Can you find area of the Trapezoid ABCD? | (Trapezoid) | (Trapezium) |

[@CloudBushyMath]

Great and Enjoyable

[@PreMath]

Excellent! Glad to hear that! Thanks ❤️

[@jimlocke9320]

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]

Excellent! Thanks for sharing ❤️

[@johnspathonis1078]

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.

[@mccloysong]

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

[@kinno1837]

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.🎉

[@Z-eng0]

I feel you there, I totally did a similar proof to what you did, before continuing the video, especially since I tried to solve it before playing the video, and was only able to prove the congruency of the right and left triangles (by SSS) and the similarity of the top and bottom triangles (by AA), although I did it slightly differently. But when he constructed the new triangle he shouldn't have assumed that if he extended AB and connected a 10unit side (CE) angle BEC would be 15deg or vice versa, he could've just put the 10unit length then stated that since CE = AC, therefore angle CEB = angle CAB

[@marcgriselhubert3915]

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]

Excellent! Thanks for sharing ❤️

[@misterenter-iz7rz]

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]

Excellent! Thanks for sharing ❤️

[@ludosmets2018]

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]

AF would simply be 10 cos 15⁰.

[@PreMath]

Thanks for sharing ❤️

[@engralsaffar]

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

[@זאבגלברד]

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.

[@hongningsuen1348]

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]

Excellent! Thanks for sharing ❤️

[@cristidan1380]

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]

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]

​@@trumpetbob15 indeed, my bad, thx!

[@hongningsuen1348]

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]

Thanks for sharing ❤️

[@unknownidentity2846]

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]

Great! Thanks for sharing ❤️

[@kimchee94112]

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 ....

[@Waldlaeufer70]

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

[@think_logically_]

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]

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.

[@kaliprasadguru1921]

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

[@JAMESYUN-e3t]

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

[@alexj8940]

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

[@nancyartiagaescoto7625]

Get well soon Intoy

[@easy2forget2ch]

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

[@jamestalbott4499]

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

[@PreMath]

Excellent! Thanks for sharing ❤️

[@tontonbeber4555]

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

[@selcukketenci884]

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

[@kaliprasadguru1921]

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

[@PreMath]

Excellent! Thanks for sharing ❤️

[@ulissepallaro8331]

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 KZbin 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.

[@fhffhff]

S=10cos15°*10sin15°=50sin30°=25

[@PreMath]

Thanks for sharing ❤️

[@sergeyvinns931]

S=1/2*10*10*sin30= 25!

[@PreMath]

Excellent! Thanks for sharing ❤️

[@johnjones8580]

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]

Thanks for sharing ❤️

[@prossvay8744]

Area trapezoid=1/2(10)(10)sin(150°)=29 square units.❤❤

[@ИванПоташов-о8ю]

25 s.u.

[@PreMath]

25 Thanks for sharing ❤️

[@wardieleppan8443]

Wouldn’t PE equal 5 Times Square root of 3 and not 5? CP=5

[@رافعالورفلي-ص5ش]

10*10*sin 150°\2

[@soli9mana-soli4953]

with trigonometry 25

[@PreMath]

Excellent! Thanks for sharing ❤️