once you have your triangle a,b,c 56,42,70 there is no need to go through Pythagorus or complimemtary angles. Simply use Herons law to find the area of the triangle a,b,c which is 1176. Then using Area = 1/2 Base * Height 1176= 35 *h therefore h= 1176/35 = 33.6
If you drop the perpendiculars DE and CF to AB, EF = 21. This creates Right Triangle AED with height h and base x, Right Triangle BFC with height h and base 70 - x, and rectangle EFCD with length 21 and width h. By combining the two triangles into one along the heights connecting D to C and E to F, the right angles mean we created a new Triangle ABC with base 70 and height h. We then use Heron's Formula to find the Area of the Triangle ABC, then plug that in to the classic base-height formula to get h. Then we calculate the area of Rectangle DEFC and add this area to that of Triangle ABC. That gives us the Area of Trapezoid ABCD.
I calculated "h" another way. I dropped perpendiculars from points D & C on to base AB. Then base length of left triangle formed, I called X. The base length of right triangle formed is ( 91 - X - 21) equal to (70 - X). Then in left triangle by Pythagoras, h^2 = 56^2 - X^2. Also, in right triangle, h^2 = 42^2 - (70 - X)^2. Equating the h^2 values:- 56^2 -X^2 = 42^2 - 70^2 + 140X - X^2. 56^2 = 42^2 - 70^2 + 140X. X = 44.8. Then from left triangle, h^2 = 56^2 - 44.8^2. h = 33.6.
To find area of triangle ADE we can also use the Heron theoreme A=sqr root of p(p-a)(p-b)(p-c). p is the semiperimetre of ADE. Thanks a lot for your lesson, dear Mr PreMath. God bless you and your family
Another way to find h from 4:10 to 7 :20 In Triangle ADE Area = ½ *AD * DE = 56 * 42 ÷ 2. = 1176 In another way. ½ * AE * h = 1176 h = 1176 * 2 ÷ 70 = 33.6 h = 33.6
I used Heron's formula to find the area of triangle ADE, and then with b = 70 worked backward to find h, which I then used to find the area of the trapezoid.
Draw a line from D parallel to CB to form triangle ADE with sides 56,42 & 70. Using heron's formula the area = 1176 = 0.5*base*height. 0.5*70*H = 1176 so height = 33.6. So area of trapezoid ABCD = 0.5*33.6*(21+91) = 1881.6
Neat! At first, I was wondering why you didn't take the fact that we were looking at a right triangle to simply use one of the known sides as the base and the other as the height, and add the area of that triangle to the area of the parallelogram, but then it occurred to me that there is not a simple way to find the area of the parallelogram without perhaps using something like the law of cosines, which I don't even know would work. I would love if you could do an alternate video that describes the solving of this problem along the lines of this suggested method! Additionally, it would be really interesting to see another trapezoidal problem with all four known sides but without an interior right triangle
Sir in-between you imagined aed as rt angled triangle, how come? and why we should we presume it to be right angled triangle?.. little confusing. By the way you are really good and I have made it aroutine to see all your clips. Thanks a lot and keep it up. God bless u! ....
Triangle ADE is right or not is immaterial. Knowing three sides of it, Heron's formula gives its area. Now area = 1/2. AD. h gives h. Trapezium area = simple to find now. Parallel side sum multiplied by h is the area.
Nice explanation. I have another approach. Draw a line from D parallel to CB to form triangle ADE. The triangle ADE sides are 42, 56 and 70 with ratios: 3:4:5 , which means that it's a Pythagorean triple right angle triangle .The sides AD and BC of the trapezoid ABCD intersect at point F. Therefore we have 3 similar Pythagorean triple right angle triangles: ADE , AFB and DFC . The areas of the right angle triangles AFB and DFC are: A(AFB) = A_0 (AB/5)^2 ; A(DFC) = A_0 (DC/5)^2 where, A_0 is the area of the basic Pythagorean triple right angle triangle (3,4,5) , which is A_0 = (3∙4)/2 = 6 The area of the trapezoid ABCD is therefore: A(ABCD) = A(AFB) - A(DFC) = 6 ((91/5)^2 - (21/5)^2 ) = 6 (14 ∙ 112/5) = 1881.6
Draw 2 perpendiculars (h) from D and C. Call the base of the left triangle y and the base of the right triangle z. Then 56^2 = h^2 + y^2 and 42^2 = h^2 + z^2. Subtract the two equations: 1372= y^2-z^2 or 1372= (y-z)(y+z). y+z= 90-21=70. So y-z= 1372/70= 19,6. If y-z= 19,6 and y+z=70, than y= 44,8. Using the pythagorian theorem for the left triangle gives: 56^2 - 44,8^2= h^2. h= 33,6.
Divide all the lengths by 7, to keep calculations simple. Here a > c are parallel sides, & area will be multiplied by 49. By the trapezoid formula, where s= ([a-c]+b+d)/2=12, Area = [(a+c)/(a-c)].√{s(s-[a-c])(s-b)(s-d)} = [(13+3)/(13-3)].√{12(12-10)(12-6)(12-8)} = (16/10)√{12(2)(6)(4)}= (8/5)(24)= 192/5. Final ans = 49(192/5) = 1,881.6 sq.units. The reader's job is to prove the Trapezoid Area formula is true using Heron's fromula. .
I took a different approach. Dropped verticles from D and C to points E and F. AE is A, FB is b. A^2 + h^2 = 56^2, B^2 + h^2 = 42^2 A^2 - B^2 = 56^2 - 42 ^2 = 1372. B = 70 - A, thus... A^2 - (70 - A)^2 = 1372 A^2 - A^2 + 140A - 4900 = 1372 140A = 6272 A = 44.8, B = 25.2 56^2 - 44.8^2 = 1128.96 = 42^2 - 25.2^2 = h^2 h = 33.6, etc.
I solved for height knowing the triangle x lengths add up to 70. The equation was cumbersome getting up to height to fourth power but it reduced to 56*42/70. Is that formula for all trapezoids? Or was this just this one?
Herein AD^2 + BC^2 = (AB - DC)^2 Hereby AD is perpendicular to BC. Height of trapezium = (AD x BC)/( AB - DC) Hereby area of Trapezium is (AD x BC)( AB + DC) /( 2(AB - DC))
Constructing DE parallel to CB, dividing the trapezoid into a triangle and a parallelogram, is an approach that many geometry students may take. Then, they may recognize that the triangle is the 3-4-5 special right triangle with sides scaled up but a factor of 14 and the problem is straightforward to solve using PreMath's method. However, let's say that the triangle is not a right triangle. Several viewers have pointed out that Heron's formula can be used to find the area and we then need to compute the area of the parallelogram. We construct DF, as PreMath did, and take the area of the triangle A = (1/2)bh, where length DF = h and length AE = b = 70. In PreMath's case, as an alternative to his similar triangles method, the area of the triangle could be computed by using DE and AD as base and height, and A = (1/2)(56)(42) = 1176 and, solving 1176 = (1/2)(70)(h), h = 33.6, so the parallelogram has area (21)(33.6) and total area is 1176 + 705.6 = 1881.6. An alternative to using Heron's formula is to consider right triangles ΔADF and ΔEDF. Let AF have length x, FE have length 70 - x and DF have length h. The Pythagorean theorem produces an equation for each triangle: x² + h² = (56)² and (70-x)² + h² = (42)². Expanding (70 - x)² and consolidating terms, we get -(140)(x) +x² + h² = -3136 or (140)(x) - x² - h² = 3136. Adding the 2 equations, the x² and h² cancel out, leaving (140)(x) = 6272 or x = 44.8. Solving x² + h² = (56)² for the positive value of h, we get h = 33.6. We now have the height of the original trapezoid and can use area of trapezoid formula A = (1/2)(b₁ + b₂)(h) = (1/2)(21 + 91)(33.6) = 1881.6
Area of ADE = 1/2 AD × DE ( trivially since angle ADE is a right angle) = 1/2 × 42 × 56 = 1/2 AE × h = 1/2 70×h Therefore h = 42×56/70 = 4.8 From this Area can be found
Since you found the right triangle measure you could find its area = 42x56/2 = 1176 sq. un. Then 70xh/2 = 1176 ; h = 33.6 un. Then the parallelogram A=21h = 705.6 sq. un. Then the total area is 1881.6 sq. un.
Draw DE parallel to CB with E on AB, draw CF parallel to DA with F on AB and name G the intersecting point of DE and CF. Observe that △ADE ≅ △FCB ∼ △FGE ∼ △DGC. Also, about areas, the trapezoid ABCD is △ADE + △FCB − △FGE + △DGC. So we have AE = FB = AB - CD = 70 ⇒ FE = 49. Then, using Heron's formula, we calculate area △ADE = area △FCB = 1176. Using areas proportions: area △FGE = 1176 (49²/70²) = 576.24 and area △DGC = 1174 (21²/70²) = 105.84. So area ABCD = 1881.6.
Let height be h. Draw two perpendiculars from upper two vertices. That forms a rectangle with dimensions(21, h). Let two adjacent sides on base of trapezium be x and (70-x). x²+h²=(42) ² __(i) (70-x)²+h²=(56) ²__(ii) (ii) -(i), (70-x) ²-x²=3136-1764 (70-x+x) (70-2x) =1372 70-2x=19.6 2x=50.2 x=25.2 Substituting this value of x in (i), (25.2) ²+h²=1764 h²=1764-635.04 h²=1138.96 h=33.6 Area of trapezium=1/2×h×(21+91) = 1/2×33.6×112 =112×16.8 =1881.6 unit²(ans)
Alternate Solution: (Note this solution is identical to that of georgebliss964 but done independently and in a slightly more concise form) draw perpendicular line to AB from D to AB at F , h=DF draw perpendicular line to AB from C to AB at G , also h=CG let AF = a then GB = 91 - 21 - a = 70 - a (56 - a) ^ 2 = 42 ^ 2 - (70 - a)^2 expanding and consolidating terms gives: 140 * a = 6272 a = 44.8 h= [56 ^ 2 - a ^ 2]^.5 or h = 33.6 Atrap = 1/2 * (91+21) * h =1881.6 sq units
what i did is i took the trapezoid 4 times to make a big square with a small square then i did the area of the big square - area of small square divided by 4 cause i took the trapezoid 4 times.But my answer is different.
Strange solution. It seems lighter to draw perpendicular downside from D and C and use the good old Pythagoras ... But why easy when a difficult way is also possible?
entweder in zwei dreiecke und ein rechteck zerlegen oder den strahlensatz anwenden: ll/21=(ll+56)/91 und lr/21=(ll+42)/91 und dann gilt sqr(ll^2-lh^2)+sqr(lr^2-lh^2)=21 bzw.sqr(56^2-lh^2)=sqr(42^2-(21-lh)^2) dann hat man ein grosses und ein kleines dreieck und zieht von der fläche des grossen die des kleinen ab. unfortunately i donot get money by doing this
Our goal is to find the height h of the trapezoid as we have all other information to find area. By observation, DA and BC have a 4:3 ratio, and AB-CD has a 5:4 ratio with DA. Therefore if we draw DE parallel to BC, AE = 70, EB = 21, and ∆EDA is a 14:1 ratio 3-4-5 Pythagorean Triple triangle and ∠EDA is 90°. If we draw DF such that F is a point on AE and DF is perpendicular to AB, as the two resultant triangles ∆AFD and ∆DFE each share an angle with ∆EDA, have a 90° angle at F, and divide the right angle ∠EDA, these triangles are similar to ∆EDA. Triangle ∆AFD: FD/DA = ED/AE h/56 = 42/70 = 3/5 h = 56(3)/5 = 168/5 Trapezoid ABCD: A = h(a+b)/2 = (168/5)(21+91)/2 A = 84(112)/5 = 9408/5 = 1881.6
Then you use a method drawing two perpendicular lines from D to side AB call that intersection point F,, and from C to side AB call that intersecion E. Now the Trapezoid is dicvided to one rectangle in middle and right angle triangle at each side of the rectangle. The H is common to all DF = CE = H
ভাই, আপনি এই ভিডিওতে মারাত্মক ভুল করেছেন। বিষয়টি হয়তো কারও চোখে পড়েনি। আপনি দুইটি ত্রিভুজের অনুপাত দেখাতে গিয়ে ভুলটি করেছেন। সেটি হল h/b = a/c কখনও হতে পারে না। কারণ h এবং b এর অন্তর্ভুক্ত কোণ α হওয়ায় h ভূমি এবং b অতিভুজ। ফলে আপনার উপস্থাপিত বিষয়টি দাড়াল ভূমি / অতিভুজ। পক্ষান্তরে a এবং c এর অন্তর্ভূক্ত কোণ α হওয়ায় a অতিভুজ এবং c ভূমি। ফলে আপনার উপস্থাপিত বিষয়টি দাড়াল অতিভুজ / ভূমি। সম্পূর্ণ তালগোল পাকিয়ে ফেললেন। যার ফলে আপনার নির্ণয়কৃত উচ্চতা ভুল হয়েছে। প্রকৃত উচ্চতা 39.191835 একক। আশা করি এই ভিডিওর সংশোধনী দিবেন যাতে শিক্ষার্থীরা ভুল শিক্ষা গ্রহণ করা থেকে বাঁচতে পারে। আর উচ্চতা নির্ণয়ের বিষয়টি এত জটিল করার দরকার ছিল না। খুব সহজ এবং সংক্ষিপ্ত।
In my opinion @nurulamin did not understand the complete logic of PreMath. The concept of similar triangles ( all 3 angles being equal ie AAA) does allow this comparison between both triangles.
You're awesome!!! I am able to understand quite well all what you explain, teacher! Thanks!
Happy to hear that!
@@PreMath Super clever.
@@PreMath Oh yeah; not UR fault , but the audio is over-modulated.
once you have your triangle a,b,c 56,42,70 there is no need to go through Pythagorus or complimemtary angles. Simply use Herons law to find the area of the triangle a,b,c which is 1176. Then using Area = 1/2 Base * Height 1176= 35 *h therefore h= 1176/35 = 33.6
If you drop the perpendiculars DE and CF to AB, EF = 21. This creates Right Triangle AED with height h and base x, Right Triangle BFC with height h and base 70 - x, and rectangle EFCD with length 21 and width h. By combining the two triangles into one along the heights connecting D to C and E to F, the right angles mean we created a new Triangle ABC with base 70 and height h. We then use Heron's Formula to find the Area of the Triangle ABC, then plug that in to the classic base-height formula to get h. Then we calculate the area of Rectangle DEFC and add this area to that of Triangle ABC. That gives us the Area of Trapezoid ABCD.
I calculated "h" another way.
I dropped perpendiculars from points D & C on to base AB.
Then base length of left triangle formed, I called X.
The base length of right triangle formed is ( 91 - X - 21) equal to (70 - X).
Then in left triangle by Pythagoras, h^2 = 56^2 - X^2.
Also, in right triangle, h^2 = 42^2 - (70 - X)^2.
Equating the h^2 values:-
56^2 -X^2 = 42^2 - 70^2 + 140X - X^2.
56^2 = 42^2 - 70^2 + 140X.
X = 44.8.
Then from left triangle, h^2 = 56^2 - 44.8^2.
h = 33.6.
That's what I would have done, too.
It is the same way I have done.
That is exactly how I did it.
And me!
Excellent!
Thank you! Cheers! 😀
To find area of triangle ADE we can also use the Heron theoreme A=sqr root of p(p-a)(p-b)(p-c). p is the semiperimetre of ADE. Thanks a lot for your lesson, dear Mr PreMath. God bless you and your family
Another way to find h
from 4:10 to 7 :20
In Triangle ADE
Area = ½ *AD * DE
= 56 * 42 ÷ 2. = 1176
In another way. ½ * AE * h = 1176
h = 1176 * 2 ÷ 70 = 33.6
h = 33.6
I used Heron's formula to find the area of triangle ADE, and then with b = 70 worked backward to find h, which I then used to find the area of the trapezoid.
Draw a line from D parallel to CB to form triangle ADE with sides 56,42 & 70. Using heron's formula the area = 1176 = 0.5*base*height.
0.5*70*H = 1176 so height = 33.6. So area of trapezoid ABCD = 0.5*33.6*(21+91) = 1881.6
Excellent!
Thanks for sharing! Cheers!
Took a similar approach to you at the start created the traingle then used law cosine to find an angle then law sine to get the height.
Super!
Thanks for sharing! Cheers!
Stay blessed, Rashid dear 😀
Neat! At first, I was wondering why you didn't take the fact that we were looking at a right triangle to simply use one of the known sides as the base and the other as the height, and add the area of that triangle to the area of the parallelogram, but then it occurred to me that there is not a simple way to find the area of the parallelogram without perhaps using something like the law of cosines, which I don't even know would work.
I would love if you could do an alternate video that describes the solving of this problem along the lines of this suggested method! Additionally, it would be really interesting to see another trapezoidal problem with all four known sides but without an interior right triangle
Very nice and useful method.
Thanks Sir to your simplify explain .
❤❤❤❤
Thanks for liking
Thank you very much, I love this question and your solution
Similar triangle formula you have applied is in correct format?
Very Nice solution. Thank you, sir.
Most welcome, dear
Sir in-between you imagined aed as rt angled triangle, how come? and why we should we presume it to be right angled triangle?.. little confusing. By the way you are really good and I have made it aroutine to see all your clips. Thanks a lot and keep it up. God bless u! ....
Triangle ADE is right or not is immaterial. Knowing three sides of it, Heron's formula gives its area. Now area = 1/2. AD. h gives h. Trapezium area = simple to find now. Parallel side sum multiplied by h is the area.
Thanks for video.Good luck sir!!!!!!!!!!!!
Nice explanation. I have another approach.
Draw a line from D parallel to CB to form triangle ADE. The triangle ADE sides are 42, 56 and 70 with ratios: 3:4:5 , which means that it's a Pythagorean triple right angle triangle .The sides AD and BC of the trapezoid ABCD intersect at point F. Therefore we have 3 similar Pythagorean triple right angle triangles: ADE , AFB and DFC .
The areas of the right angle triangles AFB and DFC are:
A(AFB) = A_0 (AB/5)^2 ; A(DFC) = A_0 (DC/5)^2
where, A_0 is the area of the basic Pythagorean triple right angle triangle (3,4,5) , which is
A_0 = (3∙4)/2 = 6
The area of the trapezoid ABCD is therefore:
A(ABCD) = A(AFB) - A(DFC) = 6 ((91/5)^2 - (21/5)^2 ) = 6 (14 ∙ 112/5) = 1881.6
Draw 2 perpendiculars (h) from D and C. Call the base of the left triangle y and the base of the right triangle z. Then 56^2 = h^2 + y^2 and 42^2 = h^2 + z^2. Subtract the two equations: 1372= y^2-z^2 or 1372= (y-z)(y+z). y+z= 90-21=70. So y-z= 1372/70= 19,6. If y-z= 19,6 and y+z=70, than y= 44,8. Using the pythagorian theorem for the left triangle gives: 56^2 - 44,8^2= h^2. h= 33,6.
I'll try two right triangles and a rectangle in the middle:
Triangle on the left: 56² = h² + x² => h² = 56² - x²
Triangle on the right: 42² = h² + (91 - 21 - x)² = h² + (70 - x)² => h² = 42² - (70 - x)² = 42² - 70² + 140x - x²
h² = h²
56² - x² = 42² - 70² + 140x - x²
140x = 56² + 70² - 42² = 6272
x = 6272 / 140 = 44.8
h² = 56² - 44.8² = (56 + 44.8)(56 - 44.8) = 100.8 * 11.2 = 1128.96
h = 33.6
A = (91 + 21)/2 * h = 56 * h = 56 * 33.6 = 1881.6
Yes, thats what i do in case of these trapezium problems
What pen-tablet are you using .....
Divide all the lengths by 7, to keep calculations simple. Here a > c are parallel sides,
& area will be multiplied by 49. By the trapezoid formula, where s= ([a-c]+b+d)/2=12,
Area = [(a+c)/(a-c)].√{s(s-[a-c])(s-b)(s-d)} = [(13+3)/(13-3)].√{12(12-10)(12-6)(12-8)}
= (16/10)√{12(2)(6)(4)}= (8/5)(24)= 192/5. Final ans = 49(192/5) = 1,881.6 sq.units.
The reader's job is to prove the Trapezoid Area formula is true using Heron's fromula.
.
I took a different approach. Dropped verticles from D and C to points E and F. AE is A, FB is b.
A^2 + h^2 = 56^2, B^2 + h^2 = 42^2
A^2 - B^2 = 56^2 - 42 ^2 = 1372.
B = 70 - A, thus...
A^2 - (70 - A)^2 = 1372
A^2 - A^2 + 140A - 4900 = 1372
140A = 6272
A = 44.8, B = 25.2
56^2 - 44.8^2 = 1128.96 = 42^2 - 25.2^2 = h^2
h = 33.6, etc.
I solved for height knowing the triangle x lengths add up to 70. The equation was cumbersome getting up to height to fourth power but it reduced to 56*42/70. Is that formula for all trapezoids? Or was this just this one?
Good information for me because I have matric level knowledge about Math .
All the best, dear
Thank you! Cheers! 😀
divide all sides by 7 to make the math clearer
Thank you! Cheers! 😀
Herein AD^2 + BC^2 = (AB - DC)^2
Hereby AD is perpendicular to BC.
Height of trapezium
= (AD x BC)/( AB - DC)
Hereby area of Trapezium is
(AD x BC)( AB + DC) /( 2(AB - DC))
Thank you! Cheers! 😀
Constructing DE parallel to CB, dividing the trapezoid into a triangle and a parallelogram, is an approach that many geometry students may take. Then, they may recognize that the triangle is the 3-4-5 special right triangle with sides scaled up but a factor of 14 and the problem is straightforward to solve using PreMath's method. However, let's say that the triangle is not a right triangle. Several viewers have pointed out that Heron's formula can be used to find the area and we then need to compute the area of the parallelogram. We construct DF, as PreMath did, and take the area of the triangle A = (1/2)bh, where length DF = h and length AE = b = 70. In PreMath's case, as an alternative to his similar triangles method, the area of the triangle could be computed by using DE and AD as base and height, and A = (1/2)(56)(42) = 1176 and, solving 1176 = (1/2)(70)(h), h = 33.6, so the parallelogram has area (21)(33.6) and total area is 1176 + 705.6 = 1881.6.
An alternative to using Heron's formula is to consider right triangles ΔADF and ΔEDF. Let AF have length x, FE have length 70 - x and DF have length h. The Pythagorean theorem produces an equation for each triangle: x² + h² = (56)² and (70-x)² + h² = (42)². Expanding (70 - x)² and consolidating terms, we get -(140)(x) +x² + h² = -3136 or (140)(x) - x² - h² = 3136. Adding the 2 equations, the x² and h² cancel out, leaving (140)(x) = 6272 or x = 44.8. Solving x² + h² = (56)² for the positive value of h, we get h = 33.6. We now have the height of the original trapezoid and can use area of trapezoid formula A = (1/2)(b₁ + b₂)(h) = (1/2)(21 + 91)(33.6) = 1881.6
Area of ADE = 1/2 AD × DE ( trivially since angle ADE is a right angle) = 1/2 × 42 × 56 = 1/2 AE × h = 1/2 70×h
Therefore h = 42×56/70 = 4.8
From this Area can be found
Nice sir
Thanks and welcome
Since you found the right triangle measure you could find its area = 42x56/2 = 1176 sq. un. Then 70xh/2 = 1176 ; h = 33.6 un. Then the parallelogram A=21h = 705.6 sq. un. Then the total area is 1881.6 sq. un.
Grea lesson!
Yay! I solved it.
Excellent!
Thank you! Cheers! 😀
Draw DE parallel to CB with E on AB, draw CF parallel to DA with F on AB and name G the intersecting point of DE and CF.
Observe that △ADE ≅ △FCB ∼ △FGE ∼ △DGC. Also, about areas, the trapezoid ABCD is △ADE + △FCB − △FGE + △DGC.
So we have AE = FB = AB - CD = 70 ⇒ FE = 49. Then, using Heron's formula, we calculate area △ADE = area △FCB = 1176.
Using areas proportions: area △FGE = 1176 (49²/70²) = 576.24 and area △DGC = 1174 (21²/70²) = 105.84. So area ABCD = 1881.6.
Thanks for sharing! Cheers!
Once you know ADE is a right triangle multiply 42 by 56 and divide by 70 giving h = 33.6
Then ((21 + 91)/2)*33.6=1881.6
Let height be h. Draw two perpendiculars from upper two vertices. That forms a rectangle with dimensions(21, h). Let two adjacent sides on base of trapezium be x and (70-x).
x²+h²=(42) ² __(i)
(70-x)²+h²=(56) ²__(ii)
(ii) -(i),
(70-x) ²-x²=3136-1764
(70-x+x) (70-2x) =1372
70-2x=19.6
2x=50.2
x=25.2
Substituting this value of x in (i),
(25.2) ²+h²=1764
h²=1764-635.04
h²=1138.96
h=33.6
Area of trapezium=1/2×h×(21+91) =
1/2×33.6×112
=112×16.8
=1881.6 unit²(ans)
merci sire , on peut utiliser le théorème d'Alkashi
Excellent!
Thank you! Cheers! 😀
That was amazing
Excellent!
Thank you! Cheers! 😀
What if ADE is not right angle triangle
Alternate Solution: (Note this solution is identical to that of georgebliss964 but done independently and in a slightly more concise form)
draw perpendicular line to AB from D to AB at F , h=DF
draw perpendicular line to AB from C to AB at G , also h=CG
let AF = a then GB = 91 - 21 - a = 70 - a
(56 - a) ^ 2 = 42 ^ 2 - (70 - a)^2
expanding and consolidating terms gives:
140 * a = 6272
a = 44.8
h= [56 ^ 2 - a ^ 2]^.5
or h = 33.6
Atrap = 1/2 * (91+21) * h
=1881.6 sq units
Solution:
h = height of the trapezoid.
I divide the lower side AB into:
(1) x+21+y = 91 ⟹ (1a) x = 91-21-y = 70-y
Then according to the Pythagorean theorem:
(2) x²+h² = 56² = 3136
(3) y²+h² = 42² = 1764
(2) - (3) = (4) x²-y² = 1372 |(1a) in (4) ⟹
(4a) (70-y)²-y² = 1372 ⟹ (4b) 4900-140y+y²-y² = 1372 |+140y-1372 ⟹
(4c) 3528 = 140y |/140 ⟹ (4d) y = 25.2 |in (3) ⟹
(3a) 25.2²+h² = 1764 |-25.2²⟹
(3b) h² = 1764 - 25.2² = 1128.96 | √( ) ⟹
(3c) h = 33.6
Area of the trapezoid = (91+21)/2*33.6 = 1881.6
what i did is i took the trapezoid 4 times to make a big square with a small square then i did the area of the big square - area of small square divided by 4 cause i took the trapezoid 4 times.But my answer is different.
Strange solution. It seems lighter to draw perpendicular downside from D and C and use the good old Pythagoras ... But why easy when a difficult way is also possible?
entweder in zwei dreiecke und ein rechteck zerlegen oder den strahlensatz anwenden:
ll/21=(ll+56)/91 und lr/21=(ll+42)/91
und dann gilt sqr(ll^2-lh^2)+sqr(lr^2-lh^2)=21
bzw.sqr(56^2-lh^2)=sqr(42^2-(21-lh)^2)
dann hat man ein grosses und ein kleines dreieck und zieht von der fläche des grossen die des kleinen ab. unfortunately i donot get money by doing this
Ah, the parallelogram , the magic ingredient.
🤓👍🏻
Thanks!
You are awesome. Keep it up 👍
Our goal is to find the height h of the trapezoid as we have all other information to find area.
By observation, DA and BC have a 4:3 ratio, and AB-CD has a 5:4 ratio with DA. Therefore if we draw DE parallel to BC, AE = 70, EB = 21, and ∆EDA is a 14:1 ratio 3-4-5 Pythagorean Triple triangle and ∠EDA is 90°.
If we draw DF such that F is a point on AE and DF is perpendicular to AB, as the two resultant triangles ∆AFD and ∆DFE each share an angle with ∆EDA, have a 90° angle at F, and divide the right angle ∠EDA, these triangles are similar to ∆EDA.
Triangle ∆AFD:
FD/DA = ED/AE
h/56 = 42/70 = 3/5
h = 56(3)/5 = 168/5
Trapezoid ABCD:
A = h(a+b)/2 = (168/5)(21+91)/2
A = 84(112)/5 = 9408/5 = 1881.6
ADE is also a triplet.
Use cosine rule
Thank you! Cheers! 😀
Assuming the top and bottom lines are parallel, I get A=1881.6
Thanks for sharing! Cheers!
Altitude =33.6 unit.Area of trapizoid=1881.6sq unit
Who used Heron's formula to calculate 'h' from △ADE 😄😄
You can use algebra
What if triangle ADE was not a right-angled triangle?
Yes, definitely you could use Heron's formula to calculate the area and the height of the triangle. I failed to notice that. Mistake on my part.
Need to look for another way.
Heron's formula for the area, then divide that area by 70.
Then you use a method drawing two perpendicular lines from D to side AB call that intersection point F,, and from C to side AB call that intersecion E. Now the Trapezoid is dicvided to one rectangle in middle and right angle triangle at each side of the rectangle. The H is common to all DF = CE = H
In this case the formula is
h² = a² - (( a² - b² + c²)/c/2)²
there are actually 5 ways to do it but 5 different answer why? this problem doesnt make sense.
In some rabbit holes there are Rabbits! 🙂
Thank you! Cheers! 😀
33.6×56×1881.6
Why not just create 2 triangles to begin with?
That would work too.
Here the given information hints at a more specific case, making a way to solve the problem.
❤🥂👍
Thank you! Cheers! 😀
Not right triangel db 56
33.6×56=1881.6
Formula di Erone , nota l'area ti calcoli altezza.
ভাই, আপনি এই ভিডিওতে মারাত্মক ভুল করেছেন। বিষয়টি হয়তো কারও চোখে পড়েনি। আপনি দুইটি ত্রিভুজের অনুপাত দেখাতে গিয়ে ভুলটি করেছেন। সেটি হল h/b = a/c কখনও হতে পারে না। কারণ h এবং b এর অন্তর্ভুক্ত কোণ α হওয়ায় h ভূমি এবং b অতিভুজ। ফলে আপনার উপস্থাপিত বিষয়টি দাড়াল ভূমি / অতিভুজ। পক্ষান্তরে a এবং c এর অন্তর্ভূক্ত কোণ α হওয়ায় a অতিভুজ এবং c ভূমি। ফলে আপনার উপস্থাপিত বিষয়টি দাড়াল অতিভুজ / ভূমি। সম্পূর্ণ তালগোল পাকিয়ে ফেললেন। যার ফলে আপনার নির্ণয়কৃত উচ্চতা ভুল হয়েছে। প্রকৃত উচ্চতা 39.191835 একক। আশা করি এই ভিডিওর সংশোধনী দিবেন যাতে শিক্ষার্থীরা ভুল শিক্ষা গ্রহণ করা থেকে বাঁচতে পারে। আর উচ্চতা নির্ণয়ের বিষয়টি এত জটিল করার দরকার ছিল না। খুব সহজ এবং সংক্ষিপ্ত।
He doesn't understood your bengali language.
@@landmmeasurement3905Say " He does not understand ... "
In my opinion @nurulamin did not understand the complete logic of PreMath. The concept of similar triangles ( all 3 angles being equal ie AAA) does allow this comparison between both triangles.