Congratulations on your videos! They are great questions and good solutions! 🎉🎉🎉 *Contribution to the solution:* Now, since GHC is an isosceles right triangle, we have: GH = √2 (9 - 4√2) = 9√2 - 8 The yellow area S is given by: S = 8×(9√2 - 8) *S = 72√2 - 64 Square units*
Second method: You know the lengths of HC & GC. The hypotenuse (GH) of this right triangle is the end of the rectangle. Calculate the length of GH and multiply by 8 (the other side length of the rectangle).
At 3:35, we have CD = 9 and DH = 4√2, so CH = CD - DH = 9 - 4√2. Similarly, we can find that CG = 9 - 4√2. Apply the Pythagorean theorem to ΔCGH. (9 - 4√2)² + (9 - 4√2)² = GH², (81 - 72√2 + 32) + (81 - 72√2 + 32) = GH², 226 - 144√2 = GH², and GH = √(226 - 144√2). Area of square = (EH)(GH) = 8√(226 - 144√2), approximately 37.82 square units. Reconciling to PreMath's answer: Our answer does have an undesirable radical inside a radical where PreMath's does not. PreMath's answer 72√2 - 64 can be factored into 8(9√2 - 8). If we square (9√2 - 8), we get 162 - 144√2 + 64 = 226 - 144√2, so 226 - 144√2 factors into (9√2 - 8)(9√2 - 8) and our area of square = 8√(226 - 144√2) = 8√((9√2 - 8)(9√2 - 8)) = 8((9√2 - 8) = 72√2 - 64, as PreMath also found. Actually, when I found that (9 - 4√2)² + (9 - 4√2)² = GH², I should have combined terms so GH² = 2(9 - 4√2)² and GH = (√2)(9 - 4√2) = 9√2 - (√2)(4√2) = 9√2 - 8, then taken area = (EH)(GH) = 8(9√2 - 8) = 72√2 - 64, and found PreMath's answer directly.
On a different note, how would you know if some random radical inside radical, such as this can actually be made into a normal radical as you did without knowing the end radical
@@Z-eng0 If you have a square root inside a square root, you need to factor the radicand to include 2 equal terms inside the radicand. Both terms can be removed from under the square root sign and one term placed outside. In this case, 226 - 144√2 factors into (9√2 - 8)(9√2 - 8) so √(226 - 144√2) = √((9√2 - 8)(9√2 - 8)) = (9√2 - 8). There are some strategies for factoring when the factors are not known. Many times, the radicand can not be factored this way. In a 15°-75°-90°, if the long leg has length 1, the hypotenuse has length 2√(2 - √3). The 2 - √3 does not factor into 2 equal terms of form (a + √b) or (a - √b).
@jimlocke9320 thanks for the help, I actually found a way to turn that into equations just now using what you just mentioned, I noticed that when the inside radical is of the form a-b(c)^0.5, we know that if it CAN be factored into 2 equal terms it would definitely have the form (x(c)^0.5-y)² [the radical insde the main root couldn't have changed after we expanded the binomial, it was only that the first and last terms were added or subtracted after they both became rational numbers so we can no longer see original form intuitively], In this case it doesn't matter which is the x and which is the y, we need to have the c known anyway so I assume it'll always be known when we try to reach our x and y, I expanded and equated the rational and irrational parts [if there was a minus it'd remain a minus after the final expansion form and a plus would remain a plus], I obtained 2 equations of 2x²+y² = a, and 2xy = b, where x and y NEED to be rational numbers, if there's a solution where they're both rational it can be expanded, otherwise it can't, same with the plus case
Go Pythagorean. Area of total square = 81. Subtract out triangles FBG and EDH which total 32. Then subtract out triangles GCH and EAF which total 11.1552. This leaves an area of 37.884 for the yellow rectangle. Very close to suggested solution.
Let's find the area: . .. ... .... ..... First of all we calculate the side length s of the square from the known length d of its diagonal: s = d/√2 = 9√2/√2 = 9 Let a=EH=FG and b=EF=GH be the side lengths of the square. For reasons of symmetry the triangles BFG and DEH on one hand and the triangles AEF and CGH on the other hand are isosceles triangles and pairwise congruent. Therefore we obtain: BF = BG = DE = DH = a/√2 = 8/√2 = 4√2 ⇒ AE = AF = CG = CH = s − BF = 9 − 4√2 ⇒ b = AE*√2 = 9√2 − 8 Now we are able to calculate the area of the yellow rectangle: A(EFGH) = a*b = 8*(9√2 − 8) = 72√2 − 64 ≈ 37.82 Best regards from Germany
The square's sides are 9, due to the diagonal being 9*sqrt(2). As the angles involved are 45deg, FB, BG, ED, and DH are all 8/sqrt(2). so 4*sqrt(2). AF, AE, HC, and GC are all 9-4*sqrt(2). Therefore, EF and GH are both (9-4*sqrt(2)) * sqrt(2). Yellow area is 8(9 - 4*sqrt(2)) * (sqrt(2) Rewrite: (8*sqrt(2))(9 - 4*sqrt(2)) Calculate: 72*sqrt(2) - 64 37.82un^2 (rounded).
The side length of the square is (9.sqrt(2))/sqrt(2) = 9. We use an orthonormal center D and first axis (DC), and we note a = AE = AF = CG = CH. We have F(a; 9) G(9; a) E(0; 9 -a), then VectorFG(9 -a; a -9) and FG^2 = 2.((9 -a)^2) and FG = sqrt(2).(9 -a) = 8, so 9 -a = 4.sqrt(2) and a = 9 -4.sqrt(2) Now HG = sqrt(2).a in triangle HCG, so HG = 9.sqrt(2) - 8. The side lengths of the rectangle EFGH are 8 and 9.sqrt(2) - 8, so its area is 72.sqrt(2) - 64.
How do we know that the rectangle's length 8 sides are parallel to the square's diagonal? It's not stated in the given information during the opening moments.
@@johankotze42 Trying to construct a rectangle with its long side not parallel to the square's diagonal seems to fail: the slopes of the rectangle's adjacent sides not being negative reciprocals, thus not forming 90° angles. Theorem about rectangles inscribed in squares?
Demonstrating that FG//AC//EH: The two diagonals of the rectangle intersect at point O. We have: OF= OH= OE=OG Because O is the midpoint of the line segment FH-> point O is on the perpendicular bisector of AD and BC Similarly, O is also the midpoint of EG --> point O is also on the perpendicular bisector of AB and CD, which means that point O is the intersecting point of the two perpendicular bisectors mentioned above -> point O is also the intersecting point of the two diagonals of the square 😅 --> EF//= HG and perpendicular to AC and FG//AC😅😅😅
@@phungpham1725 You merely prove that your point O is the midpoint of all diagonals of the square and rectangle as well as the perpendicular bisectors of the sides of the square. I think one can draw a similar diagram where the inscribed rectangle does not have its perpendicular bisectors parallel to, or on the diagonals of the square. The problem as stated by PreMath does not state that these lines are parallel nor does it give enough data to deduce that. If, for instance, it was stated that triangle AEF is isosceles, then it can be proven by showing that AC is perpendicular to EF and HE. The reverse is also true. By stating that EF and HG are perpendicular to AC, it can be shown that the lines in question are parallel. Also, by stating that one of the acute angles of any one of the triangles is 45 degrees, one can also deduce the required condition. To parphrase @PreMath "His drawing is not true to scale"
@@phungpham1725 And to prove that "if O is the midpoint of FH, then O is on the perpendicular bisector of DC" I constructed 2 lines parallel to DC: through G and through O, and also drawing a perpendicular line from O to DC, and then used similar triangles' proportions. But I'm confused by your last step: what does "EF //= HG" mean? (Remember you can paste standard Unicode math symbols into UA-cam comments)
Let EF = GH = x. Let the intersection point between AC and EF be M and between AC and GH be N. As FG and HE are parallel to AC and EFGH is fully inscribed in ABCD, then MN = FG = HE = 8. As EF = GH = x, then by symmetry EM = MF = GN = NH = x/2. As ∠GCN = ∠NCH = 45° and ∠CNG = ∠HNC = 90°, then ∠NGC = ∠CHN = 180°-(90°+45°) = 45° and ∆CNG and ∆HNC are congruent isosceles right triangles, and CN = HN = NG = x/2. By symmetry, ∆FMA and ∆AME are congruent to these triangles as well. AC = AM + MN + NC 9√2 = x/2 + 8 + x/2 x = 9√2 - 8 Yellow Rectangle EFGH: A = lw = 8(9√2-8) A = 72√2 - 64 ≈ 37.823 sq units
Did almost the same, but once I found 9-4√2, it is easy to find the “height” of the yellow rectangle multiplying it by √2, because the value found is the leg of the HCG (and AEF) triangle, and its longest Leg is found multiplying it by √2 (isoceles triangle) (9-4√2)√2 = 9√2-8 Then, (9√2-8)8 = 72√2-64
I found a simple way, with the math simple enough to do in your head. To see this, add diagonal BD which is 9 * sqrt(2) long and perpendicular to the other diagonal lines. This diagonal splits FBG and EDH into identical 45-degree isosceles triangles. The distance from B to FG can be shown to be 4. Also the distance to from D to EH. The amount of BD which passes through the yellow rectangle is 9 * sqrt(2) - 8. BD is parallel to EF and GH, so the length of EF and GH equals 9 * sqrt(2). The rectangle area is then (9 * sqrt(2) - 8) * 8 or 72 * sqrt(2) - 64.
Solution: Yellow Rectangle Area = 8 × GH ... ¹ Then, our task is to find the value of GH Since we are dealing with a square and the diagonal is 9√2, then the side of the square is 9 In the triangle DEH the diagonal is 8, therefore DH = DE = 8/√2 DH + HC = 9 HC + 8/√2 = 9 HC = 9 - 8/√2 GH = (9 - 8/√2) √2 GH = 9√2 - 8 Replacing in equation ¹, we have: Yellow Rectangle Area = 8 × (9√2 - 8) = 72√2 - 64 Yellow Rectangle Area = 72√2 - 64 Square Units ✅ Yellow Rectangle Area ≈ 37.8233 Square Units ✅
My way of solution ▶ Here we can see that [FG] // [AC] ∠BFG = ∠BAC = 45° ∠FGB = ∠ACB= 45° ⇒ [FB]= [BG]= x [AF]= [GC]= y or the right triangle ΔFGB, according to the Pyhagorean theorem, we can write : x² + x²= 8² x²= 32 x= 4√2 ⇒ [FB]= [BG]= 4√2 for the triangle ΔACB by applying the Pythagorean theorem : (x+y)²+(x+y)²= (9√2)² (x+y)²= 81 x+y= 9 x= 4√2 ⇒ y= 9- 4√2 Consider the point O : O ∈ [EF] [AO] ⊥ [EF] [AO] = [EF] = b ⇒ b√2= y b√2= 9- 4√2 ⇒ b= (9-4√2)/√2 b= (9√2 - 8)/2 ⇒ [EF]= 2b [EF]= (9√2 - 8) Ayellow= [EF]*[FG] [FG]= 8 Ayellow= (9√2 - 8)*8 Ayellow= 72√2 - 64 square units ✅
STEP-BY-STEP RESOLUTION PROPOSAL : 01) Diagonal Formula = Side * sqrt(2) 02) AC = 9 * sqrt(2) ; Sides : (AB = AD = BC = CD) = 9 lin un 03) EH = FG = 8 lin un 04) So : FB = BG = ED = DH = 4sqrt(2) 05) AE = AF = CG = CH = [9 - 4sqrt(2)] lin un ; AE = AF = CG = CH ~ 3,343 lin un 06) EF = GH ~ 4,728 07) Yellow Rectangle Area [EFGH] ~ 37,823 sq un Thus, OUR BEST ANSWER : The Yellow Rectangle Area is approx. equal to 38 Square Units.
Since you said this is a square, then the calculating EF and HG become obvious. The the length of A to the intersection of AC and EF and the length of C to the intersection AC and HC is clear to be (9*sqrt(2) -8)/2. And due to being a square, the angle AEF is 45 degrees, then you have the length of E to the intersection of EF and AC matching the said length in previous sentence and likewise for the opposite end. So the bottom line area is just (9*sqrt(2) -8) * 8.
I just looked at it and came up with the answer. I thought that I made some logic flaw when I saw it was 10 minutes. I jumped to the end and saw my answer.
Over-complicated. The leg of the larger triangle is just 8/root2 = 4root2. The side of the square is 9root2/root2 = 9. The leg of the little triangle is then 9-4root2. The hypotenuse of the small triangle is then (9-4root2)*root2. Then just multiply (9-4root2)*root2*8 and out falls 72root 2-64.
A simple answer is always desirable. It saves time and from calculation mistake. I am sorry to say, in most of the cases the solutions appear to be lengthy and tedious.
Sometimes, but I think the point of PreMath's videos are more about teaching how to 'think outside the box', add unexpected construction lines, see non-obvious relationships among elements, etc. These are very difficult skills to learn! I thought I was good at geometry until I started watching these videos, but being amazed at his ingenious answers to seemingly impossible questions, I realized I actually had a very poor geometry education in high school.
I agree that the simple answer helps prevent mistakes. That is certainly true in my experience. Still, more often than not, PreMath's solutions are shorter and cleverer than my solution. Even when that isn't the case, as @d-hat-vr2002 says, it can be illuminating to see a different way to solve the problem.
Congratulations on your videos! They are great questions and good solutions! 🎉🎉🎉
*Contribution to the solution:*
Now, since GHC is an isosceles right triangle, we have:
GH = √2 (9 - 4√2) = 9√2 - 8
The yellow area S is given by:
S = 8×(9√2 - 8)
*S = 72√2 - 64 Square units*
Great work with finding the area of the yellow rectangle! 🙏
Thanks for the feedback ❤️
Second method: You know the lengths of HC & GC. The hypotenuse (GH) of this right triangle is the end of the rectangle. Calculate the length of GH and multiply by 8 (the other side length of the rectangle).
Thanks for the feedback ❤️
That’s what I was thinking too.
That's what I was gonna comment!
EF=2(AC-EH)
EF=9*2^(1/2)-8
Yellow triangle Area=EF*EH
=8(9*2^(1/2)-8)
=72*2^(1/2)-64
Excellent!
Thanks for sharing ❤️
surely first line is wrong although the next step seems to have fixedit.
At 3:35, we have CD = 9 and DH = 4√2, so CH = CD - DH = 9 - 4√2. Similarly, we can find that CG = 9 - 4√2. Apply the Pythagorean theorem to ΔCGH. (9 - 4√2)² + (9 - 4√2)² = GH², (81 - 72√2 + 32) + (81 - 72√2 + 32) = GH², 226 - 144√2 = GH², and GH = √(226 - 144√2). Area of square = (EH)(GH) = 8√(226 - 144√2), approximately 37.82 square units.
Reconciling to PreMath's answer: Our answer does have an undesirable radical inside a radical where PreMath's does not. PreMath's answer 72√2 - 64 can be factored into 8(9√2 - 8). If we square (9√2 - 8), we get 162 - 144√2 + 64 = 226 - 144√2, so 226 - 144√2 factors into (9√2 - 8)(9√2 - 8) and our area of square = 8√(226 - 144√2) = 8√((9√2 - 8)(9√2 - 8)) = 8((9√2 - 8) = 72√2 - 64, as PreMath also found.
Actually, when I found that (9 - 4√2)² + (9 - 4√2)² = GH², I should have combined terms so GH² = 2(9 - 4√2)² and GH = (√2)(9 - 4√2) = 9√2 - (√2)(4√2) = 9√2 - 8, then taken area = (EH)(GH) = 8(9√2 - 8) = 72√2 - 64, and found PreMath's answer directly.
Excellent!
Thanks for sharing ❤️
On a different note, how would you know if some random radical inside radical, such as this can actually be made into a normal radical as you did without knowing the end radical
@@Z-eng0 If you have a square root inside a square root, you need to factor the radicand to include 2 equal terms inside the radicand. Both terms can be removed from under the square root sign and one term placed outside. In this case, 226 - 144√2 factors into (9√2 - 8)(9√2 - 8) so √(226 - 144√2) = √((9√2 - 8)(9√2 - 8)) = (9√2 - 8). There are some strategies for factoring when the factors are not known.
Many times, the radicand can not be factored this way. In a 15°-75°-90°, if the long leg has length 1, the hypotenuse has length 2√(2 - √3). The 2 - √3 does not factor into 2 equal terms of form (a + √b) or (a - √b).
@jimlocke9320 thanks for the help, I actually found a way to turn that into equations just now using what you just mentioned, I noticed that when the inside radical is of the form a-b(c)^0.5, we know that if it CAN be factored into 2 equal terms it would definitely have the form (x(c)^0.5-y)² [the radical insde the main root couldn't have changed after we expanded the binomial, it was only that the first and last terms were added or subtracted after they both became rational numbers so we can no longer see original form intuitively],
In this case it doesn't matter which is the x and which is the y, we need to have the c known anyway so I assume it'll always be known when we try to reach our x and y,
I expanded and equated the rational and irrational parts [if there was a minus it'd remain a minus after the final expansion form and a plus would remain a plus],
I obtained 2 equations of 2x²+y² = a, and 2xy = b, where x and y NEED to be rational numbers, if there's a solution where they're both rational it can be expanded, otherwise it can't, same with the plus case
Go Pythagorean. Area of total square = 81. Subtract out triangles FBG and EDH which total 32. Then subtract out triangles GCH and EAF which total 11.1552. This leaves an area of 37.884 for the yellow rectangle. Very close to suggested solution.
EF=GH=AC-FG=9vʼ2-8
[EFGH]=EF•FG=(9vʼ2-8)8=72vʼ2-64
@PreMath Why do you assume that ED = DH ? I think we should not assume this because the diagram looks like it, and probably need to prove it
EF=HG=(9-4√2)*√2=9√2-8
Yellow Rectangle area = 8*(9√2-8) = 72√2-64
Excellent!
Thanks for sharing ❤️
Wow
Very nice and useful by using many principles
Thanks Sir
Thanks PreMath
Good luck
❤❤❤❤
Yellow rectangle area= 8(9√2-8 )square units =37.82 square units.❤❤❤
Excellent!
Thanks for sharing ❤️
Woot! Nice way to complete a Friday!
Thanks for the feedback! 😊🙏
Let's find the area:
.
..
...
....
.....
First of all we calculate the side length s of the square from the known length d of its diagonal:
s = d/√2 = 9√2/√2 = 9
Let a=EH=FG and b=EF=GH be the side lengths of the square. For reasons of symmetry the triangles BFG and DEH on one hand and the triangles AEF and CGH on the other hand are isosceles triangles and pairwise congruent. Therefore we obtain:
BF = BG = DE = DH = a/√2 = 8/√2 = 4√2
⇒ AE = AF = CG = CH = s − BF = 9 − 4√2
⇒ b = AE*√2 = 9√2 − 8
Now we are able to calculate the area of the yellow rectangle:
A(EFGH) = a*b = 8*(9√2 − 8) = 72√2 − 64 ≈ 37.82
Best regards from Germany
Excellent!
Thanks for sharing ❤️
The square's sides are 9, due to the diagonal being 9*sqrt(2). As the angles involved are 45deg, FB, BG, ED, and DH are all 8/sqrt(2). so 4*sqrt(2).
AF, AE, HC, and GC are all 9-4*sqrt(2).
Therefore, EF and GH are both (9-4*sqrt(2)) * sqrt(2).
Yellow area is 8(9 - 4*sqrt(2)) * (sqrt(2)
Rewrite: (8*sqrt(2))(9 - 4*sqrt(2))
Calculate: 72*sqrt(2) - 64
37.82un^2 (rounded).
The side length of the square is (9.sqrt(2))/sqrt(2) = 9. We use an orthonormal center D and first axis (DC), and we note a = AE = AF = CG = CH.
We have F(a; 9) G(9; a) E(0; 9 -a), then VectorFG(9 -a; a -9) and FG^2 = 2.((9 -a)^2) and FG = sqrt(2).(9 -a) = 8, so 9 -a = 4.sqrt(2) and a = 9 -4.sqrt(2)
Now HG = sqrt(2).a in triangle HCG, so HG = 9.sqrt(2) - 8. The side lengths of the rectangle EFGH are 8 and 9.sqrt(2) - 8, so its area is 72.sqrt(2) - 64.
Excellent!
Thanks for sharing ❤️
How do we know that the rectangle's length 8 sides are parallel to the square's diagonal? It's not stated in the given information during the opening moments.
Also trying to work that one out. Intuitively, it looks like it cannot be otherwise.
@@johankotze42 Trying to construct a rectangle with its long side not parallel to the square's diagonal seems to fail: the slopes of the rectangle's adjacent sides not being negative reciprocals, thus not forming 90° angles. Theorem about rectangles inscribed in squares?
Demonstrating that FG//AC//EH:
The two diagonals of the rectangle intersect at point O.
We have: OF= OH= OE=OG
Because O is the midpoint of the line segment FH-> point O is on the perpendicular bisector of AD and BC
Similarly, O is also the midpoint of EG --> point O is also on the perpendicular bisector of AB and CD, which means that point O is the intersecting point of the two perpendicular bisectors mentioned above
-> point O is also the intersecting point of the two diagonals of the square 😅
--> EF//= HG and perpendicular to AC
and FG//AC😅😅😅
@@phungpham1725 You merely prove that your point O is the midpoint of all diagonals of the square and rectangle as well as the perpendicular bisectors of the sides of the square.
I think one can draw a similar diagram where the inscribed rectangle does not have its perpendicular bisectors parallel to, or on the diagonals of the square. The problem as stated by PreMath does not state that these lines are parallel nor does it give enough data to deduce that. If, for instance, it was stated that triangle AEF is isosceles, then it can be proven by showing that AC is perpendicular to EF and HE. The reverse is also true. By stating that EF and HG are perpendicular to AC, it can be shown that the lines in question are parallel. Also, by stating that one of the acute angles of any one of the triangles is 45 degrees, one can also deduce the required condition. To parphrase @PreMath "His drawing is not true to scale"
@@phungpham1725 And to prove that "if O is the midpoint of FH, then O is on the perpendicular bisector of DC" I constructed 2 lines parallel to DC: through G and through O, and also drawing a perpendicular line from O to DC, and then used similar triangles' proportions. But I'm confused by your last step: what does "EF //= HG" mean? (Remember you can paste standard Unicode math symbols into UA-cam comments)
Let EF = GH = x. Let the intersection point between AC and EF be M and between AC and GH be N.
As FG and HE are parallel to AC and EFGH is fully inscribed in ABCD, then MN = FG = HE = 8.
As EF = GH = x, then by symmetry EM = MF = GN = NH = x/2. As ∠GCN = ∠NCH = 45° and ∠CNG = ∠HNC = 90°, then ∠NGC = ∠CHN = 180°-(90°+45°) = 45° and ∆CNG and ∆HNC are congruent isosceles right triangles, and CN = HN = NG = x/2. By symmetry, ∆FMA and ∆AME are congruent to these triangles as well.
AC = AM + MN + NC
9√2 = x/2 + 8 + x/2
x = 9√2 - 8
Yellow Rectangle EFGH:
A = lw = 8(9√2-8)
A = 72√2 - 64 ≈ 37.823 sq units
Excellent!
Thanks for sharing ❤️
🔺️AEF is a 45/45/90 🔺️, so side of rectangle,
AE=(9-4•Sqrt(2))×Sqrt(2)
AE=9•Sqrt(2)-8
Area=8×(9•Sqrt(2)-8)
Area=72•Sqrt(2)=64
Thanks for sharing ❤️
(8)^2(8)^2={64+64}={128 81}=209 {90°A90°B+90°C+90°D}=360°ABCD/209=1.159 3^17 3^8^9 3^8^3^2 1^2^3^3^1 1^1^2^3^1 2^3 (ABCD ➖ 3ABCD+2).
Did almost the same, but once I found 9-4√2, it is easy to find the “height” of the yellow rectangle multiplying it by √2, because the value found is the leg of the HCG (and AEF) triangle, and its longest Leg is found multiplying it by √2 (isoceles triangle)
(9-4√2)√2 = 9√2-8
Then, (9√2-8)8 = 72√2-64
I found a simple way, with the math simple enough to do in your head. To see this, add diagonal BD which is 9 * sqrt(2) long and perpendicular to the other diagonal lines. This diagonal splits FBG and EDH into identical 45-degree isosceles triangles. The distance from B to FG can be shown to be 4. Also the distance to from D to EH. The amount of BD which passes through the yellow rectangle is 9 * sqrt(2) - 8. BD is parallel to EF and GH, so the length of EF and GH equals 9 * sqrt(2).
The rectangle area is then (9 * sqrt(2) - 8) * 8 or 72 * sqrt(2) - 64.
Excellent!
Thanks for sharing ❤️
Solution:
Yellow Rectangle Area = 8 × GH ... ¹
Then, our task is to find the value of GH
Since we are dealing with a square and the diagonal is 9√2, then the side of the square is 9
In the triangle DEH the diagonal is 8, therefore
DH = DE = 8/√2
DH + HC = 9
HC + 8/√2 = 9
HC = 9 - 8/√2
GH = (9 - 8/√2) √2
GH = 9√2 - 8
Replacing in equation ¹, we have:
Yellow Rectangle Area = 8 × (9√2 - 8)
= 72√2 - 64
Yellow Rectangle Area = 72√2 - 64 Square Units ✅
Yellow Rectangle Area ≈ 37.8233 Square Units ✅
Excellent!
Thanks for sharing ❤️
My way of solution ▶
Here we can see that [FG] // [AC]
∠BFG = ∠BAC = 45°
∠FGB = ∠ACB= 45°
⇒
[FB]= [BG]= x
[AF]= [GC]= y
or the right triangle ΔFGB, according to the Pyhagorean theorem, we can write :
x² + x²= 8²
x²= 32
x= 4√2
⇒
[FB]= [BG]= 4√2
for the triangle ΔACB by applying the Pythagorean theorem :
(x+y)²+(x+y)²= (9√2)²
(x+y)²= 81
x+y= 9
x= 4√2
⇒
y= 9- 4√2
Consider the point O :
O ∈ [EF]
[AO] ⊥ [EF]
[AO] = [EF] = b
⇒
b√2= y
b√2= 9- 4√2
⇒
b= (9-4√2)/√2
b= (9√2 - 8)/2
⇒
[EF]= 2b
[EF]= (9√2 - 8)
Ayellow= [EF]*[FG]
[FG]= 8
Ayellow= (9√2 - 8)*8
Ayellow= 72√2 - 64 square units ✅
Excellent!
Thanks for sharing ❤️
EF=9√2-8---> EFGH =EF*8 =72√2 -64 =37,82...ud².
Gracias y saludos
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Diagonal Formula = Side * sqrt(2)
02) AC = 9 * sqrt(2) ; Sides : (AB = AD = BC = CD) = 9 lin un
03) EH = FG = 8 lin un
04) So : FB = BG = ED = DH = 4sqrt(2)
05) AE = AF = CG = CH = [9 - 4sqrt(2)] lin un ; AE = AF = CG = CH ~ 3,343 lin un
06) EF = GH ~ 4,728
07) Yellow Rectangle Area [EFGH] ~ 37,823 sq un
Thus,
OUR BEST ANSWER :
The Yellow Rectangle Area is approx. equal to 38 Square Units.
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Since you said this is a square, then the calculating EF and HG become obvious. The the length of A to the intersection of AC and EF and the length of C to the intersection AC and HC is clear to be (9*sqrt(2) -8)/2. And due to being a square, the angle AEF is 45 degrees, then you have the length of E to the intersection of EF and AC matching the said length in previous sentence and likewise for the opposite end. So the bottom line area is just (9*sqrt(2) -8) * 8.
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(9-8/√2)*√2*8
From similar of triangle so easy
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EF=AC-EH=9√2-8 area=EFxEH=72√2-64. In 2 seconds, not in 10 minutes.
I just looked at it and came up with the answer. I thought that I made some logic flaw when I saw it was 10 minutes. I jumped to the end and saw my answer.
[(9✓2 - 8)/2] x 2 x 8 = (9✓2 - 8) x 8 = 72✓2 - 64
When AE is known, EF is known. Area of the rectangle is EF×EH . It is that simple
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*=read as square root
Ans : (72.*2 - 64 ).....May be2
You can paste the Unicode square root symbol into a comment like this √ rather than needing to use another symbol.
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Rather rustic but not uncultured! ...IMHO 🙂
Very true!😀
A nie prościej : ( 9√2 - 8 ) x 8 = 37,8233
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A = b.h = 8*(9√2-8) = 72√2-64
A = 37,82 cm² ( Solved √ )
Once again, too complicated video solution !!
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Over-complicated. The leg of the larger triangle is just 8/root2 = 4root2. The side of the square is 9root2/root2 = 9. The leg of the little triangle is then 9-4root2. The hypotenuse of the small triangle is then (9-4root2)*root2. Then just multiply (9-4root2)*root2*8 and out falls 72root 2-64.
Not difficult and interesting 😅
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A simple answer is always desirable. It saves time and from calculation mistake. I am sorry to say, in most of the cases the solutions appear to be lengthy and tedious.
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Sometimes, but I think the point of PreMath's videos are more about teaching how to 'think outside the box', add unexpected construction lines, see non-obvious relationships among elements, etc. These are very difficult skills to learn! I thought I was good at geometry until I started watching these videos, but being amazed at his ingenious answers to seemingly impossible questions, I realized I actually had a very poor geometry education in high school.
I agree that the simple answer helps prevent mistakes. That is certainly true in my experience. Still, more often than not, PreMath's solutions are shorter and cleverer than my solution. Even when that isn't the case, as @d-hat-vr2002 says, it can be illuminating to see a different way to solve the problem.