Can you find area of the Trapezoid ABCD? | (Trapezoid) | (Trapezium) | #math #maths | #geometry

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Learn how to find the area of the Trapezoid ABCD. Important Geometry and algebra skills are also explained: Trapezoid; Trapezium; Exterior angle theorem; Triangle area formula. Step-by-step tutorial by PreMath.com
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Can you find area of the Trapezoid ABCD? | (Trapezoid) | (Trapezium) | #math #maths | #geometry
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Комментарии : 54

@CloudBushyMath 5 месяцев назад

Great and Enjoyable

@PreMath 5 месяцев назад

Excellent! Glad to hear that! Thanks ❤️

@marcgriselhubert3915 5 месяцев назад

Be H the orthogonal projection of C on (AB), h the height of the trapezoïd h = CH, L = AB and l = DC. The area of the trapezoïd is ((L + l)/2).h In triangle AHC we have h = CH = AC.sin(15°) = 10.sin(15°), and also AH = AC.cos(15°) = 10.cos(15°), so HB = AB - AH = L - 10.cos(15°) By symetry (the trapezoïd beeing isosceles) we have: L = l +2.HB, so L = l +2.(L - 10.cos(15°)) , so L = l + 2.L -20.cos(15°) and 0 = l + L - 20.cos(15°) So we have l + L = 20.cos(15°). Now the area of the trapezoïd is ((L + l)/2).h = (10.cos(15°).(10.sin(15°) = 100.sin(15°).cos(15°) = 50.sin(30°) = 25.

@PreMath 5 месяцев назад

Excellent! Thanks for sharing ❤️

@misterenter-iz7rz 5 месяцев назад

Simply we can assume the exemete case, C,D coincides, thus there is an isosceles with common equal length 10, therefore the area is 1/2 10×10×sin 30=25.😅 Second way to solve for general case. Let 10=a+b, then we have four triangles, and the answer is exactly the sum of areas of all triangles=1/2×(axa+a×b+b×b+b×a)sin 30=1/2×(a+b)^2. sin 30=1/2×10^2 ×1/2=25.😊

@PreMath 5 месяцев назад

Excellent! Thanks for sharing ❤️

@kinno1837 5 месяцев назад

In 5:40, the steps here are not rigorous enough because you haven't prove that CD=BE. We can't actually determine that the area of the trapezoid equals to the area of triangle ACE since there are only SSA but not SAS. Therefore,our mission is to prove angle DAC=angle BCE. Suppose O is the intersection point of blue lines. First, we can prove DO=CO since angle OCD=angle ODC=15°. Then, AO=10-CO=10-DO, BO=10-DO.So AO=BO. Next, AD=BC. Thus, triangle AOD is congruent to triangle BOC. Then, angle DAO=angle CBO.(corr. angles,~∆s) So angle ADC=180°-15°-angle DAC angle CBE=180°-15°-angle DBC=180°-15°-angle DAC(proved) Therefore, angle ADC=angle CBE. Finally, angle BCE=180°-15°-(180°-15°-angle ADC)=angle DAC Since AD=BC, angle DAC=angle BCE, AC=CE, triangle DAC is congruent to triangle BCE(SAS) and we can prove that their area are the same. Anyway, that's interesting math question. Your videos are amazing and clear.🎉

@engralsaffar 5 месяцев назад

The way I did it is that I reconstructed the trapezoid into rectangle AB'CD', which has the same area because of symmetry, where, h = CB', b=AB' sin15=h/10, cos15=b/10 sin(2*15)=2sin15cos15=2*h/10*b/10=hb/50=sin30=1/2 Therefore Area=hb=50/2=25 square units

@jimlocke9320 5 месяцев назад

At 5:40 PreMath has determined that the area of ΔACE is equal to the area of the trapezoid, so the problem becomes one of determining the area of ΔACE. Another way to compute the area is to drop a perpendicular from C to AE, dividing ΔACE into 2 15°-75°-90° right triangles. When the hypotenuse of a 15°-75°-90° right triangle is treated as the base, the height is 1/4 the length of the hypotenuse, or 2.5 in this case. The area is (1/2)bh = (1/2)(10)(2.5) = 12.5 and must be doubled because there are 2 congruent triangles. So, the area of ΔACE is 25 and this is also the area of the trapezoid. Our answer is the same as PreMath's. If you don't recall the above property of a 15°-75°-90° right triangle but do recall that the ratio of sides, short, long, hypotenuse, is (√3 -1):(√3 + 1):(2√2), you can compute the area as follows. The short side has length ((√3 - 1)/(2√2))(10) and long side ((√3 + 1)/(2√2))(10). So, for each triangle, using the long horizontal side as the base, the area is (1/2)((√3 + 1)/(2√2))(10)((√3 - 1)/(2√2))(10). Regrouping, area = (1/2)(√3 + 1)(√3 - 1)(10)(10)/((2√2)(2√2)) = (1/2)(3 - 1)(100)/(8) = 12.5. Since there are 2 triangles, we must double the area, getting 25, which is also the area of the trapezoid. If the diagram is not 100% true to the scale, we can make a couple observations. These observations depend on how true the diagram is to the scale! In the extreme, I have seen problems where an angle was actually obtuse when the diagram showed it as acute, so am not completely "off the wall". First observation:

@PreMath 5 месяцев назад

Excellent! Thanks for sharing ❤️

@johnspathonis1078 5 месяцев назад

I used a similar method but arrived at the solution using the expansion of Sin(2A) = 2.Sin (A).Cos(A) where A = 15 degrees. The base equation for the area is 100. Sin(15).Cos(15) By substituting the answer reduces to 25.

@זאבגלברד 5 месяцев назад

A Trapezoid has one and only one pair of parallell sides. Very nice out-of-the box approach. Area of ACE is 10x10xsin150 /2 ... we should know sin 150 .... Thank you for great videos.

@ludosmets2018 5 месяцев назад

Extend DC to the left and drop a perpendicular EA. Drop a perpendicular CF to AB. The area of the rectangle AECF = 2 times the area of triangle ACF. The area of the trapezium = the area of the rectangle AECF + area of triangle BCF - area of triangle ADE. By symmetry these triangles are equal. So the area of the trapezium is simply 2 times the area of triangle ACF. CF is 10sin15°. Use pythagorean theorem to calculate AF and multiply the result by 10sin15° to calculate the area of rectangle AECF which is 25.

@johnjones8580 5 месяцев назад

AF would simply be 10 cos 15⁰.

@PreMath 5 месяцев назад

Thanks for sharing ❤️

@cristidan1380 5 месяцев назад

CP is 5 in the 30-60-90 triangle. PE which is used as height is 10 root 3. Am I the only one that sees it?

@trumpetbob15 5 месяцев назад

I had the same thought, but if you look, the issue is the triangle isn't drawn properly. The angle PCE is 30 degrees so the side opposite would be the length 5 (half of the hypotenuse), which is side PE.

@cristidan1380 5 месяцев назад

@@trumpetbob15 indeed, my bad, thx!

@hongningsuen1348 5 месяцев назад

Don't be misguided by the picture. CP is opposite to 60 angle ( in picture it looks like 30). Hence CP is 10 root 3, not 5.

@PreMath 5 месяцев назад

Thanks for sharing ❤️

@Waldlaeufer70 5 месяцев назад

h = 10 * sin (15°) (height of trapezoid) m = 10 * cos (15°) (average width of trapezoid) A = h * m My thought was: If you look at the diagonal from the bottom left-hand corner to the top right-hand corner, its projection onto one of the two parallel sides (cosine function) must correspond exactly to the average of these two parallel lines. If you move this projection along one of the sides to half its height, its ends will exactly touch both side walls of the trapezoid, i.e. it will correspond to the average of the two parallel sides. Therefore: A = h * m = 10 * sin (15°) * 10 * cos (15°) = 100 * sin (15°) * cos (15°) = 25 square units

@JAMESYUN-e3t 5 месяцев назад

A great question and a great solution. I prefer these kinds of math questions to hackneyed ones

@think_logically_ 4 месяца назад

O - intersection point of AC and BD, AO=BO=a, CO=DO=b. Then ∠AOD=∠COB=30°, ∠AOB=∠COD=150°, sin(150°)= sin(30°)=1/2. [AOB]=1/2 * a² * sin(150°) = a²/4, similarly [COD] = b²/4, [AOD] = [BOC] = ab/4. [ABCD] = [AOB]+[COD]+[AOD]+[BOC]= a²/4 + b²/4 + 2 * ab/4 = 1/4 (a²+b²+2ab) = 1/4 (a+b)² = 10²/4 = 25.

@LuisdeBritoCamacho 5 месяцев назад

I used Algebraic Geometry (2D) to solve this Problem. 1) x^2 + y^2 = 100 and y = tan(15º)x 2) The Point of Interception is I -> (9,659 ; 2,588) 3) The height of the Trapezoid is 2,588 lin un. But it can be also calculated by : h = 10*sin(15º) ; h ~ 2,588 lin un; wich is the very same. 4) Now is understandable that AC = AB = 10 5) Making easy Interceptions I calculated the Minor Base to be : b ~ 9,659 - 0,341 ; b ~ 9,32 lin un 6) Trapezoid Area = B + b * (h/2) ; TA = 19,32 * (2,588/2) ; TA = 19,32 * 1,294 ; TA = 25,00008 sq un 7) Answer : The Trapezoid [ABCD] Area is Equal to 25 Square Units.

@hongningsuen1348 5 месяцев назад

Draw CC' perpendicular to AB with C' on AB. Draw AA' perpendicualr to CD with A' on CD. AA'CC' is a rectangle. (All angles are right angles by construction.) Triangles AA'D is congruent to triangle CC'B (RHS) hence they have equal areas. Hence area of trapezoid ABCD = area of rectangle AA'CC'. (replacing CC'B by AA'D) Height of rectangle AA'CC' = CC' = 10 sin15 Width of rectangle AA'CC' = AC' = 10 cos 15 Hence area of rectangle AA'CC' = 100 sin 15 cos 15 = 25 = area of trapezoid ABCD.

@PreMath 5 месяцев назад

Excellent! Thanks for sharing ❤️

@alexj8940 5 месяцев назад

Make the rectangle, it doesn’t affect conditions. Area=10xsin15x10cos15=10x10x1/2sin30=25. Sin30=2sin15cos15

@unknownidentity2846 5 месяцев назад

Let's find the area: . .. ... .... ..... First of all we add the points E and F on AB in order to obtain the rectangle CDEF with CD=EF=a and CF=DE=h. Due to the sketch we can conclude that the trapezoid ABCD is a symmetric trapezoid. Therefore we obtain AE=BF=b. Since the triangles ACF and BDE are congruent right triangles, we can conclude: A(ABCD) = (1/2)*(AB + CD)*h(AB) = (1/2)*(AE + EF + BF + CD)*h(AB) = (1/2)*(b + a + b + a)*h = (1/2)*(2*a + 2*b)*h = (a + b)*h = (EF + AE)*CF = AF*CF = [AC*cos(∠FAC)]*[AC*sin(∠FAC)] = AC²*sin(∠FAC)*cos(∠FAC) = AC²*(1/2)*sin(2*∠FAC) = 10²*(1/2)*sin(2*15°) = 100*(1/2)*sin(30°) = 100*(1/2)*(1/2) = 25 Best regards from Germany

@PreMath 5 месяцев назад

Great! Thanks for sharing ❤️

@kaliprasadguru1921 5 месяцев назад

Sir , kindly explain why and how lines AB and BE will lie in one line .

@kimchee94112 5 месяцев назад

At 7:50 the triangle ECP are way out of proportion where 30 degrees looks like 60 and 60 looks like 30. It was mentioned not 100% true to scale, but ....

@jamestalbott4499 5 месяцев назад

Good challenge, manipulating the trapezoid, then finding a 30-60-90 triangle.

@PreMath 5 месяцев назад

Excellent! Thanks for sharing ❤️

@nancyartiagaescoto7625 5 месяцев назад

Get well soon Intoy

@kaliprasadguru1921 5 месяцев назад

Area of triangle ACE =1/2 xAC x CE x sinACE = 1/2 x 10 x 10 x sin 150°=25

@PreMath 5 месяцев назад

Excellent! Thanks for sharing ❤️

@easy2forget2ch 5 месяцев назад

Same area as an isosceles triangle with an apex angle of 30° and equal sides of 10.

@tontonbeber4555 5 месяцев назад

I'm always going brute force with this kind of problem ... A(0,0) B(x,0) C(10 cos 15, 10 sin 15) D(x - 10 cos 15, 10 sin 15) AB = x CD = 20 cos 15 - x (AB+CD)/2 = 10 cos 15 height = 10 sin 15 Area = 100 sin 15 cos 15 = 50 sin 30 = 25

@ulissepallaro8331 5 месяцев назад

Hi i'm an 15 years old boy from italy and i always watch your video; they ate fandastic and very well made. I and one of my friend are going to attempt IMO at the italian final stage and we'd like to write a book with lots of math exercises in preparaction yo Cesenatico (italian stage).We think that the ezercise you post on RU-vid fit what are we going to do. So,we are asking if for you is okey id we use some of your exercise and if you'd like to revise the argomentsof the soluction we are going to write based on the video.and we'd also like yo know if the exercise are taken form some olimpiad book and if there is copyright on them.

@mccloysong 5 месяцев назад

I got confused 7:15 … 30º and 60º angles seem to be swapped, no?

@selcukketenci884 5 месяцев назад

CE=10 CP=5 PE=5 root 3. Area=10*5root3/2=43,30…..

@johnjones8580 5 месяцев назад

Cut off one end of the trapezoid, flip it, and attach it to the other end to turn the diagram into a rectangle with the same area. The height will be 10 sin 15⁰. The width will be 10 cos 15⁰. A=(10 sin 15⁰)(10 cos 15⁰) A = 25

@PreMath 5 месяцев назад

Thanks for sharing ❤️