..depends, in the initial definition of the problem it isent specified that the square is a 45 degree tilted square.... wich makes it very much harder.. if its as simple as a 45 degre tilted square this is correct... if its not, we have to consider the square can be rotated in any angle, if we assume the squere can only rotate around its center, its fairly easy calculate if it also must fit inside the triangle, the are the aquare can occupy is therfor a circle, arnd the are is an circle with an inscribed square Are of square becomes ( 2√2(r) )² if r=12/7 Area = ( 2√2(12/7) )² if however the square can rotaated so it take upp as much space as possible, it gets trickier (we get a max and min area) .. im not going to solve it, but just consider stating nessesary information .. that the squeare is tilted in a certain angle say 45 deg) ... as u can never assume it is
If I have to do this by construction, there are two degrees of freedom: the distance of E from B and the angle BED. But we have only one equation (area of the right triangle) with these two unknowns. Hence it must be stated that FE is parallel to AB, otherwise there are multiple solutions.
If we don't assume EF is parallel to AB, finding the smallest such square becomes a nice minimisation problem. The smallest such square has side 12/5 (slightly smaller than the 12*√2/7 you get when you add the missing parallel assumption), with BE = 36/25.
It needs to be stated initially that AB is perpendicular to EF, otherwise there is no solution and the squares’area is not defined by the given information.
@@PreMathokay now you’re just making me mad. Why just tell everyone you’re sorry you weren’t clear but then not answer the question or modify the video to make it a given that EF is perpendicular to BC?!?
Not this time, Professor. Since it is not a given that BE= BD, nor is that equality necessarily true based on what we are given, FE is not necessarily parallel to AB, triangle CEF is not necessarily a right triangle and therefore triangles CEF and ABC are not similar. Your proof thus falls apart.
∆ ABC → AB = 4 = BD + DR + AR = k + k + 4 - 2k; BC = BE + CE = k + (b - k) = b k = BE = BD = DR = EF/2 = DG/2; sin(ABC) = 1 2b = 24 → b = 12 → AC = 4√10 → tan(ϑ) = 4/12 = 1/3 = (4 - 2k)/k → k = 12/7 → area ∎DEGF = (12√2/7)^2
+1 ... for using trigonometry, when not necessarily the easiest way! See my little write-up for a non-trigonometric way that uses intersecting line equations. It was fun. GoatGuy
This problem is very simple if we use an adapted orthonormal: B(0;0) A(0;4) C(12;0) (easy to calculate BC with the Pythagorean theorem just as you did) Now let's note BE = BD =a. Then we have E(a;0) D(0;a) F(a;2.a) We have VectorAC(12;-4) which is colinear to VectorU(3;-1). The equation of (AC) is then (x).(-1) - (y-4).(3) = 0 or -x - 3.y +12 = 0. Point F is on (AC) then -a -3.(2.a) +12 = 0 which gives that a =12/7, so the length of the square is sqrt(2). (12/7) and its area is 2.(144/49) = 288/49.
I went a slightly different. I called DB and EB, a, but it took me a short while to be sure that FE and AB are parallel. As the sides are a 3:1 ratio, 12-a = 2a (the diagonal). Rearrange: 12 - a = 3(2a) 12 - a = 6a 12 = 7a a = 12/7 diagonal is 24/7 Squares sides are 24/7)/sqrt(2) or (24/(7*sqrt(2)) Square is for 576/98 288/49 Squares area is 288/49 approximates to 5.88 un^2
I dare say that it should be expressed clearly in the terms of the problem that [FE] is parallel to [AB], or angle BED=GEC=D45 otherwise we face nonsense.
EF = 2 BE. Similar triangles ABC & FEC. 4 / 12 = EF / (12 - BE ). 4 / 12 = 2 BE / ( 12 - BE ). Cross multiplying. 24 BE = 48 - 4 BE. 28 BE = 48. BE = 48 / 28 = 12 / 7. The side length of square will be ( 12 / 7 ) x sq. rt. of 2. Area of square = ( 144 / 49 ) x 2. 5.88.
At no point in the question was it given that the square was not skew in the triangle. This was assumed to make the math work but was not given as a statement in the question.
1/BC=12 😀 Drop the height FH to AB. The triangles AHF and ABC are similar so AH/HF= AB/BC= 4/12=1/3. Label AH= x so HF=3x. 2/ We have: HB=EF = 2 HF ----> HB= 6x and---> AB= 7x = 4----> x=4/7 The area of the rectangle HFEB = The area of the green square. -----> Area of the green square = 3x . 6x = 18 sq x= (18 .16)/49= 5.877 sq units😀
This problem is so easy!! Just solving a System of 2 Equations with 2 Unknowns. BC = (48 / 4) = 12 lin un Now... Analytic Geometry Solution. 1) Equation of Straight Line passing through point A and C: Slope = (- 4)/12 = (- 1)/3 ; so the First equation is y = -4x/3 + 4 Considering O the center of the Red Square, EF = 2*OD; so the Slope of a Straight Line passing through point B and F is equal to 2; m = 2. 2) Second Linear Equation: y = 2x 3) As y= -1x/3 + 4 and y = 2x we must have: 4) - x/3 + 4 = 2x ; - x/3 + 12/3 = 6x/3 ; 12/3 = (6x + x)/3 ; 12/3 = 7x/3 ; 12 = 7x ; x = 12/7 5) OD = 12/7 ~ 1,714 li un and EF = 24/7 ~ 3,429 li un As we know that a diagonal of a Square is equal to D = Side * sqrt(2) In ou case Diagonal = 24/7 ; Side * sqrt(2) = 24/7 ; Side = 24 / (7 * sqrt(2)) ; Side = 24*sqrt(2)/14 ; Side = 12*sqrt(2)/7 ; Side ~ 4,424 lin un Green Square Area = [12*sqrt(2)/7] = 144*2 / 49 = 288 / 49 square units or approx. equal to 5,9 square units. Answer: Green Square Area is equal to (288/49) square units.
A different solution, by way of intersecting line functions! Game plan: find line function for hypotenuse, and for the upper left diagonal side of the square. Make 'em equal, find the intersection point, and then use that to determine side [𝒔]. The area is of course 𝒔² Area△ = ½ base height 24 = ½ base 4 12 = base △ABC hypotenuse drops 4 units, at [base = 12], so 𝒚 = -⁴⁄₁₂𝒙 ⊕ 4 With [⊕ 4] to account for the fact that the height of hypotenuse is [4] at [𝒙 = 0]. I then used same reasoning for the □ upper left diagonal: 𝒚 = 𝒙 + 𝒔/√2 Its a square, so by definition (when on its side) a 45° line has slope [1]. Don't know the size of the side though, so substituted in 𝒔/√2. Again because of 45° angle. Set these two equations equal to each other (to find the 𝒙 where they intersect at the top): 𝒙 + 𝒔/√2 = -⅓𝒙 ⊕ 4 … rearrange finding 𝒙 ⁴⁄₃𝒙 = 4 - 𝒔/√2 … multiply both sides by ¾ 𝒙 = 3 - (3𝒔)/4√2 Now in particular, we're interested finding 𝒔, and we know that the interceptiong point must also be at 𝒔/√2, so 𝒔/√2 = 3 - (3𝒔)/4√2 … solving for 𝒔 (multiply all terms by 4√2) 4𝒔 ⊕ 3𝒔 = 3•4√2 𝒔 = 12(√2)/7 𝒔 = 2.4244 And the area is that, squared 𝒔² = (12√(2)/7)² 𝒔 = 288 ÷ 49 𝒔 = 5.8776 Completing the goal of the problem, by way of intersecting lines! Yay! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
Why can you not split the 4 in half = 2. You now have the two lengths BD and BE. so use Pythagorean theory to get ED. 2 squared + 2 squared equals ED squared. Get the root and multiply to get the area.
... Good day, I BC I = 12 (easy to figure out) , TAN(C) = 4/12 = 1/3 , naming 4 sides green square T , angle (GEC) = 45 deg. , triangle (DBE) is an isosceles 45 - 45 - 90 triangle , so I BE I = SQRT(2)*T/2 , I EC I = 12 - SQRT(2)*T/2 , I FE I = SQRT(2)*T ... finally TAN(C) = (SQRT(2)*T) / (12 - SQRT(2)*T/2) = 1/3 ... solving for T, and after a few basic algebraic steps we obtain ... T = 12*SQRT(2) / 7 ... Area (Green square) = T^2 = (12*SQRT(2) / 7)^2 = 144 * 2 / 49 = 288 / 49 u^2 ... thanking you for your as always clear presentation ... Happy weekend, Jan-W
Triangle ∆ABC: A = bh/2 24 = b(4)/2 = 2b b = 24/2 = 12 a² + b² = c² 4² + 12² = CA² CA² = 16 + 144 = 160 CA = √160 = 4√10 Let x be the length of DB and BE and s be the length of the side of the square. Triangle ∆DBE: a² + b² = c² x² + x² = s² s² = 2x² s = √(2x²) = √2x Triangle ∆FEC FE/EC = AB/BC 2x/12-x = 4/12 = 1/3 6x = 12 - x 7x = 12 x = 12/7 s = (√2)12/7 = (12/7)√2 Square DEGF: A = s² = ((12/7)√2)² = 2(144/49) A = 288/49 ≈ 5.88
Lovely use of extended characters! One tiny thing you might want to change in the future: s = √(2x²) = √2x ... is ambiguous. Is the x on the inside, or outside of the sqrt? It would be less ambiguous as s = √(2x²) = x√2 Although that is a slightly less 'normal math book' form, it does clearly denote the [x] being outside the sqrt. GoatGuy
Interesting problem! Also if, instead of EF ⊥ BC (or BD = BE), we assume that DA = DB or DA = DF or AD = AF🤔 In the video, if we draw a square with AB as diagonal, its vertex outside the triangle is collinear with C and D 😉
Really nice question! Definitely the channel to go to increase you IQ level
Thanks ❤️
..depends, in the initial definition of the problem it isent specified that the square is a 45 degree tilted square.... wich makes it very much harder..
if its as simple as a 45 degre tilted square this is correct...
if its not, we have to consider the square can be rotated in any angle, if we assume the squere can only rotate around its center, its fairly easy calculate if it also must fit inside the triangle, the are the aquare can occupy is therfor a circle, arnd the are is an circle with an inscribed square
Are of square becomes ( 2√2(r) )²
if r=12/7
Area = ( 2√2(12/7) )²
if however the square can rotaated so it take upp as much space as possible, it gets trickier (we get a max and min area) .. im not going to solve it, but just consider stating nessesary information .. that the squeare is tilted in a certain angle say 45 deg) ... as u can never assume it is
I agree with you@@Patrik6920
Professor, I though the segment FE was not, necessarily perpendicular to the segment EC. It was not given in the problem. I did not undestand...
My apologies if I were not clear...
Thanks sharing sir❤❤❤
Thanks ❤️
If I have to do this by construction, there are two degrees of freedom: the distance of E from B and the angle BED. But we have only one equation (area of the right triangle) with these two unknowns. Hence it must be stated that FE is parallel to AB, otherwise there are multiple solutions.
If we don't assume EF is parallel to AB, finding the smallest such square becomes a nice minimisation problem. The smallest such square has side 12/5 (slightly smaller than the 12*√2/7 you get when you add the missing parallel assumption), with BE = 36/25.
It needs to be stated initially that AB is perpendicular to EF, otherwise there is no solution and the squares’area is not defined by the given information.
Exactly
Why EF is perpendicular to BC?
Yeah! Y?
Yes, my question too. Square can be tilted while points E, D & F remain touching the sides
My apologies if I were not clear...
How FE is perpendicular to BC.
Is it part of given facts?
My apologies if I were not clear...
@@PreMathokay now you’re just making me mad. Why just tell everyone you’re sorry you weren’t clear but then not answer the question or modify the video to make it a given that EF is perpendicular to BC?!?
@@mumps59 This was not his first incident of this nature.
Not this time, Professor. Since it is not a given that BE= BD, nor is that equality necessarily true based on what we are given, FE is not necessarily parallel to AB, triangle CEF is not necessarily a right triangle and therefore triangles CEF and ABC are not similar. Your proof thus falls apart.
My apologies if I were not clear...
Ok so.....
If P is the point of intersection of 2 diagonals then
DP=PE (Diagonals of sq. Bisect each other )
EF is not stated as being parallel to AB.
I think the fact that FE is parallel to AB probably needs to be in the problem statement.
∆ ABC → AB = 4 = BD + DR + AR = k + k + 4 - 2k; BC = BE + CE = k + (b - k) = b
k = BE = BD = DR = EF/2 = DG/2; sin(ABC) = 1
2b = 24 → b = 12 → AC = 4√10 → tan(ϑ) = 4/12 = 1/3 = (4 - 2k)/k → k = 12/7 → area ∎DEGF = (12√2/7)^2
+1 ... for using trigonometry, when not necessarily the easiest way! See my little write-up for a non-trigonometric way that uses intersecting line equations. It was fun. GoatGuy
Nice sharing
Thanks ❤️
Area of the green square=(12√2/7)^2=288/49=5.88square units .❤❤❤ Thanks sir.
Excellent!
You are very welcome!
Thanks ❤️
Very Very nice video sir 🎉🎉🎉🎉🎉
So nice of you
Thanks ❤️
This problem is very simple if we use an adapted orthonormal: B(0;0) A(0;4) C(12;0) (easy to calculate BC with the Pythagorean theorem just as you did)
Now let's note BE = BD =a. Then we have E(a;0) D(0;a) F(a;2.a)
We have VectorAC(12;-4) which is colinear to VectorU(3;-1). The equation of (AC) is then (x).(-1) - (y-4).(3) = 0 or -x - 3.y +12 = 0.
Point F is on (AC) then -a -3.(2.a) +12 = 0 which gives that a =12/7, so the length of the square is sqrt(2). (12/7) and its area is 2.(144/49) = 288/49.
I went a slightly different.
I called DB and EB, a, but it took me a short while to be sure that FE and AB are parallel.
As the sides are a 3:1 ratio, 12-a = 2a (the diagonal).
Rearrange: 12 - a = 3(2a)
12 - a = 6a
12 = 7a
a = 12/7
diagonal is 24/7
Squares sides are 24/7)/sqrt(2) or (24/(7*sqrt(2))
Square is for 576/98
288/49
Squares area is 288/49 approximates to 5.88 un^2
I dare say that it should be expressed clearly in the terms of the problem that [FE] is parallel to [AB], or angle BED=GEC=D45 otherwise we face nonsense.
My apologies if I were not clear...
Thank you!
EF = 2 BE.
Similar triangles ABC & FEC.
4 / 12 = EF / (12 - BE ).
4 / 12 = 2 BE / ( 12 - BE ).
Cross multiplying.
24 BE = 48 - 4 BE.
28 BE = 48.
BE = 48 / 28 = 12 / 7.
The side length of square will be ( 12 / 7 ) x sq. rt. of 2.
Area of square = ( 144 / 49 ) x 2.
5.88.
The condition of EF perpendicular to BC was not given.
how did you know that EF is perpendicular to BC?
At no point in the question was it given that the square was not skew in the triangle. This was assumed to make the math work but was not given as a statement in the question.
1/BC=12 😀
Drop the height FH to AB. The triangles AHF and ABC are similar so AH/HF= AB/BC= 4/12=1/3.
Label AH= x so HF=3x.
2/ We have: HB=EF = 2 HF ----> HB= 6x and---> AB= 7x = 4----> x=4/7
The area of the rectangle HFEB = The area of the green square.
-----> Area of the green square = 3x . 6x = 18 sq x= (18 .16)/49= 5.877 sq units😀
Got it!!
How is it concluded thqt FE is perpendicular to BC?
My apologies if I were not clear...
This problem is so easy!! Just solving a System of 2 Equations with 2 Unknowns.
BC = (48 / 4) = 12 lin un
Now...
Analytic Geometry Solution.
1) Equation of Straight Line passing through point A and C: Slope = (- 4)/12 = (- 1)/3 ; so the First equation is y = -4x/3 + 4
Considering O the center of the Red Square, EF = 2*OD; so the Slope of a Straight Line passing through point B and F is equal to 2; m = 2.
2) Second Linear Equation: y = 2x
3) As y= -1x/3 + 4 and y = 2x we must have:
4) - x/3 + 4 = 2x ; - x/3 + 12/3 = 6x/3 ; 12/3 = (6x + x)/3 ; 12/3 = 7x/3 ; 12 = 7x ; x = 12/7
5) OD = 12/7 ~ 1,714 li un and EF = 24/7 ~ 3,429 li un
As we know that a diagonal of a Square is equal to D = Side * sqrt(2)
In ou case Diagonal = 24/7 ; Side * sqrt(2) = 24/7 ; Side = 24 / (7 * sqrt(2)) ; Side = 24*sqrt(2)/14 ; Side = 12*sqrt(2)/7 ; Side ~ 4,424 lin un
Green Square Area = [12*sqrt(2)/7] = 144*2 / 49 = 288 / 49 square units or approx. equal to 5,9 square units.
Answer:
Green Square Area is equal to (288/49) square units.
4＊BC/2=24 BC=12
FE=DG=2x
4/12=2x/(12-x) 48-4x=24x
28x=48 x=12/7
Green Square area :
24/7＊24/7＊1/2=288/49
Thanks ❤️
How FE is perpendicular to BC?
A different solution, by way of intersecting line functions! Game plan: find line function for hypotenuse, and for the upper left diagonal side of the square. Make 'em equal, find the intersection point, and then use that to determine side [𝒔]. The area is of course 𝒔²
Area△ = ½ base height
24 = ½ base 4
12 = base
△ABC hypotenuse drops 4 units, at [base = 12], so
𝒚 = -⁴⁄₁₂𝒙 ⊕ 4
With [⊕ 4] to account for the fact that the height of hypotenuse is [4] at [𝒙 = 0]. I then used same reasoning for the □ upper left diagonal:
𝒚 = 𝒙 + 𝒔/√2
Its a square, so by definition (when on its side) a 45° line has slope [1]. Don't know the size of the side though, so substituted in 𝒔/√2. Again because of 45° angle.
Set these two equations equal to each other (to find the 𝒙 where they intersect at the top):
𝒙 + 𝒔/√2 = -⅓𝒙 ⊕ 4 … rearrange finding 𝒙
⁴⁄₃𝒙 = 4 - 𝒔/√2 … multiply both sides by ¾
𝒙 = 3 - (3𝒔)/4√2
Now in particular, we're interested finding 𝒔, and we know that the interceptiong point must also be at 𝒔/√2, so
𝒔/√2 = 3 - (3𝒔)/4√2 … solving for 𝒔 (multiply all terms by 4√2)
4𝒔 ⊕ 3𝒔 = 3•4√2
𝒔 = 12(√2)/7
𝒔 = 2.4244
And the area is that, squared
𝒔² = (12√(2)/7)²
𝒔 = 288 ÷ 49
𝒔 = 5.8776
Completing the goal of the problem, by way of intersecting lines! Yay!
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Thanks
شكرا لكم على المجهودات
اذاكان DEB=45 فان
7/(جذر2 )DE=12
S=DE^2=288/49
Why can you not split the 4 in half = 2. You now have the two lengths BD and BE. so use Pythagorean theory to get ED. 2 squared + 2 squared equals ED squared. Get the root and multiply to get the area.
Thanks ❤️
Btw. I love these videos. I share them with my sons.
Dude just says "thnx" lol tf
... Good day, I BC I = 12 (easy to figure out) , TAN(C) = 4/12 = 1/3 , naming 4 sides green square T , angle (GEC) = 45 deg. , triangle (DBE) is an isosceles 45 - 45 - 90 triangle , so I BE I = SQRT(2)*T/2 , I EC I = 12 - SQRT(2)*T/2 , I FE I = SQRT(2)*T ... finally TAN(C) = (SQRT(2)*T) / (12 - SQRT(2)*T/2) = 1/3 ... solving for T, and after a few basic algebraic steps we obtain ... T = 12*SQRT(2) / 7 ... Area (Green square) = T^2 = (12*SQRT(2) / 7)^2 = 144 * 2 / 49 = 288 / 49 u^2 ... thanking you for your as always clear presentation ... Happy weekend, Jan-W
(4)^2=16° {180°+16}=196° ((360°-196°)=√164° √4^√4:2^2 √2^√2 √2^√2 √2^2 √1^√1√ 1^√1 1^2 (x+1x-2)
Thanks ❤️
If DB=2, DE = sq. Rt. Of 8 = 2.828. Then the area of the square = 4.756, How is this wrong?
Sorry, the area would be 8.
Triangle ∆ABC:
A = bh/2
24 = b(4)/2 = 2b
b = 24/2 = 12
a² + b² = c²
4² + 12² = CA²
CA² = 16 + 144 = 160
CA = √160 = 4√10
Let x be the length of DB and BE and s be the length of the side of the square.
Triangle ∆DBE:
a² + b² = c²
x² + x² = s²
s² = 2x²
s = √(2x²) = √2x
Triangle ∆FEC
FE/EC = AB/BC
2x/12-x = 4/12 = 1/3
6x = 12 - x
7x = 12
x = 12/7
s = (√2)12/7 = (12/7)√2
Square DEGF:
A = s² = ((12/7)√2)² = 2(144/49)
A = 288/49 ≈ 5.88
How could u say BE = BD??
@@saiakashanjalisakaray2865 Because
△BED is a 45-45-90 △
Lovely use of extended characters! One tiny thing you might want to change in the future:
s = √(2x²) = √2x ... is ambiguous.
Is the x on the inside, or outside of the sqrt? It would be less ambiguous as
s = √(2x²) = x√2
Although that is a slightly less 'normal math book' form, it does clearly denote the [x] being outside the sqrt.
GoatGuy
Thanks ❤️
@@robertlynch7520any proof?..
It could be but that's only one case..
BY moving the square it could make some other anle too
Rien ne dit que FE soit perpendiculaire à BC !!!!!
Mes excuses si je n'ai pas été clair...
There are infinite number of squares.
My apologies if I were not clear...
Interesting problem! Also if, instead of EF ⊥ BC (or BD = BE), we assume that DA = DB or DA = DF or AD = AF🤔
In the video, if we draw a square with AB as diagonal, its vertex outside the triangle is collinear with C and D 😉
not too difficult
Gt i!!