Let R is a Radius of sector quarter Connect O to B In ∆ OBC OB^2=OC^2+BC^2 R^2=(R-14)^2+(R-7)^2 R=35; R=7 rejected So Pink area=1/4(π)(35^2)-(21)(28)=49(25π/4-12)=374.11 square units.
Guess I'm gonna learn how to play the banjo. What I'm told it's like learning math. The only way to learn math is too do math. This problem is insightful exposing a little joy on applying Theorems like Pythagorean or Algebra...🙂
In triangle OCB: R^2 = (R - 14)^2 + (R - 7)^2, with R the radius of the quater circle. Then R^2 - 42.R + 245 = 0. Deltaprime = 21^2 - 245 = 196 R = 21 + 14 = 35 or R = 21 - 14 = 7 (rejected) So the area of the rectangle OABC is 21.28 = 588 The area of the quater circle is (Pi/4).(35^2) = (1225/4).Pi Finally the pink area is (1225/4).Pi - 588. (Very easy)
As OABC is a rectangle, OC = BA, OA = CB, and ∠AOC = ∠OCB = ∠CBA = ∠BAO = 90°. Let r be the radius of quarter circle O. Draw radius OB. Triangle ∆OCB: OC² + CB² = OB² (r-14)² + (r-7)² = r² r² - 28r + 196 + r² - 14r + 49 = r² r² - 42r + 245 = 0 (r-35)(r-7) = 0 r = 35 | r = 7 ❌ --- r > 14 The pink area is equal to the area of quarter circle O minus the area of rectangle OABC. Pink area: Aᴘ = πr²/4 - hw Aᴘ = π(35)²/4 - (35-7)(35-14) Aᴘ = 1225π/4 - 28(21) Aᴘ = 1225π/4 - 588 ≈ 374.11 sq units
306.25 pi - 588 or 374.12 Draw a diagonal inside the rectangle. This the radius Let the radius = r, then r^2 = (r - 14)^2 + (r -7)^2 ........................... ......................... 0 = r^2 -42r + 245 0 = (r-35)(r-7) r = 35 Hence, the radius of the quarter circle = 35, and its area = 35^2 pi *1/4 = 306.25 pi Let's find the length and width of the rectangle. 35 - 7 = 28 35-14 = 21 Hence, its area = 588 Hence, area of purple region = 306.25pi - 588 or 374.12
Let's find the area: . .. ... .... ..... The triangle OAB is a right triangle, so we can apply the Pythagorean theorem. With r being the radius of the quarter circle we obtain: OB² = OA² + AB² = OA² + OC² r² = (r − 7)² + (r − 14)² r² = r² − 14*r + 49 + r² − 28*r + 196 0 = r² − 42*r + 245 r = 21 ± √(21² − 245) = 21 ± √(441 − 245) = 21 ± √196 = 21 ± 14 According to r−14>0 ⇒ r>14 the only useful solution is r=21+14=35. Now we are able to calculate the area of the pink region: A(pink) = A(quarter circle) − A(rectangle) = πr²/4 − OA*OC = πr²/4 − (r − 7)*(r − 14) = π*35²/4 − (35 − 7)*(35 − 14) = (1225/4)π − 28*21 = (1225/4)π − 588 ≈ 374.11
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Nice! r^2 = (r - 7)^2 + (r - 14)^2 → r = 35 → pink shaded area = (49/4)(25π - 48)
btw: ∆ 𝑂𝐵𝐶 = 𝑝𝑦𝑡ℎ. 𝑡𝑟𝑖𝑝𝑙𝑒 = 7(3 − 4 − 5); 𝑂𝐵𝐶 = 𝛿 → sin(𝛿) = 4/5
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Make some videos on AP GP series, challenging problems
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Let R is a Radius of sector quarter
Connect O to B
In ∆ OBC
OB^2=OC^2+BC^2
R^2=(R-14)^2+(R-7)^2
R=35; R=7 rejected
So Pink area=1/4(π)(35^2)-(21)(28)=49(25π/4-12)=374.11 square units.
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r^2=(r-14)^2+(r-7)^2...r^2-42r+245=0..r=21+14=35...r=21-14=7(no)...Apurple=π1225/4-(35-14)(35-7)=π1225/4-588=374,11
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I got this one!
OB=r (r-7)²+(r-14)²=r² r²-14r+49+r²-28r+196=r²
r²-42r+245=0 (r-7)(r-35)=0 r>14 , r=35
AB=OC=35-14=21 AO=BC=35-7=28
Pink area = 35*35*π*1/4 - 21*28 = 1225π/4 - 588
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306.25p -588 or 374.12
Got it!
Guess I'm gonna learn how to play the banjo. What I'm told it's like learning math. The only way to learn math is too do math. This problem is insightful exposing a little joy on applying Theorems like Pythagorean or Algebra...🙂
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S=(1225π-2352)/4≈374,5
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I've seen this pattern enough to instantly recognize the 3-4-5 Pythagorean triangle.
(r-14)^2 + (r-7)^2 = r^2
r^2 - 28r + 196 + r^2 - 14r + 49 = r^2
2r^2 - 42r + 245 = r^2
r^2 - 42r + 245 = 0
(42+or-sqrt(1764 - 4*1*245))/2 = r
(42+or-sqrt(784))/2
42+or-2*sqrt(196))/2 = r
(42+or-28)/2 = r
Retain the + result for r=35
Rectangle is 28*21 = 588
Full quadrant area would be 35^2 * pi= 962.115
962 - 588 = 374 + 0.115
374.115 un^2
In triangle OCB: R^2 = (R - 14)^2 + (R - 7)^2, with R the radius of the quater circle.
Then R^2 - 42.R + 245 = 0. Deltaprime = 21^2 - 245 = 196
R = 21 + 14 = 35 or R = 21 - 14 = 7 (rejected)
So the area of the rectangle OABC is 21.28 = 588
The area of the quater circle is (Pi/4).(35^2) = (1225/4).Pi
Finally the pink area is (1225/4).Pi - 588.
(Very easy)
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374.5 .Sq.units
Better solve a system of equations for OA and OC. Then you come to a quadratic equation x^2-28x=0 which solves itself at once.
As OABC is a rectangle, OC = BA, OA = CB, and ∠AOC = ∠OCB = ∠CBA = ∠BAO = 90°. Let r be the radius of quarter circle O. Draw radius OB.
Triangle ∆OCB:
OC² + CB² = OB²
(r-14)² + (r-7)² = r²
r² - 28r + 196 + r² - 14r + 49 = r²
r² - 42r + 245 = 0
(r-35)(r-7) = 0
r = 35 | r = 7 ❌ --- r > 14
The pink area is equal to the area of quarter circle O minus the area of rectangle OABC.
Pink area:
Aᴘ = πr²/4 - hw
Aᴘ = π(35)²/4 - (35-7)(35-14)
Aᴘ = 1225π/4 - 28(21)
Aᴘ = 1225π/4 - 588 ≈ 374.11 sq units
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306.25 pi - 588 or 374.12
Draw a diagonal inside the rectangle. This the radius
Let the radius = r, then
r^2 = (r - 14)^2 + (r -7)^2
...........................
.........................
0 = r^2 -42r + 245
0 = (r-35)(r-7)
r = 35
Hence, the radius of the quarter circle = 35,
and its area = 35^2 pi *1/4 = 306.25 pi
Let's find the length and width of the rectangle.
35 - 7 = 28
35-14 = 21
Hence, its area = 588
Hence, area of purple region = 306.25pi - 588 or 374.12
374.5 Sq.nuits
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My way of solution ▶
the radius of this quarter circle is r
[OC]=[BA]= r-14
[OA]= [CB]= r-7
[OB]= r
according to the Pythagorean theorem for the right triangle ΔOCB :
[OB]²= [OC]²+[CB]²
r²= (r-14)²+(r-7)²
r²= r²-28r+196 + r²-14r+49
r²-42r+245 = 0
Δ= 42²-4*1*245
Δ= 784
√Δ= √2²*14²
√Δ= 28
r₁= (42+28)/2
r₁= 35
r₂= (42-28)/2
r₂= 7
if r= 7, r-14 < 0 ❌
⇒
r= 35 length units
a= r-14
a= 35-14
a= 21
b= 35-7
b= 28
Arectangle= a*b
Arectangle= 21*28
Arectangle= 588 square units
Apink= πr²/4 - Arectangle
Apink= π*35²/4 - 588
Apink= (1225 π - 2352)/4
Apink ≈ 374,11 square units
Solution:
a + 14 = b + 7 = r
b - a = 7
a² + b² = r²
a² + b² = (b + 7)²
a² + b² = b² + 14b + 49
a² - 14 (a + 7) - 49 = 0
a² - 14a - 98 - 49 = 0
a² - 14a - 147 = 0
a = 14 ± 28/2
a = 21 Accepted
a = -7 Rejected
a = 21
b = 28
r = 35
A = π (35)²/4 - (21.28)
A = 1225π/4 - 588
A = 306,25π - 588
Square Units
Or
A ~= 374,1 Square Units
Let's find the area:
.
..
...
....
.....
The triangle OAB is a right triangle, so we can apply the Pythagorean theorem. With r being the radius of the quarter circle we obtain:
OB² = OA² + AB² = OA² + OC²
r² = (r − 7)² + (r − 14)²
r² = r² − 14*r + 49 + r² − 28*r + 196
0 = r² − 42*r + 245
r = 21 ± √(21² − 245) = 21 ± √(441 − 245) = 21 ± √196 = 21 ± 14
According to r−14>0 ⇒ r>14 the only useful solution is r=21+14=35. Now we are able to calculate the area of the pink region:
A(pink)
= A(quarter circle) − A(rectangle)
= πr²/4 − OA*OC
= πr²/4 − (r − 7)*(r − 14)
= π*35²/4 − (35 − 7)*(35 − 14)
= (1225/4)π − 28*21
= (1225/4)π − 588
≈ 374.11
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) OB = Radius = R lin un
02) OA = (R - 7) li un
03) OC = (R - 14) li un
04) (R - 7)^2 + (R - 14)^2 = R^2
05) R^2 - 14R + 49 + R^2 - 28R + 196 = R^2
06) 2R^2 - 42R + 245 = 0
07) Quadratic Equation with Two Positive Integer Solutions : R = 7 and R = 35
08) R = 35, because R > 14
09) Quarter of Circle Area = 35^2 * Pi / 4 ; QCA = 1.225Pi / 4 ; QCA = 306,25Pi sq un QCA ~ 962 sq un
10) Rectangle [ABCO] Area = (35 - 7) * (35 - 14) = 28 * 21 = 588 sq un
11) Pink Area = 962 - 588 ; Pink Area = 374 sq un
Therefore,
OUR ANSWER :
Pink Area approx. equal to 374 Square Units.
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