So nice of you Elisen! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Add the points G & H directly opposite E & F; & let J be the intersection of the lines EG & FH. This will give us a partition of ABCD into four rectangles. Let area(CGJH)= z. Then area(AEJF)=10, area (DFJG)=6-z, area(BEJH)=8-z, and z
I am from india and not all but some people thinks that we can't understand the english of the foreigners but we can understand and i like to watch your videos to learn new things
I enjoyed the video, but I think the reasoning for why one possible value of x^2 was rejected could have been clearer. At that point, there wasn't a clear criterion for why one value should be accepted over the other. Perhaps the video should have started with that the area of the square is x^2, and that the area of the triangle EFC would be the area of the square minus the sum of the areas of the other three triangles, and thus x^2-12. Then go into finding the sides of the triangles in terms of X. Then when you get to the value of x^2 you have a reason to reject the value that is less than 12 because that would make the area of triangle EFC negative, and you can't have a negative value for the area.
Dear Patrick, thanks for your nice feedback. Please keep giving your honest feedback. That would help us to server our audience much better. Actually, u=x^2, and X^2 is the area of the square. If I pick u=2.2 value => X^2 = 2.2 = the area of the square. And that's not possible because the square is way bigger than 2.2. You are awesome 👍 Take care dear and stay blessed😃
Patrick's comments are exactly right. I didn't follow in my first viewing why 2.2 should be rejected. But more important than that, as Patrick also noted, the solution should start not with "Step 1," but rather with a logical explanation of the strategy to be used to solve the problem. In this case, "we're going to find the area of the internal triangle by first finding the area of the square and then subtracting the total area of the three triangles for which the areas are given." "The area of the square, X>2, should be a hint that this is a quadratic equation problem, and one of the first steps will be to express each segment of the sides of the square in terms of a single variable, X." It is not always easy to communicate these strategies, but I think therein lies the challenge for the student/teacher (and thus the learning opportunity), more so than in the algebraic manipulations of the solution steps (though those are important too). You're doing a great job, Professor; keep up the good work in getting our youth excited about math. 👍
@@jamesb2675 Thanks my friend for your kind feedback. I really appreciate that. This feedback would help me shaping future videos. Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊
General case for all rectangles (not only squares)... Having: (Y)ellow; (B)lue;(G)reen;(W)hite. W = √( (Y+B+G)^2 - 4(B)(Y) ) Then W = √( (5+3+4)^2 - 4(3)(4) ) = √96 =4√6
At 0:37, Let AB=BC=CD=DA=t say From the orange triangle we worked out DF=a say From the blue triangle we worked out EB=4a/3 From the green triangle we worked out FA*AE/2=5 (t-a)(t-4a/3)=10 a=6/t and substitute into above equation, That ended up as a quadratic in t^2, (t^2)^2-24(t)^2-48=0 We worked out the area of ABCD =t^2 =12+4sqrt6 Then subtract the area of the three triangles of 3+4+5 from area ABCD =12+4sqrt6-12 =4sqrt6 square units. Same as last time with different symbols used.
Thank u for teaching or explaining in simple ways.. You take the time to break things down so that we understand the basic and the details of a certain problem... Thank u for being patient and clear..
Wow, a phenomenally phenomenal phenomena, a work of absolute algebraic artistry. And... no way did I do this one , but just so enlightening, particularly like the last bit using "u", that is new knowledge for me, thank you 🤓
It's always satisfying when you work through one of these elaborate problems, then watch the video and see the equations you just wrote down on paper start appearing on the screen. :) Thanks.
Answer 9.797 of 9.8 did a similar problem some time ago. let A = length of square B= base of the triangle with area 3, and C=length of the triangle with the area, 5. Ended up with the equation "A^4 - 24A^2 + 48 =0" let X = A^2 (A x A or area of square) therefore X^2-24X+48=0 Solving for X (using the quadratic formula or quadratic calculator) will yield 12 (+ -) 4 sqrt 6. Since theTOTAL area of the three triangles is 12 (3+4+5), then the total area of the triangle inside is 4 sqrt 6 or 9.7979 Answer
Great feedback my friend. I appreciate that. Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
Here is my answer to the same problem around a year ago. The notation on the figure was different . My square is described as BCEDFA in a clockwise direction from the top LHS of the square. Area of triangle EDF=5,Area of triangle FAB=4 and Area of triangle BCE=3. Let AF=d and CE= 3d/4 since the areas of triangle BAF=4 and triangle BCE =3 and both had the same height=x. Let side of square ABCD =x and area of square is therefore x^2 Consider triangle BAF, 4= xd/2 d=8/x...............(1) Consider triangle FDE, 5=(x-d)(x-3d/4)/2............(2) Substitute d=8/x into (2), We end up with 10=x^2-14+48/x^2 Let y=x^2, 0=y-24+48/y Multiply through by y, y^2-24y+48=0 y= 12+4sqrt(6) (-ve sqrt rejected in quadratic formula) Area of square ABCD =x^2 = y =12+4sqrt(6), Area of triangle BEF= area of ABCD-area of ABF-area of BCE-area of FED =(12+4sqrt(6)}4-3-5 =12+4sqrt(6)-12 = 4sqrt(6) square units as required. Note the -ve value of the sqrt in the quadratic formula was rejected because the area of BEF would have been -ve.
Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
Let's label the side of the square "a" and the bases of the orange and the blue triangels x and y respectively. Than x=6/a and y=8/a.The area of the green triangel =(a-x) (a-y) /2 Hence :(a-6/a)(a-8/a)=10 We receive an biquadratic equation :a^4-24a^2+48 =0.Puting in an auxiliar unknown t=a^2 we get a square equation t^2-24t+48=0. D=384 sqrtD=sqrt16*sqrt 4*sqrt6=4*2*sqrt 6=8 sqrt 6. Therefor we get two walues of t=a^2 :12-4sqrt 6 - to be rejected and 12 +4sqrt 6. Now we have to substract the sum of the given triangels from the calculated area of the whole square :(12 +4sqrt6)-(3+4+5)=12+4sqrt6-12=4sqrt6. Greetings.
So nice of you Beeru! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Please don't gloss over details which are absolutely important, like 10:27 to 10:33 (time) in your video... (i.e., without giving the reason why you rejected u = 2.2).
It's because u=x^2 represents the area of the square, it cannot be smaller than the area of any of the enclosed triangles, the smallest one of which is 3.
I mean I kinda solve it and I got the equation however I didn’t know how to solve quadratic equations other wise pretty good Btw keep up making these videos
I want to prove the Area of triangle FCE is 4sqrt6 square units on the basis that the area of the square is (12+4sqrt6) square units as proved by the lecturer.(x^2= (12+4sqrt6) Let FC^2=a^2=((x^4+36)/x^2)=23.44948974 approx =(21+sqrt6) exact value is the same as the decimal value on computer up to 25 places Let EC^2=b^2=( (x^4+64)/x^2=24.73401368 approx =(4/3(21-sqrt6)) exact value is the same as the decimal value on computer up to 25 places. Let FE^2=c^2=AF^2+AE^2= 21.1850341 approx and the ties up with FE^2= 1/3(63-sqrt6) to 25 decimal places for 25 decimal places. The area of triangle CFE =sqrt(4a^2*b^2-(a^2+b^2-c^2)^2)/4 we know x^2=(12+4sqrt6) Plug in values of a^2,b^2and c^2 into formula above; Area of triangle FCE. IFCEI=sqrt( 4*(21+sqrt6)(4/3)(21-sqrt6)-((21+sqrt6)+(4/3)(21-sqrt6)-(1/3)(63-sqrt6))^2/4 = sqrt(2320-(28)^2/4 = sqrt(2320-784)/4 = sqrt(1536)/4 = sqrt(256*6)/4 = sqrt(16^2*6)/4 = (16*sqrt6)/4 = 4*sqrt6 square units using surd values in the structure. Assisted by Wolfram Alpha and hand calculator to work out the exact values in the structure.
saudações meu grande professor respondo pelo nome Juma Cassimo , estudante de Engenharia Moçambique ( Africa) e sou muito fã dos seus trabalhos. Existem vários geômetras mas o Sr e o Melhor em todo universo Eu, gostaria de deixa uma proposta de resolução a cada exercícios ou seja, uma outra forma de resolução.
So nice of you Jagannath! Thanks for your feedback on audio quality. You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
I like the challenges on your channel, but I really don't like the misrepresented images you give most of the time. In this one, it's obvious the white triangle is not a 345 triangle. Just ask the question with an actual representation.
Very nice sharing, sending my full support
So nice of you Elisen! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Add the points G & H directly opposite E & F; & let J be the intersection of the lines EG & FH. This will give us a partition of ABCD into
four rectangles. Let area(CGJH)= z. Then area(AEJF)=10, area (DFJG)=6-z, area(BEJH)=8-z, and z
I am from india and not all but some people thinks that we can't understand the english of the foreigners but we can understand and i like to watch your videos to learn new things
Dear Mahendra, you guys are very smart and hard working people! You are the best 👍
Keep smiling😊
I enjoyed the video, but I think the reasoning for why one possible value of x^2 was rejected could have been clearer. At that point, there wasn't a clear criterion for why one value should be accepted over the other. Perhaps the video should have started with that the area of the square is x^2, and that the area of the triangle EFC would be the area of the square minus the sum of the areas of the other three triangles, and thus x^2-12. Then go into finding the sides of the triangles in terms of X. Then when you get to the value of x^2 you have a reason to reject the value that is less than 12 because that would make the area of triangle EFC negative, and you can't have a negative value for the area.
Dear Patrick, thanks for your nice feedback. Please keep giving your honest feedback. That would help us to server our audience much better.
Actually, u=x^2, and X^2 is the area of the square. If I pick u=2.2 value => X^2 = 2.2 = the area of the square. And that's not possible because the square is way bigger than 2.2.
You are awesome 👍 Take care dear and stay blessed😃
Patrick's comments are exactly right. I didn't follow in my first viewing why 2.2 should be rejected. But more important than that, as Patrick also noted, the solution should start not with "Step 1," but rather with a logical explanation of the strategy to be used to solve the problem. In this case, "we're going to find the area of the internal triangle by first finding the area of the square and then subtracting the total area of the three triangles for which the areas are given." "The area of the square, X>2, should be a hint that this is a quadratic equation problem, and one of the first steps will be to express each segment of the sides of the square in terms of a single variable, X." It is not always easy to communicate these strategies, but I think therein lies the challenge for the student/teacher (and thus the learning opportunity), more so than in the algebraic manipulations of the solution steps (though those are important too). You're doing a great job, Professor; keep up the good work in getting our youth excited about math. 👍
@@jamesb2675 Thanks my friend for your kind feedback. I really appreciate that. This feedback would help me shaping future videos. Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊
General case for all rectangles (not only squares)...
Having: (Y)ellow; (B)lue;(G)reen;(W)hite.
W = √( (Y+B+G)^2 - 4(B)(Y) )
Then
W = √( (5+3+4)^2 - 4(3)(4) ) = √96 =4√6
Where did you find this formula?
@@davidbrisbane7206
I solved this problem...
ua-cam.com/video/6K_j4Cj7mVo/v-deo.html
Got the same answer using another method but the original is ok.
@@restablex
Great. Thanks. MindYourDecisons demonstrates how to derive the formula very simply.
At 0:37,
Let AB=BC=CD=DA=t say
From the orange triangle we worked out DF=a say
From the blue triangle we worked out EB=4a/3
From the green triangle we worked out FA*AE/2=5
(t-a)(t-4a/3)=10
a=6/t and substitute into above equation,
That ended up as a quadratic in t^2,
(t^2)^2-24(t)^2-48=0
We worked out the area of ABCD =t^2 =12+4sqrt6
Then subtract the area of the three triangles of 3+4+5 from area ABCD
=12+4sqrt6-12
=4sqrt6 square units.
Same as last time with different symbols used.
Thank u for teaching or explaining in simple ways.. You take the time to break things down so that we understand the basic and the details of a certain problem... Thank u for being patient and clear..
When you're venturing into that quadratic formula, you know there's A LOT of book keeping to do!!
Too much fo me!
Wow, a phenomenally phenomenal phenomena, a work of absolute algebraic artistry.
And... no way did I do this one ,
but just so enlightening, particularly like the last bit using "u", that is new knowledge for me, thank you 🤓
It's always satisfying when you work through one of these elaborate problems, then watch the video and see the equations you just wrote down on paper start appearing on the screen. :) Thanks.
So nice of you J R! You are awesome 👍 I'm glad you liked it! Take care dear and stay blessed😃
Thank you for the info, I needed this! 👍🏽
Thanks Brotha dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Answer 9.797 of 9.8
did a similar problem some time ago.
let A = length of square B= base of the triangle with area 3, and C=length
of the triangle with the area, 5.
Ended up with the equation "A^4 - 24A^2 + 48 =0"
let X = A^2 (A x A or area of square) therefore X^2-24X+48=0
Solving for X (using the quadratic formula or quadratic calculator) will yield 12 (+ -) 4 sqrt 6. Since theTOTAL area of the three triangles is 12 (3+4+5), then the total area of the triangle inside is 4 sqrt 6 or 9.7979 Answer
Great feedback my friend. I appreciate that.
Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
Here is my answer to the same problem around a year ago.
The notation on the figure was different .
My square is described as BCEDFA in a clockwise direction from the top LHS of the square.
Area of triangle EDF=5,Area of triangle FAB=4 and Area of triangle BCE=3.
Let AF=d and CE= 3d/4 since the areas of triangle BAF=4 and triangle BCE =3 and both had the same height=x.
Let side of square ABCD =x and area of square is therefore x^2
Consider triangle BAF,
4= xd/2
d=8/x...............(1)
Consider triangle FDE,
5=(x-d)(x-3d/4)/2............(2)
Substitute d=8/x into (2),
We end up with 10=x^2-14+48/x^2
Let y=x^2,
0=y-24+48/y
Multiply through by y,
y^2-24y+48=0
y= 12+4sqrt(6) (-ve sqrt rejected in quadratic formula)
Area of square ABCD =x^2 = y =12+4sqrt(6),
Area of triangle BEF= area of ABCD-area of ABF-area of BCE-area of FED
=(12+4sqrt(6)}4-3-5
=12+4sqrt(6)-12
= 4sqrt(6) square units as required.
Note the -ve value of the sqrt in the quadratic formula was rejected because the area of BEF would have been -ve.
Amazing video 👍
Thank you so much sir😀
Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
Thank you for interesting tasks.
why u = 2.2 is not realistic (10:24)?
Let's label the side of the square "a" and the bases of the orange and the blue triangels x and y respectively. Than x=6/a and y=8/a.The area of the green triangel =(a-x) (a-y) /2 Hence :(a-6/a)(a-8/a)=10 We receive an biquadratic equation :a^4-24a^2+48 =0.Puting in an auxiliar unknown t=a^2 we get a square equation t^2-24t+48=0. D=384 sqrtD=sqrt16*sqrt 4*sqrt6=4*2*sqrt 6=8 sqrt 6. Therefor we get two walues of t=a^2 :12-4sqrt 6 - to be rejected and 12 +4sqrt 6. Now we have to substract the sum of the given triangels from the calculated area of the whole square :(12 +4sqrt6)-(3+4+5)=12+4sqrt6-12=4sqrt6. Greetings.
Thank you sir
Very good question
So nice of you Beeru! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Good work teacher , God bless you
Why you rejected ( 12-4√6 ) please 🙏
Thank you sir, Keep it up👍👍
Thank you so much Faisal for your continued love and support. Take care dear and stay blessed😃 You are awesome. Enjoy every moment of your life
It's interesting to note that ABCD does not need to be a square. The result will be the same for any rectangle.
@Solomon Grundy You are working with areas, and the area of the square/rectangle is only one parameter.
Please don't gloss over details which are absolutely important, like 10:27 to 10:33 (time) in your video... (i.e., without giving the reason why you rejected u = 2.2).
Ya.... rejecting by saying not realistic is not a logic.
If we use that value.. Then area EFC will become - ve... So its not realistic.
It's because u=x^2 represents the area of the square, it cannot be smaller than the area of any of the enclosed triangles, the smallest one of which is 3.
Good one. I almost got it.
Thx for your effort :)
I mean I kinda solve it and I got the equation however I didn’t know how to solve quadratic equations other wise pretty good
Btw keep up making these videos
Good techer
I want to prove the Area of triangle FCE is 4sqrt6 square units on the basis that the area of the square is (12+4sqrt6) square units as proved by the lecturer.(x^2= (12+4sqrt6)
Let FC^2=a^2=((x^4+36)/x^2)=23.44948974 approx =(21+sqrt6) exact value is the same as the decimal value on computer up to 25 places
Let EC^2=b^2=( (x^4+64)/x^2=24.73401368 approx =(4/3(21-sqrt6)) exact value is the same as the decimal value on computer up to 25 places.
Let FE^2=c^2=AF^2+AE^2= 21.1850341 approx and the ties up with FE^2= 1/3(63-sqrt6) to 25 decimal places for 25 decimal places.
The area of triangle CFE =sqrt(4a^2*b^2-(a^2+b^2-c^2)^2)/4 we know x^2=(12+4sqrt6)
Plug in values of a^2,b^2and c^2 into formula above;
Area of triangle FCE.
IFCEI=sqrt( 4*(21+sqrt6)(4/3)(21-sqrt6)-((21+sqrt6)+(4/3)(21-sqrt6)-(1/3)(63-sqrt6))^2/4
= sqrt(2320-(28)^2/4
= sqrt(2320-784)/4
= sqrt(1536)/4
= sqrt(256*6)/4
= sqrt(16^2*6)/4
= (16*sqrt6)/4
= 4*sqrt6 square units using surd values in the structure.
Assisted by Wolfram Alpha and hand calculator to work out the exact values in the structure.
The chance I would’ve figured out these calculations on my own is
My answer exactly!
Thank you sir
You are very welcome Gowri!
Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊
Thanks a lot
Very understandable! („Let‘s fill in the blanks...!“)
saudações meu grande professor respondo pelo nome Juma Cassimo , estudante de Engenharia Moçambique ( Africa) e sou muito fã dos seus trabalhos. Existem vários geômetras mas o Sr e o Melhor em todo universo Eu, gostaria de deixa uma proposta de resolução a cada exercícios ou seja, uma outra forma de resolução.
EFC is a right-angled triangle. Hence its area is half3x4=6.
All ur videos have low sound, pls increase the volume... Thnx
So nice of you Jagannath! Thanks for your feedback on audio quality. You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Why does Heron's formula say A=6?
Actually, 3 ,4 and 5 are areas of triangle and not side length... Hope it's clear now.
👍👍👍
Ar=6sq unit.
base × height ÷ 2
base × height
I like the challenges on your channel, but I really don't like the misrepresented images you give most of the time. In this one, it's obvious the white triangle is not a 345 triangle. Just ask the question with an actual representation.
Thanks Kirk for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard
👌👍👏❤
🙏🙏🙏🙏
Too easy, 3 4 5 is a pythagorean triple, the area is 3×4=12
I agree, It is a Pythagorean: A=1/2(3x4)=6
bh = b × h
Um,,, 3-4-5 is a right angle triangle…
What?!!
Never mind… 🙄
🦉