Cool Problem. I attacked it too complicatedly. While I figured out the Pythagoras properties of the right triangles and the fact that b-a=h, I had chosen the variables in an unfortunate way and ended up with some sqrt-expressions. Knowing the formula for the area of a trapezoid would have helped! I was very close to the solution, but „couldn‘t see it“. But I again learned something!
Great example, I learned much from this, i need practice at these types that have multi stage solutions..soo very interesting to see and follow the logical release of the solution, I got as far as seeing the two right triangles then missed the difference of two squares, more like this I love these complicated ones, where one needs a strategy, thanks again 👍🏻
Simply draw a perpendicular line from D, then a = b + h. Area = [(b + b + h) * h]/2 = (2bh + h) / 2 From the two diagonals, Substract the above: So area = (2bh + h) / 2 = 136 /2 = 68
Did it in slightly different way but very similar. I started by drawing the line DE, dividing the figure in a rectangle (base = x) and a right triangle (base = y) so the large base of the trapezoid equals to x+y and the small base is equal to x. Same approach on the right triangle which appears to be an isosceles triangle : DE equals to y. So the area of the trapezoid ((small base+large base)*height/2) becomes : (x+y+x)*y/2=(2x+y)*y/2=(y²+2xy)/2=A. Using the Pythagorean theorem, I had two equations : (x+y)²+y²=19² and x²+y²=15² => x²+2xy+y²+y²=19². x²+y²=15² so 2xy+y²=2*A=19²-15²=(19-15)*(19+15)=4*(34) => A=2*34=68 square units.
by drawing a symmetric trapezoid along the (BC) line, we can find the area of the trapezoid is also equal to (a²-b²)/2. by soustracting the 2 first pythagorean equations from this video, we find (a²-b²)/2 = (19²-15²)/2 = (361-225)/2=68... And that's the answer ! !! ^^ (sorry for my english)
This is a very interesting problem. Professor often has problems where you can't solve for the individual dimensions because there is a range of solutions. Typically something like you can solve for a+b but can't solve for a and b individually because there are a range of solutions. And yet, there is just one solution for a+b and the parameter the problem is asking for is a function strictly of a+b, hence there is a unique solution. But this problem is different. Based on the geometry it appears that there is definitely a single solution for b and h, and if you got those you then are able to calculate the area of the trapezoid. However, it seems to be a major pain in the butt to actually find b and h. That's the way I tried to solve it. By using pythagorean on the two known diagonals to get b^2+h^2=15^2 and (b+h)^2+h^2=19^2. Since you have two equations and two unknowns it is reasonable to believe there is a single solution to the problem. But then if you try to solve these two equations and two unknowns it is a mess because of the 2*b*h term in the second equation. So Professor's solution is a clever way where you don't need to deal with that. So this is an excellent problem in that it looks pretty easy but it is not quite so easy. I couldn't solve it. I should have looked at the trapezoid formula, then I would have realized that I had the solution without knowing it.
Your videos are always quite enjoyable. I love the moment when I suddenly see where you are going. Great fun. What input devices do you use, and what software? TIA -- Dan
From the figure: a=b+h, and so area = hb+h^2/2. Pythagoras: h^2+b^2=225 and h^2+(h+b)^2=361. Epanding the last and subtracting the first yields h^+2bh=361-225. Left side is twice the area formular, so the area is (361-225)/2 = 68
construct a perpendicular line from D to BC, call the point E. let’s lable some lenghts: AB=DE=h (the h stands for height), AD=BE=b (b stands for base). the area of the trapezoid will be bh+((h^2)/2) look at triangle CDE. we have one 90* angle (E) and one 45* angle (C), so the other angle (D) is also 45*. this means that it is a 1:1:sqrt(2) special triangle. lets fill in; 1:1:sqrt(2)=CE:DE:CD, we know that DE=h, CE:DE:CD=1•h:h:sqrt(2)•h, so CE is h aswell. note that triangle ABC and BDE are triangles where pythagorean works. in ABC:AB^2+BC^2=h^2+(b+h)^2=2h^2+b^2+2bh=19^2=361. and in BDE: b^2+h^2=15^2=225. now we have a system of equations: {2h^2+b^2+2bh =361 h^2+b^2=225 if we substract the equations, we get h^2+2bh=136 recall that the area of the trapezoid is bh+((h^2)/2). if we divide h^2+2bh=136 by 2 we get exactly bh+((h^2)/2) and what will be equal to 136/2=68 kind of proud of being able to solve it like this
Hey premath ... Please tell me how did you shot this video. Softwares, Hardwares, and anything else if needed. Your videos are very simple yet effective.
Brute force approach: calculate a, b and h and solve using equation for area of a trapezoid. Equation 1: a² + h² = 361 Equation 2: b² + h² = 225 Equation 4: h = a - b so a = b + h Substituting in equation 1: (b + h)² + h² = 361 which expands to: b² + 2hb + 2h² = 361 From equation 2: b² = 225 - h² and b = √(225- h²) so: 225 - h² + 2h√(225- h²) + 2h² = 361 Combining terms: h² + 2h√(225- h²) - 136 = 0 and 2h√(225- h²) = 136 - h² Squaring both sides: 4 h²(225 - h²) = 18496 - 272h² + h⁴ Let x = h² so we can reduce this to a quadratic: 4x(225 - x) = 18496 - 272x + x² which simplifies to: 5x² - 1172x + 18496 = 0 Solving: x = 117.2 +/- 0.4√(62729) Squaring both sides has introduced an invalid solution, only x = 117.2 - 0.4√(62729) is valid. So: h² = 117.2 - 0.4√(62729) b² = 107.8 + 0.4√(62729) a² = 243.8 + 0.4√(62729) From which: h ~= 4.1252 b ~= 14.422 a ~= 18.547 Area = ((a + b)h)/2 ~= 68.0018594 This method provides an approximate value which is within 0.003% of PreMath's exact value 68. Carrying the calculation to more places yields a value closer to 68. Leaving the radicals and simplifying, there must be a way to get an exact value of 68.
note that if the diagonals were b and c, then the area is (c^2-b^2)/2 , note also in this case and ingeneral, the linear dimensions are complicated involving nested square roots.
1.draw perpendicular DO on BC 2.you will get a rectangle with AD=Bo=x , AB=DO=y and OC=OD/tan45=y (tan45=1) On applying Pythagoras 1. X²+y²=225 2.y²+x²+y²+2xy=361 3.y²+2xy=361-225=136 4.(y²+2xy) /2 = 68 On looking carefully we find that area of trapezoid ABCD=(y²+2xy) /2
My aproach is a little bit different. In the picture we have a rectangle ABED and a right isosceles triangle DEC. So AD=BE= b (small base of the trapezoid) and AB=DE=CE=h ( the height of the trapezoid). Consider the triangle BED we have sq DE+sqBE=sq 15=225 ; and in the right triangle ABC we have: sq(BE+CE) + sqCE = sq19=361-----> sq BE + sq CE + 2xBExCE + sq CE = sq BE+ sq DE + 2 BExCE + sq DE. because sq DE+sqBE = 225 -----> 225 + 2xBExh + sq h = 361------> 2 BExh + sq h = 136------> BE xh + sq h/2 = 68. Because BExh is the value of the area of the rectangle ABED and sq h/2 the area of the triangle DEC so the area of the trapezoid is 68 units
Variation on PreMath's solution: We observe that the trapezoid can be divided into a rectangle with height h and base b and an isosceles right triangle with sides of length h ( 6:30 ). The rectangle has area bh and the triangle 0.5h², for total area of bh + 0.5h². The two equations for the right triangles are: 1) (b + h)² + h² = 361 2) b² + h² = 225 Expanding 1): b² + 2bh + 2h² = 361 Subtracting 2): 2bh + h² = 136 Dividing both left and right sides by 2: bh + 0.5h² = 68 However, the left side is the formula for the trapezoid's area as derived above, so the area of the trapezoid is 68.
I labelled AD = BE = x. EC = y. and AB = h. Hence, Area of Trapezium = (h/2)(2x+y). By Pythagoras Theorem, I obtained BC^2 - AD^2 = 136. (BC + AD)(BC - AD) = 136 (2x + y)(y) = 136. ---> (2x+y) = (136/y) ---> Area of Trapezium = (h/2)(136/y) = (68z/y). tan (z/y) = 45. --> z/y = 1. --> length of z = length of y. Therefore, Area of Trapezium = 68.
Let AB = x and AD = y. Then, the Perpendicularity Lemma gives 136 = (x+y)^2 - y^2 = x(x + 2y). Since the sum of the bases of the given trapezoid is precisely x + 2y, its area is just 136/2 = 68.
teor. Pitagora for ABC e ABD … ABsquared=ACsquared-BCsquared=BDsquared-ADsquared … or ACsquared-BDsquared=BCsquared-ADsquared=(19+15)*(19-15)= (BC+AD)*(BC-AD) … (1) da BCD=45°, e DH⊥BC , DH=CH
CH=BC-AD , therefore (1) becomes 34*4=(somma basi trapezio)*(altezza trapezio)= 2*area trapezio where (area trapezio)=68
Обозначим АВ как а, АD как b, тогда ВС будет равно a+b. Площадь трапеции равна произведению полусуммы оснований на высоту P=a(a+2b)/2. a и b находим из прямоугольных треугольников ВАС и ВАD. a^2+b^2=225, a^2+(a+b)^2=361. a^2+a^2+2ab+b^2=361. b^2=225-a^2. a^2+a^2+2ab+225-a^2=361. a^2+2ab=136. a^2+2ab/2=68, а это и есть площадь трапеции Р=68!
Let AD = a AB = b Since angle C= 45 BC = a+b In triangle ABD a^2+b^2 = 225..eq1 In triangle BCD (a+b^2+b^2 = 361..eq2 eq2-eq1 b^2 + 2ab = 136...eq3 Dividing both sides by 2 we get area Because b^2 area of two triangles(square part) And 2ab is area of two rectangle part Hence Area = 68
Although the process of arriving at the solution appeared to be sound, there is a huge flaw with the whole exercise if analyzed in detail. If all the given data have to be maintained the trapezoid is a nonexistent trapezoid If AC is really 19 units long and the angle at point C is 45 degrees then it is a false trapezoid since you cannot form a true one without changing the 45 degree dimension. Drawing the trapezoid to scale using the lines dimensions given it would turn out that the angle at point C is NOT 45 degrees - it is actually greater by a huge amount. That is significant and important because that given degree was used to calculate the relationship between ( a-b ) and h to claim an isosceles triangle. Also the arrived-at final answer is way smaller when you consider that the rectangle ABED contains two triangles that are similar and equal in size. Since one has a hypotenuse of 15 units and a right triangle it can be surmised that both triangles (ABD and BDE) have a 3-4-5 format. Using proportion, one triangle and its twin also has the same dimensions: 9, 12 and 15. Using the triangle formula of bh/2 = 9x12/2 = 54 ---> this is the area of just one. Two of them will be 108 square units. If you want to use the area formula of the rectangle ABED, A=height x base) 9 x 12 = 108. This is still way greater than the 68 square units that was arrived at and concurred by some commenters even without including the area of the "isosceles" triangle at the right side of the trapezoid. Did I miss something? What happened?
Your conclusion that both triangles (ABD and BDE) have a 3-4-5 format is wrong. For example the right triangle with h=5, b=10*sqr(2) has also a hypotenuse of 15 units
@@thomasklinkhammer782 I do realize that I made a wrong conclusion when I posted my comment. The presented scaling of the trapezoidal figure somehow created a false illusion with me to a point of careless haste in thought. Thank you.
There's something strange here in the figure. BD is the hypotenuse of the BAD right-angled triangle. Its length is 15. The point is that 15 is a multiple of 5. It's 5x3. So the right angle sides must be 4x3=12 for the longest and 3x3=9 for the shortest. Because of the 3,4,5 rule. And the short sides of the isoceles right-angled triangle are 9 each one too, since DE=EC=AB, on the right of the figure (DEC). If ABC is a right-angled triangle too, with its 19 hypotenuse, it means that BC=12+9=21>19. So the AC hypotenuse can't be 19. And the surface becomes: (AD+BC).h/2=(12+21)x9/2=148.5 u² and not 68. Anyway, 9x12 already=108>68. Where am I wrong?
@@javier7831 If the hypotenuse length of a right triangle is a multiple of 5, the length of the 2 other sides are automatically the same multiple, but of 3 and 4. Here we have 5x3 for the hypotenuse. So the 2 sides of the right angle are automatically 3x3 and 4x3 too, and the angles are automatically 53° and 37°. It's called homothety. 3,4,5; 6,8,10; 9,12,15; 12,16,20, etc... are all similar triangles and the angles remain identical regardless the lengthes. Only the lengthes ratio changes. Else, you just can't get any right triangle anymore. It's impossible. So AC is not 19 but sqrt(21²+9²)=22.85, or 3 x sqrt(58). Cheers and happy new year! :)
@@herves7727 With hypotenuse at length 15 the triangle does not necessarily mean ratio 3, 4, 5. Imagine moving a 90 degree angle around in an arc. It eventually becomes an isosceles triangle with angles 45, 90, 45 and hypotenuse = 15. That is not a 3, 4, 5 ratio. There are many solutions in between to a triangle and it is easy to fall into the 3, 4, 5 trap by assuming it. I do it sometimes with thinking the teacher is being clever. The 15 and 19 are real lengths with h = 4.125 and 45 degree angle.
@@chrisrowland2255 Yes Chris, indeed. I realize you're right. So I fell in this trap ! If I have an angle of 10° in a right triangle, then I still can have a 15 hypotenuse, and a 15*cos 10 long side and a 15*sin 10 short side. 15cos10=14.77 15sin10=0.966 So it's not necessarily a 3,4,5 situation indeed because all 15 hypotenuse right triangles don't necessarily have 3,4,5 ratios. It's just a bad habit I got! Mea culpa! and thx you and Javier too:)
i dont understand what i did wrong triangle BAD is a 3-4-5 triangle so side AD = 12 and AB = 9 then you take triangle ABC 9^2 + BC^2 = 19^2 BC^2 = 19^2 - 9^2 BC^2 = 280 BC = sqrt(280) BC = 16,7 and then the formula 1/2 * (12 + 16,7) * 9 = 129,15 can someone help me?
If ABD is a 345 right angled triangle, AB = 9, AD = 12. So area of rectangle = 108. Triangle on right is 45, 45, 90 degree triangle with height and base = AB = 9, so its area = (9*9)/2 = 40.5. So total area = 108 + 40.5 = 148.5. One of us is wrong!
Hello, while the hypotenuse of ABD of 15 is consistent with a 9 12 15 right triangle, it is consistent with other shapes of right triangles as well. In this case I think it is a 4.13 14.42 15 triangle.
@@scorpio7524 Hello, A right triangle with hypotenuse 15 of course can be a 345 triangle with sides 9 and 12 but can also have infinitely many other shapes. But there is only one triangle ABD that fits the other given conditions given by PreMath; namely that the other triangle ABC has hypotenuse 19, and that angle BCD is 45 degrees. And so there is only one area for this trapezoid as described. The triangle ABC as described by Mike would have sides 9 and 21 and so its hypotenuse is greater than 19. And if triangle ABC had side 9 and hypotenuse 19, then angle BCD would not be 45 degrees. You are correct that the dimensions that I gave are approximate. I used Desmos to get them. I would be interested if someone has the exact algebraic solution to that. I tried.
@@waheisel 4.125 and 14.422. I zoomed in very close to get the circle and ellipse intersects. Then used AutoCad to draw the figure. It is a very interesting question as calculations were needed first to draw the trapezoid.
Super ... like usual.
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Cool Problem. I attacked it too complicatedly. While I figured out the Pythagoras properties of the right triangles and the fact that b-a=h, I had chosen the variables in an unfortunate way and ended up with some sqrt-expressions. Knowing the formula for the area of a trapezoid would have helped! I was very close to the solution, but „couldn‘t see it“. But I again learned something!
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Great example, I learned much from this, i need practice at these types that have multi stage solutions..soo very interesting to see and follow the logical release of the solution, I got as far as seeing the two right triangles then missed the difference of two squares, more like this I love these complicated ones, where one needs a strategy, thanks again 👍🏻
Simply draw a perpendicular line from D, then a = b + h. Area = [(b + b + h) * h]/2 = (2bh + h) / 2
From the two diagonals,
Substract the above:
So area = (2bh + h) / 2 = 136 /2 = 68
Cool
Keep it up😀
Did it in slightly different way but very similar. I started by drawing the line DE, dividing the figure in a rectangle (base = x) and a right triangle (base = y) so the large base of the trapezoid equals to x+y and the small base is equal to x. Same approach on the right triangle which appears to be an isosceles triangle : DE equals to y. So the area of the trapezoid ((small base+large base)*height/2) becomes : (x+y+x)*y/2=(2x+y)*y/2=(y²+2xy)/2=A.
Using the Pythagorean theorem, I had two equations : (x+y)²+y²=19² and x²+y²=15² => x²+2xy+y²+y²=19². x²+y²=15² so 2xy+y²=2*A=19²-15²=(19-15)*(19+15)=4*(34) => A=2*34=68 square units.
Good one
by drawing a symmetric trapezoid along the (BC) line, we can find the area of the trapezoid is also equal to (a²-b²)/2.
by soustracting the 2 first pythagorean equations from this video, we find (a²-b²)/2 = (19²-15²)/2 = (361-225)/2=68... And that's the answer ! !! ^^
(sorry for my english)
Sir please continue forwarding such question this channel is some thing very productive for me as this gains my knowledge
Sure Anita dear
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This is a very interesting problem. Professor often has problems where you can't solve for the individual dimensions because there is a range of solutions. Typically something like you can solve for a+b but can't solve for a and b individually because there are a range of solutions. And yet, there is just one solution for a+b and the parameter the problem is asking for is a function strictly of a+b, hence there is a unique solution. But this problem is different. Based on the geometry it appears that there is definitely a single solution for b and h, and if you got those you then are able to calculate the area of the trapezoid. However, it seems to be a major pain in the butt to actually find b and h. That's the way I tried to solve it. By using pythagorean on the two known diagonals to get b^2+h^2=15^2 and (b+h)^2+h^2=19^2. Since you have two equations and two unknowns it is reasonable to believe there is a single solution to the problem. But then if you try to solve these two equations and two unknowns it is a mess because of the 2*b*h term in the second equation. So Professor's solution is a clever way where you don't need to deal with that. So this is an excellent problem in that it looks pretty easy but it is not quite so easy. I couldn't solve it. I should have looked at the trapezoid formula, then I would have realized that I had the solution without knowing it.
I understand what you mean on the individual dimensions but, to me, it feels incomplete and a little "cheaty" to not have them.
Your videos are always quite enjoyable. I love the moment when I suddenly see where you are going. Great fun. What input devices do you use, and what software? TIA -- Dan
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Please send me an email at premathchannel@gmail.com and I'll give you info on that.
From the figure: a=b+h, and so area = hb+h^2/2.
Pythagoras: h^2+b^2=225 and h^2+(h+b)^2=361.
Epanding the last and subtracting the first yields h^+2bh=361-225.
Left side is twice the area formular, so the area is (361-225)/2 = 68
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lol my name is also Peter and that's exactly what I did c;
Very helpful channel. Interesting mathematics questions 📚📚📚📚.Good luck!!!!!!
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Today I started from the line DE..and then (h=BC-AD)...calculations..Thaks a lot again
construct a perpendicular line from D to BC, call the point E. let’s lable some lenghts: AB=DE=h (the h stands for height), AD=BE=b (b stands for base).
the area of the trapezoid will be bh+((h^2)/2)
look at triangle CDE. we have one 90* angle (E) and one 45* angle (C), so the other angle (D) is also 45*. this means that it is a 1:1:sqrt(2) special triangle. lets fill in; 1:1:sqrt(2)=CE:DE:CD, we know that DE=h, CE:DE:CD=1•h:h:sqrt(2)•h, so CE is h aswell. note that triangle ABC and BDE are triangles where pythagorean works. in ABC:AB^2+BC^2=h^2+(b+h)^2=2h^2+b^2+2bh=19^2=361. and in BDE: b^2+h^2=15^2=225. now we have a system of equations: {2h^2+b^2+2bh =361
h^2+b^2=225
if we substract the equations, we get h^2+2bh=136
recall that the area of the trapezoid is bh+((h^2)/2).
if we divide h^2+2bh=136 by 2 we get exactly bh+((h^2)/2) and what will be equal to 136/2=68
kind of proud of being able to solve it like this
Very well done
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Hey premath ... Please tell me how did you shot this video. Softwares, Hardwares, and anything else if needed.
Your videos are very simple yet effective.
I subscribed your channel because of your simple and effective way of presentation.
Sir you are the treasure of math knowledge. We are proud of you sir 😊👍
So nice of you Sameer dear
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Awesome problem sir.And a very innovative solution too.Thank you.
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This is a good problem with a very good slution👍, thank you teacher 🙏.
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Its actually pretty easy to solve with trigometr,
It can be done (though involved in radicals) using coordinate axes and analytic geometry. I did it.
Brute force approach: calculate a, b and h and solve using equation for area of a trapezoid.
Equation 1: a² + h² = 361
Equation 2: b² + h² = 225
Equation 4: h = a - b so a = b + h
Substituting in equation 1:
(b + h)² + h² = 361
which expands to:
b² + 2hb + 2h² = 361
From equation 2: b² = 225 - h² and b = √(225- h²) so:
225 - h² + 2h√(225- h²) + 2h² = 361
Combining terms:
h² + 2h√(225- h²) - 136 = 0
and
2h√(225- h²) = 136 - h²
Squaring both sides:
4 h²(225 - h²) = 18496 - 272h² + h⁴
Let x = h² so we can reduce this to a quadratic:
4x(225 - x) = 18496 - 272x + x² which simplifies to:
5x² - 1172x + 18496 = 0
Solving:
x = 117.2 +/- 0.4√(62729)
Squaring both sides has introduced an invalid solution, only x = 117.2 - 0.4√(62729) is valid. So:
h² = 117.2 - 0.4√(62729)
b² = 107.8 + 0.4√(62729)
a² = 243.8 + 0.4√(62729)
From which:
h ~= 4.1252
b ~= 14.422
a ~= 18.547
Area = ((a + b)h)/2 ~= 68.0018594
This method provides an approximate value which is within 0.003% of PreMath's exact value 68. Carrying the calculation to more places yields a value closer to 68. Leaving the radicals and simplifying, there must be a way to get an exact value of 68.
great job, thanks for sharing, fully understand your step-by-step, happy holidays
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note that if the diagonals were b and c, then the area is (c^2-b^2)/2 , note also in this case and ingeneral, the linear dimensions are complicated involving nested square roots.
Amazing. Math Guru Ji Made it look so simple !!
Thanks
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Perp. DE is drawn upon BC.
DE=EC
Sides AB=a, AD=b, BC=a+b,
a²+(a+b)²=361
a²+b²=225
Subtracting, a²+2ab=136
Trapezium ABCD=ab+½a²=68
Super
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Yup. I did it that way too.
@@davidbrisbane7206 - Same.
Wonderful!. I love watching you're videos.
1.draw perpendicular DO on BC
2.you will get a rectangle with AD=Bo=x , AB=DO=y and OC=OD/tan45=y (tan45=1)
On applying Pythagoras
1. X²+y²=225
2.y²+x²+y²+2xy=361
3.y²+2xy=361-225=136
4.(y²+2xy) /2 = 68
On looking carefully we find that area of trapezoid ABCD=(y²+2xy) /2
My aproach is a little bit different.
In the picture we have a rectangle ABED and a right isosceles triangle DEC. So AD=BE= b (small base of the trapezoid) and AB=DE=CE=h ( the height of the trapezoid).
Consider the triangle BED we have sq DE+sqBE=sq 15=225 ;
and in the right triangle ABC we have:
sq(BE+CE) + sqCE = sq19=361-----> sq BE + sq CE + 2xBExCE + sq CE = sq BE+ sq DE + 2 BExCE + sq DE.
because sq DE+sqBE = 225 -----> 225 + 2xBExh + sq h = 361------> 2 BExh + sq h = 136------> BE xh + sq h/2 = 68.
Because BExh is the value of the area of the rectangle ABED and sq h/2 the area of the triangle DEC so the area of the trapezoid is 68 units
Without DCE = 45° this problem would have been much more challenging.
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Yes, you can see instantly that the triangle on the right-hand side of the trapezoid is a square cut in half. No need for SOHCAHTOA here!
You have explained vey well
Variation on PreMath's solution: We observe that the trapezoid can be divided into a rectangle with height h and base b and an isosceles right triangle with sides of length h ( 6:30 ). The rectangle has area bh and the triangle 0.5h², for total area of bh + 0.5h².
The two equations for the right triangles are:
1) (b + h)² + h² = 361
2) b² + h² = 225
Expanding 1): b² + 2bh + 2h² = 361
Subtracting 2): 2bh + h² = 136
Dividing both left and right sides by 2: bh + 0.5h² = 68
However, the left side is the formula for the trapezoid's area as derived above, so the area of the trapezoid is 68.
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I labelled AD = BE = x. EC = y. and AB = h. Hence, Area of Trapezium = (h/2)(2x+y).
By Pythagoras Theorem, I obtained
BC^2 - AD^2 = 136.
(BC + AD)(BC - AD) = 136
(2x + y)(y) = 136. ---> (2x+y) = (136/y) ---> Area of Trapezium = (h/2)(136/y) = (68z/y).
tan (z/y) = 45. --> z/y = 1. --> length of z = length of y. Therefore, Area of Trapezium = 68.
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@@PreMath thank you. your solution is simpler!
Sir if we take tan45=1
So................(..h/a-b)=1
..............................h=a-b
So..A=(a+b)h/2=136/2=68
.
That was my post-Christmas win lol. Solved it and feeling good that I could have been something else other than a healthcare worker lol!
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With preMath mathematics is always easy 👍👍👍👍🙋 keep it up
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Sir you help a lot in developing me
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Very helpful. Enjoyed this
i got the answer of 68 while trying (unsuccessfully) to find side lengths.
2(h^2)+(h×b)=68
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I solved it in 1 minute
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Amazing video
Let AB = x and AD = y. Then, the Perpendicularity Lemma gives 136 = (x+y)^2 - y^2 = x(x + 2y). Since the sum of the bases of the given trapezoid is precisely x + 2y, its area is just 136/2 = 68.
teor. Pitagora for ABC e ABD …
ABsquared=ACsquared-BCsquared=BDsquared-ADsquared … or ACsquared-BDsquared=BCsquared-ADsquared=(19+15)*(19-15)=
(BC+AD)*(BC-AD) … (1)
da BCD=45°, e DH⊥BC , DH=CH
CH=BC-AD , therefore (1) becomes
34*4=(somma basi trapezio)*(altezza trapezio)= 2*area trapezio where (area trapezio)=68
Обозначим АВ как а, АD как b, тогда ВС будет равно a+b. Площадь трапеции равна произведению полусуммы оснований на высоту P=a(a+2b)/2. a и b находим из прямоугольных треугольников ВАС и ВАD. a^2+b^2=225, a^2+(a+b)^2=361. a^2+a^2+2ab+b^2=361. b^2=225-a^2. a^2+a^2+2ab+225-a^2=361. a^2+2ab=136. a^2+2ab/2=68, а это и есть площадь трапеции Р=68!
Let AD = a
AB = b
Since angle C= 45
BC = a+b
In triangle ABD
a^2+b^2 = 225..eq1
In triangle BCD
(a+b^2+b^2 = 361..eq2
eq2-eq1
b^2 + 2ab = 136...eq3
Dividing both sides by 2 we get area
Because b^2 area of two triangles(square part)
And 2ab is area of two rectangle part
Hence
Area = 68
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Excellent again
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Although the process of arriving at the solution appeared to be sound, there is a huge flaw with the whole exercise if analyzed in detail. If all the given data have to be maintained the trapezoid is a nonexistent trapezoid
If AC is really 19 units long and the angle at point C is 45 degrees then it is a false trapezoid since you cannot form a true one without changing the 45 degree dimension.
Drawing the trapezoid to scale using the lines dimensions given it would turn out that the angle at point C is NOT 45 degrees - it is actually greater by a huge amount.
That is significant and important because that given degree was used to calculate the relationship between ( a-b ) and h to claim an isosceles triangle.
Also the arrived-at final answer is way smaller when you consider that the rectangle ABED contains two triangles that are similar and equal in size.
Since one has a hypotenuse of 15 units and a right triangle it can be surmised that both triangles (ABD and BDE) have a 3-4-5 format.
Using proportion, one triangle and its twin also has the same dimensions: 9, 12 and 15.
Using the triangle formula of bh/2 = 9x12/2 = 54 ---> this is the area of just one.
Two of them will be 108 square units.
If you want to use the area formula of the rectangle ABED, A=height x base) 9 x 12 = 108.
This is still way greater than the 68 square units that was arrived at and concurred by some commenters even without including the area of the "isosceles" triangle at the right side of the trapezoid.
Did I miss something? What happened?
Your conclusion that both triangles (ABD and BDE) have a 3-4-5 format is wrong. For example the right triangle with h=5, b=10*sqr(2) has also a hypotenuse of 15 units
@@thomasklinkhammer782 I do realize that I made a wrong conclusion when I posted my comment. The presented scaling of the trapezoidal figure somehow created a false illusion with me to a point of careless haste in thought. Thank you.
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What a good question thnx a lot
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Superb
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Can we solve for the dimensions??
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How will find it the perimeter?! 😀😉
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Great!
That was fun. What a good problem.
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There's something strange here in the figure. BD is the hypotenuse of the BAD right-angled triangle. Its length is 15. The point is that 15 is a multiple of 5. It's 5x3. So the right angle sides must be 4x3=12 for the longest and 3x3=9 for the shortest. Because of the 3,4,5 rule. And the short sides of the isoceles right-angled triangle are 9 each one too, since DE=EC=AB, on the right of the figure (DEC). If ABC is a right-angled triangle too, with its 19 hypotenuse, it means that BC=12+9=21>19. So the AC hypotenuse can't be 19.
And the surface becomes:
(AD+BC).h/2=(12+21)x9/2=148.5 u² and not 68.
Anyway, 9x12 already=108>68.
Where am I wrong?
Your reasoning is correct. But it can only apply in a 37, 53 degree right triangle. ABD triangle needs to be demonstrated as the mentioned triangle.
@@javier7831
If the hypotenuse length of a right triangle is a multiple of 5, the length of the 2 other sides are automatically the same multiple, but of 3 and 4. Here we have 5x3 for the hypotenuse. So the 2 sides of the right angle are automatically 3x3 and 4x3 too, and the angles are automatically 53° and 37°. It's called homothety. 3,4,5; 6,8,10; 9,12,15; 12,16,20, etc... are all similar triangles and the angles remain identical regardless the lengthes. Only the lengthes ratio changes. Else, you just can't get any right triangle anymore. It's impossible. So AC is not 19 but sqrt(21²+9²)=22.85, or 3 x sqrt(58).
Cheers and happy new year! :)
Or perhaps BD is not 15
@@herves7727 With hypotenuse at length 15 the triangle does not necessarily mean ratio 3, 4, 5. Imagine moving a 90 degree angle around in an arc. It eventually becomes an isosceles triangle with angles 45, 90, 45 and hypotenuse = 15. That is not a 3, 4, 5 ratio. There are many solutions in between to a triangle and it is easy to fall into the 3, 4, 5 trap by assuming it. I do it sometimes with thinking the teacher is being clever. The 15 and 19 are real lengths with h = 4.125 and 45 degree angle.
@@chrisrowland2255
Yes Chris, indeed. I realize you're right. So I fell in this trap ! If I have an angle of 10° in a right triangle, then I still can have a 15 hypotenuse, and a 15*cos 10 long side and a 15*sin 10 short side.
15cos10=14.77
15sin10=0.966
So it's not necessarily a 3,4,5 situation indeed because all 15 hypotenuse right triangles don't necessarily have 3,4,5 ratios. It's just a bad habit I got! Mea culpa! and thx you and Javier too:)
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Surprising!
My favorite refrain from these videos is, " , , and here's our much nicer looking diagram."
Is this Japanese Olympiad puestion?
Different sum
Logical solution
Super
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Easy one
I don’t understand why we subtract equation 2 from equation 1.
Good tecther،,,,,
i dont understand what i did wrong
triangle BAD is a 3-4-5 triangle so side AD = 12 and AB = 9
then you take triangle ABC
9^2 + BC^2 = 19^2
BC^2 = 19^2 - 9^2
BC^2 = 280
BC = sqrt(280)
BC = 16,7
and then the formula 1/2 * (12 + 16,7) * 9 = 129,15
can someone help me?
Sure.
Triangle BAD is not 3-4-5.
2 side lengths are needed to prove 3-4-5.
@@montynorth3009 its a right triangle so what would other possible values be?
@@jiglo2385 4.125 and 14.422. It is not a 3-4-5 triangle this time.
@@chrisrowland2255 ohh OK thx :)
Very nice, thank you!
The title needs to be edited to something like: "Find the Area of Trapezoid ABCD when only diagonals ***and angles*** are known"
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Area=136/2=68 sq unit
What's the reason deducting equation 2 out of equation 1? Im Being dumb
The answer is 68 square units.
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Interesante.
Area trapezoid 82,5
If ABD is a 345 right angled triangle, AB = 9, AD = 12. So area of rectangle = 108. Triangle on right is 45, 45, 90 degree triangle with height and base = AB = 9, so its area = (9*9)/2 = 40.5. So total area = 108 + 40.5 = 148.5. One of us is wrong!
Hello, while the hypotenuse of ABD of 15 is consistent with a 9 12 15 right triangle, it is consistent with other shapes of right triangles as well. In this case I think it is a 4.13 14.42 15 triangle.
@@scorpio7524 Hello, A right triangle with hypotenuse 15 of course can be a 345 triangle with sides 9 and 12 but can also have infinitely many other shapes. But there is only one triangle ABD that fits the other given conditions given by PreMath; namely that the other triangle ABC has hypotenuse 19, and that angle BCD is 45 degrees. And so there is only one area for this trapezoid as described. The triangle ABC as described by Mike would have sides 9 and 21 and so its hypotenuse is greater than 19. And if triangle ABC had side 9 and hypotenuse 19, then angle BCD would not be 45 degrees. You are correct that the dimensions that I gave are approximate. I used Desmos to get them. I would be interested if someone has the exact algebraic solution to that. I tried.
@@waheisel 4.125 and 14.422. I zoomed in very close to get the circle and ellipse intersects. Then used AutoCad to draw the figure. It is a very interesting question as calculations were needed first to draw the trapezoid.
Super duper bumper easy question indeed!!
@@northeastchronicles9311 Sorry ???
993//3.01.22.
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68
🙂
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You lost me when you subtracted equation #2 from equation #1 ........ without any explanation why.
Ans : 68 square unit. (*Interesting question)
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wrong! a=b!
I prefer maths over English.
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