Can you find area of the Yellow shaded Rectangle? | (Square) | #math #maths | #geometry

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Can you find area of the Yellow shaded Rectangle? | (Square) | #math #maths | #geometry

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[imetroangola]

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*

[wmcomprev]

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).

[jimlocke]

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.

[LeoniloBajao]

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

[lasalleman]

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.

[himo]

EF=HG=(9-4√2)*√2=9√2-8
Yellow Rectangle area = 8*(9√2-8) = 72√2-64

[d-hat-vr]

I want to share with everyone a related construction ("square-within-a-square") I discovered while pondering this video.

Suppose we are given a square ABCD with a line segment JK. JK can be at any angle and location relative to the square.

What is an algorithm to inscribe a smaller square PQRS in ABCD where 2 sides of PQRS are parallel to JK?

Answer:
1. Draw lines AC and DB to locate the center of □ABCD; label the center point as O.
2. Extend JK (by approximately 3/2 times the JO distance) in both directions
3. Construct a perpendicular line to JK through O. Label the intersection point L (that is, OL ⊥ JK with L being somewhere on extended line JK)
4. Copy the length OL to locate points M and N along line JK, so that OL = LM = LN.
5. Draw a line through OM to intersect □ABCD in two points; call them P and R
6. Draw a line through ON to intersect □ABCD in two points; call them Q and S

PQRS is the desired inscribed square!

Proof of correctness available upon request.

[murdock]

Nice! EF = ET + FT = a + a → AE = a√2; AC = AM + CM = 9√2/2 + 9√2/2 →
a = 9√2/2 - 4 = (1/2)(9√2 - 8) → area ∎EFGH = 16a = 8(9√2 - 8)

[prossvay]

Yellow rectangle area= 8(9√2-8 )square units =37.82 square units.❤❤❤

[MrPaulc]

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).

[d-hat-vr]

Not long after this video was posted, I asked how we know that the yellow rectangle's length 8 sides are parallel to the square's diagonal, since it's not stated in the given information at the beginning of the video.

This led to some interesting discussion in the comments, and it seemed from tinkering around a bit that it must be that way.

So I soon got to work on the conjecture, stated formally as:

"Suppose there is a non-square rectangle inscribed in a square, where inscribed means that each edge of the square contains exactly one corner of the rectangle, and no corner of the rectangle is coincident with a corner of the square. Must each of the rectangle's sides be parallel to a diagonal of the square?"

A proof-by-contradiction strategy seemed most promising. So I worked on and off for a couple weeks, along the way discovering the square-within-a-square construction that I mention in another comment. But I made little progress. Finally on December 8, 2024, I discovered a proof of this conjecture.

So there was no need for Mr. PreMath to state that the yellow rectangle's length 8 sides are parallel to the square's diagonal before using this in his work.

To outline the proof of the conjecture above, we begin by considering one side of the supposed inscribed non-square rectangle, let's call it "side 1", and assume it is not parallel to either of the square's diagonals. Using the algorithm presented in my square-within-a-square comment, we inscribe a smaller square (call it S₂) with side parallel to "side 1". Next, let's construct the adjacent side of the supposed rectangle "side 2" which is of course perpendicular to "side 1". Finally let's construct another side of the supposed rectangle as "side 3" perpendicular to "side 2" and opposite "side 1". It may look obvious that "side 1" and "side 3" are of different lengths, so the supposed rectangle doesn't look like a rectangle at all! To prove that "side 1" and "side 3" are in fact of different lengths, we can use the unequal angles and similarity to show that the small triangles formed by the sides of the supposed rectangles and sides of S₂ are similar but of decreasing size. Finally we observe that "side 1" and "side 3" are unequal, by analyzing the sums or differences of the various line segments that add up to "side 1" and side 3".

That's just the basic idea; the full proof is available upon request.

[marcgriselhubert]

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.

[LucasBritoBJJ]

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

[MuffyA]

@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

[unknownidentity]

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

[alfastur]

The triangle made by AE and the point where the red line exits the yellow rectangle is a right triangle with 45 degrees. So, the part of the red line outside the rectangle is equal to half EF. So, the difference between the red line and the lng side of the rectangle is equal to the short side of the rectangle (9sqr(2)-8). Multiply by 8 to get the rectangle area.
PS: I see @LeoniloBajao published the same solution with other wording. Maybe others have found as well. I didn't read all of the comments.

[allanflippin]

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.

[yalchingedikgedik]

Wow
Very nice and useful by using many principles
Thanks Sir
Thanks PreMath
Good luck
❤❤❤❤

[nandisaand]

🔺️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

[quigonkenny]

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