You only found one of two possible solutions. You arrived at x=8, y=6. However, a valid solution also happens when x=6, y=8. In that case, the yellow area is smaller than the unshaded white area to the right. Theta would be the complement of your 53.13deg. Basically this second solution would visually ‘flip’ your solution around the line OB. A quick calculation is to subtract the blue rectangle area + yellow region region from the total area of the quarter circle pi * 10 * 10 / 4.
I agree. The point where the original workings go wrong is where he jumps from (x-y)^2 = 4 to the conclusion that x -y = 2, thereby missing the other possible solution, x-y = -2.
Exactly. I was looking at the thumbnail and noticed that the diagram is symmetric about the diagonal of the rectangle, so there must be 2 solutions. Unless the rectangle turned out to be a square, in which case the 2 solutions are the same.
Your solution assumes that (x-y) = 2 and not (x-y) = -2, which would have resulted, instead, with x = 6 and y = 8. If we assume the diagram is not drawn to scale, as we are warned at 0:42, this second solution must also be considered. This means we should locate two possible solutions and not just one. So I get the area is either A ~ 22.36 cm^2 or A ~ 8.176cm^2.
Given the usual warning about the scale of the diagram, let's make it clear that x > y. Then note that 48 is 6 × 8, a perfect match to 10 for a 3 : 4 : 5 Pythagorean triangle. So, with angles in radians, the yellow area will be ½ (10² arcsin ⅘ − 48) ≈ 22.36 cm².
Rectangle área : A = x.y = 48 cm² R sin α . R cos α = 48 sin α cos α = 48 / R² ½ sin(2α) = 48/10² α = 36,8699° / 53,1301° Area angular sector A = ½.R².α A = ½ R² 53,1301° π/180 A = 46,365 cm² Area right triangle: A = 48/2 = 24 cm² Yellow shaded area: A = A₂ - A₁ A = 22,365 cm² ( Solved √ ) Is not necessary to calculate 'x' and 'y', base and height of rectangle.
Your "action plan" in the beginning doesnt mention "sector area" or angles at all, just x and y. So I solved it with brute for: y=48/x x²+(48/x)²=(10)² ... x⁴-100x²+2304=0 x²=36 or 64 x=6 or 8 My only question was how do we assign which side 6 or 8? I went with: by visual inspection, x>y.
22.364 Easy The rectangle can be divided into a two 3-4-5 triangle with degrees 53.13. and 38.87 Since radius is 10, and the degree is 53.13, the area of the sector is 53.13/360 * 10^2. 3.14 = 46.34 subtract half of the retangle 48/2= 24 from the above 46.34- 24 =22.364
Here's how I did it. For the blue rectangle, I used a hypotenuse of 10, a height of L, and a width of 48/L. I created triangle BOA. I called angle BOA Θ. I used the double angle formula, sin(2Θ) = 2sin(Θ)cos(Θ), to nicely eliminate the "L" variable and get sin(2Θ) = 73.74°. That was the main trick. My final answer was A = 22.36°.
I got x and y using bisecting chords theorem. (10+y) * (10-y) = x^2; we know xy=48 so substitute 48/y for x. Get 100-y^2 = 2304/y^2 ; substitute t for y^2 ; Get 100t - t^2 = 2304 ; Use quadratic formula to solve for t and get t=36, so y = 6. So x=8. I knew we'd have to find the area of sector OBQ, given angle theta. I got sin(theta) = 0.8. So inverse sine gets theta. I got stuck here because I didn't think we were allowed to use a calculator.
Let x be the width of the blue rectangle and y the height. then, invoking Pythagoras: x^2 + y^2 = 10^2 = 100. BUT y = 48/x. Therefore, performing this substitution: x^2 + 48^2/x^2 = 100; carrying out the square operation: x^2 + 2304/x^2 = 100; multiplying through by x^2: x^4 + 2304 = 100(x^2); rearranging: x^4 -- 100(x^2) + 2304 = 0; solving for x^2 by the quadratic formula: x^2 = (100 +/-- sqrt(100^2 -- (4)(2304)))/2; carrying out the arithmetic operations: x^2 = (100 +/-- sqrt(784))/2 = (100 +/-- 28)/2 = either 128/2 = 64; or 72/2 = 36; taking the square roots, we find that x = 8 and y = 6. Now consider the arc OQB. Angle OQB = arctan(8/6) = 53.13 degrees; so the area of this pie slice is (100 pi)(53.13)/(360) = 14.76 pi. Subtract half the area of the rectangle (24 square units) and we have 14.76 pi -- 24 = approximately 22.37 for the yellow region. Veni, vidi, vici! 🤠
let AB = a and OA = b. Draw radius OB. Rectangle OABC: A = OA(AB) = ab ab = 48 ---- (1) b = 48/a Triangle ∆OAB: AB² + OA² = OB² a² + b² = 10² ---- (2) a² + (48/a)² = 100 a² + 48²/a² - 100 = 0 a⁴ - 100a² + 48² = 0 a⁴ - 8²a² - 6²a² + 48² = 0 a²(a²-8²) - 6²(a²-8²) = 0 (a²-8²)(a²-6²) = 0 a² = 8² | a² = 6² a = 8 ❌ | a = 6 a b = 48/8 | b = 48/6 b = 6 ❌ | b = 8 a Sector BOP: sin(θ) = 6/10 = 3/5 θ = sin⁻¹(3/5) ≈ 36.87° A = (θ/360)πr² A ≈ (36.87/360)π10² A ≈ 10.24π By observation, the yellow area is the area of quarter circle O, minus rectangle OABC, minus sector BOP, plus triangle ∆OAB. A ≈ πr²/4 - 48 - 10.24π + 24 A ≈ 25π - 24 - 10.24π A ≈ 14.76π - 24 ≈ 22.36 cm²
👍🍻❤
Many thanks ❤️🌹
You only found one of two possible solutions.
You arrived at x=8, y=6. However, a valid solution also happens when x=6, y=8.
In that case, the yellow area is smaller than the unshaded white area to the right. Theta would be the complement of your 53.13deg.
Basically this second solution would visually ‘flip’ your solution around the line OB.
A quick calculation is to subtract the blue rectangle area + yellow region region from the total area of the quarter circle pi * 10 * 10 / 4.
Yes
Addition et multiplication son des operations commutatives donc : xy =yx et y^2+x^2 = x^2+y^2
I agree. The point where the original workings go wrong is where he jumps from (x-y)^2 = 4 to the conclusion that x -y = 2, thereby missing the other possible solution, x-y = -2.
Exactly. I was looking at the thumbnail and noticed that the diagram is symmetric about the diagonal of the rectangle, so there must be 2 solutions. Unless the rectangle turned out to be a square, in which case the 2 solutions are the same.
Nice, many thanks, Sir! For small θ → sin(θ) ≈ θ (radians) → no calculator needed:
φ = 30°; r = QO = PO = BO = 10; ∎ABCO → AO = BC = b → AP = r - b; CO = a → CQ = r - a
ab = 48 → b = 48/a → r^2 = 100 = a^2 + (48/a)^2 → a = 6 → b = 8
∆ BCO → COB = δ → sin(δ) = b/r = 4/5 → cos(δ) = a/r = 3/5
sin(3φ/2) = cos(3φ/2) = √2/2
sin(θ) = sin(δ - 3φ/2) = sin(δ)cos(3φ/2) - sin(3φ/2)cos(δ) = √2/10
6φ/π = rad ≈ 57,3° → (rad)(√2/10) = 8,1° → 8,1° + 3φ/2 = 53,1° → 100π(53,1°/360°) - 24 ≈ 22,34 🙂
Nice sharing 🌹👌🌹👌
Your solution assumes that (x-y) = 2 and not (x-y) = -2, which would have resulted, instead, with x = 6 and y = 8.
If we assume the diagram is not drawn to scale, as we are warned at 0:42, this second solution must also be considered. This means we should locate two possible solutions and not just one.
So I get the area is either A ~ 22.36 cm^2 or A ~ 8.176cm^2.
Given the usual warning about the scale of the diagram, let's make it clear that x > y.
Then note that 48 is 6 × 8, a perfect match to 10 for a 3 : 4 : 5 Pythagorean triangle.
So, with angles in radians, the yellow area will be ½ (10² arcsin ⅘ − 48) ≈ 22.36 cm².
Well explained.
Rectangle área :
A = x.y = 48 cm²
R sin α . R cos α = 48
sin α cos α = 48 / R²
½ sin(2α) = 48/10²
α = 36,8699° / 53,1301°
Area angular sector
A = ½.R².α
A = ½ R² 53,1301° π/180
A = 46,365 cm²
Area right triangle:
A = 48/2 = 24 cm²
Yellow shaded area:
A = A₂ - A₁
A = 22,365 cm² ( Solved √ )
Is not necessary to calculate 'x' and 'y', base and height of rectangle.
Your "action plan" in the beginning doesnt mention "sector area" or angles at all, just x and y.
So I solved it with brute for:
y=48/x
x²+(48/x)²=(10)²
...
x⁴-100x²+2304=0
x²=36 or 64
x=6 or 8
My only question was how do we assign which side 6 or 8?
I went with: by visual inspection, x>y.
Yay! I solved the problem, and pretty much the same way as in the video.
Back catalogue time :)
Diagonal (r) is 10, so rectangle sides must be 6 and 8.
By trigonometry,
22.364
Easy
The rectangle can be divided into a two 3-4-5 triangle with degrees 53.13. and 38.87
Since radius is 10, and the degree is 53.13, the area of the sector is
53.13/360 * 10^2. 3.14 = 46.34
subtract half of the retangle 48/2= 24 from the above
46.34- 24 =22.364
Thank you
Here's how I did it. For the blue rectangle, I used a hypotenuse of 10, a height of L, and a width of 48/L. I created triangle BOA. I called angle BOA Θ. I used the double angle formula, sin(2Θ) = 2sin(Θ)cos(Θ), to nicely eliminate the "L" variable and get sin(2Θ) = 73.74°. That was the main trick. My final answer was A = 22.36°.
I got x and y using bisecting chords theorem. (10+y) * (10-y) = x^2; we know xy=48 so substitute 48/y for x. Get 100-y^2 = 2304/y^2 ; substitute t for y^2 ; Get 100t - t^2 = 2304 ; Use quadratic formula to solve for t and get t=36, so y = 6. So x=8. I knew we'd have to find the area of sector OBQ, given angle theta. I got sin(theta) = 0.8. So inverse sine gets theta. I got stuck here because I didn't think we were allowed to use a calculator.
There is a way to solve it (approx.) without calculator, see above.
Amazing sir
Thanks and welcome ❤️
There are 2 solutions
Equation 4 x-y=-2 is also possible
Dai dati risulta b=8,h=6(Pitagora)...Ay=1/4πr^2-((r^2/2)arctg(h/b)+24)=25π-(50arctg6/8+24)=25π-24-50arctg3/4=22,36476
Let x be the width of the blue rectangle and y the height. then, invoking Pythagoras:
x^2 + y^2 = 10^2 = 100. BUT y = 48/x. Therefore, performing this substitution:
x^2 + 48^2/x^2 = 100; carrying out the square operation:
x^2 + 2304/x^2 = 100; multiplying through by x^2:
x^4 + 2304 = 100(x^2); rearranging:
x^4 -- 100(x^2) + 2304 = 0; solving for x^2 by the quadratic formula:
x^2 = (100 +/-- sqrt(100^2 -- (4)(2304)))/2; carrying out the arithmetic operations:
x^2 = (100 +/-- sqrt(784))/2 = (100 +/-- 28)/2
= either 128/2 = 64; or 72/2 = 36; taking the square roots, we find that x = 8 and y = 6.
Now consider the arc OQB. Angle OQB = arctan(8/6) = 53.13 degrees;
so the area of this pie slice is (100 pi)(53.13)/(360) = 14.76 pi.
Subtract half the area of the rectangle (24 square units) and we have 14.76 pi -- 24 = approximately 22.37 for the yellow region.
Veni, vidi, vici! 🤠
Or 8.17 square if you change x and y😊
It's 3 ,4 ,5 type rt triangle
let AB = a and OA = b. Draw radius OB.
Rectangle OABC:
A = OA(AB) = ab
ab = 48 ---- (1)
b = 48/a
Triangle ∆OAB:
AB² + OA² = OB²
a² + b² = 10² ---- (2)
a² + (48/a)² = 100
a² + 48²/a² - 100 = 0
a⁴ - 100a² + 48² = 0
a⁴ - 8²a² - 6²a² + 48² = 0
a²(a²-8²) - 6²(a²-8²) = 0
(a²-8²)(a²-6²) = 0
a² = 8² | a² = 6²
a = 8 ❌ | a = 6 a
b = 48/8 | b = 48/6
b = 6 ❌ | b = 8 a
Sector BOP:
sin(θ) = 6/10 = 3/5
θ = sin⁻¹(3/5) ≈ 36.87°
A = (θ/360)πr²
A ≈ (36.87/360)π10²
A ≈ 10.24π
By observation, the yellow area is the area of quarter circle O, minus rectangle OABC, minus sector BOP, plus triangle ∆OAB.
A ≈ πr²/4 - 48 - 10.24π + 24
A ≈ 25π - 24 - 10.24π
A ≈ 14.76π - 24 ≈ 22.36 cm²
Diagonal 10, area 48 - it's egyptian 6×8. 😉
Deux cas
53,13 puis 90 -53,13
Yellow region=22.36 square unit
Let's solve this problem:
.
..
...
....
.....
With a=AO=BC, b=AB=CO, A=A(blue) and R being the radius of the circle we have:
ab = A
a² + b² = R²
a² + (A/a)² = R²
a² + A²/a² = R²
a⁴ + A² = R²a²
a⁴ − R²a² + A² = 0
a² = R²/2 ± √(R⁴/4 − A²)
a² = (10cm)²/2 ± √((10cm)⁴/4 − (48cm²)²)
a² = 100cm²/2 ± √(10000cm⁴/4 − 2304cm⁴)
a² = 50cm² ± √(2500cm⁴ − 2304cm⁴)
a² = 50cm² ± √(196cm⁴)
a² = 50cm² ± 14cm²
⇒ a² = 64cm² ⇒ a = 8cm ⇒ b = 6cm
∨ a² = 36cm² ⇒ a = 6cm ⇒ b = 8cm
The size of the yellow area is the difference between the area of the circle sector OBQ and the triangle OBC:
A(yellow) = A(circle sector OBQ) − A(triangle OBC)
A(yellow) = πR²*(∠BOQ/2π) − ab/2
A(yellow) = R²*(∠BOQ/2) − ab/2
A(yellow) = R²*(∠BOC/2) − ab/2
A(yellow) = R²*(arctan(a/b)/2) − ab/2
First case: a = 8cm / b = 6cm
A(yellow) = (10cm)²*(arctan(8/6)/2) − (8cm)*(6cm)/2
A(yellow) = 50cm²*arctan(4/3) − 24cm²
A(yellow) ≈ 22.36cm²
Second case: a = 6cm / b = 8cm
A(yellow) = (10cm)²*(arctan(6/8)/2) − (6cm)*(8cm)/2
A(yellow) = 50cm²*arctan(3/4) − 24cm²
A(yellow) ≈ 8.175cm²
Best regards from Germany