEB = 60°. Since △CEB is a 30-60-90 triangle, with side opposite 30° angle = 15, then hypotenuse CE = 2 * 15 = 30 In △CED, ∠CDE = 30° = ∠DCE. So triangle is isosceles with DE = CE = 30. Also, angle between these sides (∠CED) = 120° Area = 1/2 * 30 * 30 * sin(120°) = 1/2 * 30 * 30 * √3/2 = *225√3*

Nice problem. I solved it in a similar way but to find the hight I did the following; A right triangle with an 60º angle always has the proportion 2:1:√3 Triangle BEC is a 60º right triangle and it's shortest side is 15. To contain the right proportions you need to multiply the other two sides by 15 as well. Therefore it's hypotenuse is 30 and it's hight (=line BC) is 15√3

CEB is an isosceles triangle. sin30deg = 1/2,CB=30, tan60 deg= sqrt 3, h=15sqrt 3 Area of green triangle = 1/2 (30)(15sqrt3)=225 sqrt3.

Yes i solved it mentally. An awesome question btw!

@PreMath

2 жыл бұрын

Excellent! Glad you think so! You are awesome, Sreehari. Keep it up 👍 Stay blessed 😀

Nice.

@PreMath

2 жыл бұрын

Thank you! Cheers! Thanks for your feedback! Cheers! You are awesome, Luigi. Keep it up 👍

Thnku

@PreMath

2 жыл бұрын

You are very welcome! You are awesome, Pranav. Keep it up 👍 Stay blessed 😀

Wow!!!!great!

@PreMath

2 жыл бұрын

Excellent! Glad you think so! You are awesome, Phan. Keep it up 👍 Love and prayers from the USA! 😀

@vananphan6803

2 жыл бұрын

@@PreMath thanks! From Vietnamese! Best wishes to you and your family!👍

Segment CD = segment CE by Isosceles property. Angle DCE = angle CDE.

ꓥCDE=180º-150º=30º → ꓥCED=180º-30º-30º=120º → ꓥCEB=180º-120º=60º → ꓥECB=180º-60º-90º=30º Deducting the value of these angles, we can see that the triangle ∆ECB is half of an equilateral triangle and we can deduce the length of several segments: EC=2x15=30 and CB=15√3. The value of EC could also have been deduced from the well-known plot of the double angle that allows us to deduce: If ꓥCEB=60º and ꓥCDE=30º → ꓥCEB=2ꓥCDE → DE=EC=30 On the other hand, ∆CDB is also half of an equilateral triangle and we can deduce the following: DB=CB√3=15√3√3=15x3=45 → DE=DB-EB=45-15=30 With the above data we can now calculate the area sought: Area ∆DCE=DExCB/2=30x15√3/2=15x15√3=225√3

@PreMath

2 жыл бұрын

Great Thanks for your feedback! Cheers! You are awesome, Santiago. Keep it up 👍 Stay blessed 😀

Awesome explanation

@PreMath

2 жыл бұрын

Glad you liked it Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@manis1522

2 жыл бұрын

@@PreMath I'm From India

@PreMath

2 жыл бұрын

@@manis1522 Great! You are the best.

@manis1522

2 жыл бұрын

@@PreMath Thank you so much sir

angle EDC=angleDCE=30° SO DE=CE=X then realation between "EB=15" and "x" by trigonometric ratio with angle 60° we can get as well as "h". so we can get result

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Thanks for sharing! Cheers!

nice

@PreMath

2 жыл бұрын

Great! Thanks for your feedback! Cheers! You are awesome, Susen Keep it up 👍 Stay blessed 😀

I found an easier method. BCE is a special right angle triangle. CE = 15x2 =30 and square root of 3 x15 is the height. The green triangle is isosceles triangle so DE =EC=30.

You could find that x = 30 at the beginning by saying that in front of equal angles there are equal sides.

@PreMath

2 жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍

first very great!

@PreMath

2 жыл бұрын

Yes you are! So nice of you, YoonHo You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@SuperYoonHo

2 жыл бұрын

@@PreMath awesome!

Angle ECB=30 and angle CEB=60. Sin 30=EB/EC. EC=EB/sin30. EC=15/0.5 EC=30. Sin 60=CB/EC. CB=ECxsin60. CB=30x0.8660. CB=25.98. Triangle EDC is isosceles,so DE=EC=30. Area green triangle is 0.5x30x25.98=389.7.

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Thanks for your feedback! Cheers! You are awesome, Monty. Keep it up 👍 Stay blessed 😀

6:32 cross multiply fractions on each side of an equation.

Having done the same steps determining angIes and |CE| I went a bit another way A=1/2*DE*EC*sin(DEC)=1/2*30*30*sin(D120)=225sqrt(3).

1:48 aka Supplementary angle property

@PreMath

2 жыл бұрын

Thanks for your feedback! Cheers! You are awesome, Chavo. Keep it up 👍 Stay blessed 😀

S=225√3≈389,25

I used the fact that we have 30-60-90 triangles and an Isosceles triangle. It's a much simpler approach.

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Stay blessed 😀

I think This is easier than the others

you can also use law of sines: AB/sin(30) = AC / sin(90) ==> AC = 30 units then use same logic to find DC in triangle DEC then use area rule: A = 1/2*a*b*sin(C) 225*√3

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Thanks for your feedback! Cheers! You are awesome, Martin. Keep it up 👍 Stay blessed 😀

So easy

DR.IS ANY triangle have two angles equal is isosceles? how the green triangle has two angle is 30 degree and not isodceles triangle

If the angles 1:2:3 in right angled triangle the sides 1:√3:2 so height is15✓3 and also base of biggest triangle is 15✓3×✓3 =45

How are you my favourite teacher?

@PreMath

2 жыл бұрын

I'm doing super! Thanks for checking up on me. So nice of you, Aniwar dear You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Cũng đơn giản. Sử dụng pitago 1 chút là ra kết quả mà

@PreMath

2 жыл бұрын

Cảm ơn phản hồi của bạn! Chúc mừng! Bạn thật tuyệt vời. Giữ nó lên

resolved without paper

@PreMath

2 жыл бұрын

Excellent! You are awesome. Keep it up 👍

I just used sines and cosines of the angles 30 and 60, they're enough this .

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Thanks for your feedback! Cheers! You are awesome, Vegeta. Keep it up 👍 Stay blessed 😀

@ssjbevegeta3008

2 жыл бұрын

@@PreMath thanks

I could tell early in the problem that the diagram was not an accurate representation when I found that the smaller angles in the middle triangle were 30 degrees and the sides opposite those angles were not close to equal. I noticed that the triangle on the right was a 60,30,90 triangle with proportional sides of x,2x, and root3 x. Using side EB = 15 Triangle then has proportional sides of 15,30, and 15 root 3. DE = CE = 30 because they are equal sides opposite equal angles of triangle CDE. Finally, the area of Triangle CDE is (30x15root3)/2 = 225 root3.

15^2*sqrt3

@PreMath

2 жыл бұрын

Great Thanks for your feedback! Cheers! You are awesome, Jaime. Keep it up 👍 Stay blessed 😀

I skipped the trig and used the triangle EBC as a 30/60/90 so I know the sides are in a 1/2/3^.5 relationship so the BC is 15*3^.5. We know that triangle DBC is a 30/60/90. That means DB = 15*3.^.5*3^.5 = 45. 45-15 = 30 for our base and our height is still 15*3^.5. :-) Always more than one way to skin a cat.

@PreMath

2 жыл бұрын

Very true! Many approaches are possible to find the solution of this problem! Thanks for your feedback! Cheers! You are awesome, Arthur. Keep it up 👍 Stay blessed 😀

@arthurschwieger82

2 жыл бұрын

@@PreMath - It has been watching your videos that has gotten me thinking about geometry and higher math again. I have gotten my brother working on these and even my mom on some of the simple ones or I create something for her that is at her level of relearning math. I started the journey so I could stay one step ahead of my daughter as she goes through more complex math at school. Keep up the great work and full explanation on each video. I appreciate the reminders about axioms related to the problem.

Giving the value of AD=15 is not necessary to solve the problem. An alternative method to approach it is to observe that CE=DE and CE is equal to (1/cos 60 )* 15 = 30. This leads to the required area A=1/2 * 30^2 * sin 120 = 225 sqrt(3).

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Thanks for your feedback! Cheers!

I guess we don't need to know the length of AD: CDE = 180° - 150° = 30° CED = 180° - 30° - 30° = 120° BEC = 180° - 120° = 60° BCE = 180° - 90° - 60° = 30° Triangle BCE is a 30-60-90° triangle: - BE = 15 - CE = 2 * 15 = 30 - BC = 15 * √3 Triangle BCD is a 30-60-90° triangle: - BC = 15 * √3 - CD = 2 * 15 * √3 = 30 * √3 - BD = 15 * √3 * √3 = 15 * 3 = 45 => DE = 45 - 15 = 30 A = 1/2 * DE * BC = 1/2 * 30 * 15 * √3 = 15 * 15 * √3 = 225 * √3 ≈ 389.7 square units

@PreMath

2 жыл бұрын

Many approaches are possible to find the solution of this problem! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Stay blessed 😀

225√3

@PreMath

2 жыл бұрын

Great Thanks for your feedback! Cheers! You are awesome, Atilla. Keep it up 👍 Stay blessed 😀

This is impossible, the reason is from following steps: step1. get boundary condition: AC=sqrt(AB^2+BC^2)=15sqrt(19), CD=sqrt(DB^2+BC^2)=15sqrt(12), sin(CAD)=sin(CAB)=CB/AC=sqrt(3)/sqrt(19),cos(CAB)=4/sqrt(19), sin(ACD)=sin(180-150-DAC)=sin(30-DAC)=sin(30-CAB)=2/sqrt(19) - 3/2sqrt(19)=1/2sqrt(19), step2. By triangle formula, BC/sin(A)=AC/sin(B)=AB/sin(C): in ACD: AC/sin(ADC)=15sqrt(19)/(1/2)=30sqrt(19) CD/sin(CAD)=15sqrt(12)/[4/sqrt(19)]=15sqrt(57)/2, not 30sqrt(19) AD/sin(ACD)=15/[1/2sqrt(19)]=30sqrt(19) step3. in ACD, formula BC/sin(A)=AC/sin(B)=AB/sin(C) is not exist. So, This question is impossible.

proportions are crap i didn't want to believe it was an isosceles triangle but i guess it isn't to scale

Unsanely innecesary steps... EBC is 30/60/90 so BC = 15√3 EB = 30 DEC is isosceles ∆ so DE = 30 Finally A = (DE)(BC)/2 = (30)(15√3)/(2)= 225√3

It seems that x is smaller than 30 according to the graph

@PreMath

2 жыл бұрын

Shape is not true to the scale. I should have mentioned it at the beginning. Thanks for your feedback! Cheers! You are awesome, Ortega. Keep it up 👍

Too much long method