Yes, chord DP and AB intersect. Let OA = OB = r (radius). AE = AB - EB = 2r - 2√5.. So, from intersecting chords DP and AB, (2r - 2√5)(2√5) = (8)(8). (4√5)r - 20 = 64. (4√5)r = 84 and r = (84)/(4√5) = 21/(√5) = (21√5)/5. as PreMath also found. PreMath could compute r directly by letting OD = r and OE = r - 2√5, without introducing variable b.
Sketch the whole circle. Extend DE to make a full chord. Use intersecting chord theorem to find half DE is 4. Using trig find angle B then join Cto A which forms a right angle triangle ACB. Use sine theorem to find AC and pythagoras to find AB, hence Radius is half AB.
A possible alternative: Label the angle ABC as alpha. FE=4 so sin alpha= 4/6= 2/3 so AC/AB= sin alpha= 2/3-----> AC /2R=2/3---> AC= 4R/3 so we have sq (2R)= sq 14 + sq(4R/3) ----> sq R= sq14 X9/ 20===> R= 21/sqrt5
Circle chord theorem: mirroring DE to the bottom, one has 4x as the whole chord, with division x and 3x, so x·3x = 6·8 → 3x² = 48 or x = 4. Then EB = √(6²-4²) = √20 = 2√5. Chord theorem again with (2r - 2√5)·2√5 = 8² = 64, hence 4r√5 - 20 = 64 or r = 84/(4√5) = 21/√5. It's good practices to rationalize that, so r = 21/√5 · √5/√5 = 21√5/5.
By the Law of Proportionality between Chords we have: Given the Point F as the point of intersection between the 2 chords. DF = EF = x 3x/8 = 6/x 3x^2 = 48 x^2 = 48/3 x^2 = 16 x = 4 Let angle CBA = alpha Let angle CAB = beta sin (alpha) = 4/6 = 2/3 Being C' the projection in diameter AB. EF/BF = CC'/14 CC' = 2/3 * 14 = 28/3 ~ 9,333 sin (beta) = CC'/AC AC = CC' / sin (beta) AC = 9,333 / sin (beta) AC ~ 12,522 Area of the Triangle Rectangle ([ABC] = (AC * BC) / 2 = (12,522 * 14) / 2 = 175,308 / 2 ~ 87,654 su 87,654 = (CC' * 2r) / 2 97,384 = 9,333 * 2r 175,308 / 9,333 = 2r 18,783 = 2r r ~ 18,783 / 2 r ~ 9,39 lu Answer: Radius equal ~ 9,39 lu
Extend the semicircle to a full circle and draw EG down from the diameter AB to intersect the new other side of the circumference at G, extending line segment DE into chord DG. Let DF = FE = x. By symmetry, EG = 2x. By Intersecting Chords Theorem: FC(FB) = FD(FG) 8(6) = x(3x) 48 = 3x² x² = 48/3 = 16 x = 4 Triangle ∆FEB: FE² + EB² = BF² 4² + EB² = 6² EB² = 36 - 16 = 20 EB = √20 = 2√5 As radii, OA = OB = r. As EB = 2√5, OE = r-2√5. We can now use Intersecting Chords Theorem again. DE(EG) = AE(EB) 8(8) = (2r-2√5)(2√5) 64 = 4r√5 - 20 4r√5 = 64+20 = 84 r = 84/4√5 = 21/√5 = 21√5/5
Connect C to A and you have a right triangle ABC similar to FBE. Then use proportions to compute AB which is 2r, divide that by 2 and you arrive at the same answer
When EB = 2.sqrt(5) is known we can finish using in the right triangle ADB the formula EA . EB = ED^2, So: (2R - 2.sqrt(5)). (2.sqrt(5)) = (4 +4)^2, or 4.sqrt(5).R - 20 = 64, or R = (64 + 20) / 4. sqrt(5), or R = 21 / sqrt(5) = (21. sqrt(5)) /5
Thank you so much for the tricky problem! I solved it almost the same as you , except for the last part. Because the triangle ADB is a right one so we can have: sq ED=AExEB or we can use chord theorem again.
Could you just apply the chord product rule a second time, once you had found ED, to solve for EB*AE=ED*its symmetrical other side? AE*EB=8*8. Then R = (AE+EB)/2.
Use law of intersecting chords: DF*3DF=6*8=48 yields DF=√16=4 Evaluate leg of right triangle: EB=√(6^2-4^2)=2√5 Use law of intersecting chords: EB*AE=(2DF)(2DF) yields AE=4*4^2/(2√5)=32√5/5 Radius R=(AE+EB)/2=(32√5/5+10√5/5)/2=21√5/5 units
@ 1:30 , Divergent Thinking is critical too the success of analyzing and evaluating this problem. If only Schrodinger's Cat had the same creative imagination. Meow! 🙂
Got it. Many thanks.
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Alternatively, once you have EB, you can use the intersecting chords theorem again, and write (2r - 2√5)•(2√5)=8•8, and the same result follows.
Yes, chord DP and AB intersect. Let OA = OB = r (radius). AE = AB - EB = 2r - 2√5.. So, from intersecting chords DP and AB, (2r - 2√5)(2√5) = (8)(8). (4√5)r - 20 = 64. (4√5)r = 84 and r = (84)/(4√5) = 21/(√5) = (21√5)/5. as PreMath also found.
PreMath could compute r directly by letting OD = r and OE = r - 2√5, without introducing variable b.
Thanks ❤️
А как ты пишешь здесь знак радикала? 😊
Sketch the whole circle. Extend DE to make a full chord. Use intersecting chord theorem to find half DE is 4. Using trig find angle B then join Cto A which forms a right angle triangle ACB. Use sine theorem to find AC and pythagoras to find AB, hence Radius is half AB.
Beautiful question!!
A possible alternative: Label the angle ABC as alpha. FE=4 so sin alpha= 4/6= 2/3
so AC/AB= sin alpha= 2/3-----> AC /2R=2/3---> AC= 4R/3
so we have sq (2R)= sq 14 + sq(4R/3) ----> sq R= sq14 X9/ 20===> R= 21/sqrt5
Circle chord theorem: mirroring DE to the bottom, one has 4x as the whole chord, with division x and 3x, so x·3x = 6·8 → 3x² = 48 or x = 4. Then EB = √(6²-4²) = √20 = 2√5.
Chord theorem again with (2r - 2√5)·2√5 = 8² = 64, hence 4r√5 - 20 = 64 or r = 84/(4√5) = 21/√5.
It's good practices to rationalize that, so r = 21/√5 · √5/√5 = 21√5/5.
Beautiful problem! 😊
My approach:
1) Intersecting chord theorem:
x * 3x = 8 * 6
3x² = 48
x² = 16
x = 4
2) Pythagoras:
x² + y² = 6²
y² + 16 = 36
y² = 20 = 4 * 5
y = 2√5
3) Intersecting chord theorem:
y * (d - y) = 2x * 2x = 4x²
2√5 (d - 2√5) = 4 * 16
2√5 d - 20 = 64
2√5 d = 84
d = 84 / (2√5) = 42/√5 = 42√5 / 5
4) Radius:
r = d/2 = 21√5 / 5 ≈ 9.39 units
P.S. I just see that my plan was quite similar to yours - I just used the intersecting chord theorem twice.
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▲ABC ~ ▲BFE. AB/BC = BF/BE. 2R/14 = 6/2√5. R = 21√5/5.
Thank you!
By the Law of Proportionality between Chords we have:
Given the Point F as the point of intersection between the 2 chords. DF = EF = x
3x/8 = 6/x
3x^2 = 48
x^2 = 48/3
x^2 = 16
x = 4
Let angle CBA = alpha
Let angle CAB = beta
sin (alpha) = 4/6 = 2/3
Being C' the projection in diameter AB.
EF/BF = CC'/14
CC' = 2/3 * 14 = 28/3 ~ 9,333
sin (beta) = CC'/AC
AC = CC' / sin (beta)
AC = 9,333 / sin (beta)
AC ~ 12,522
Area of the Triangle Rectangle ([ABC] = (AC * BC) / 2 = (12,522 * 14) / 2 = 175,308 / 2 ~ 87,654 su
87,654 = (CC' * 2r) / 2
97,384 = 9,333 * 2r
175,308 / 9,333 = 2r
18,783 = 2r
r ~ 18,783 / 2
r ~ 9,39 lu
Answer: Radius equal ~ 9,39 lu
Love it!!!!!!!!!!!!!!!!!!
Thanks ❤️
Extend the semicircle to a full circle and draw EG down from the diameter AB to intersect the new other side of the circumference at G, extending line segment DE into chord DG. Let DF = FE = x. By symmetry, EG = 2x.
By Intersecting Chords Theorem:
FC(FB) = FD(FG)
8(6) = x(3x)
48 = 3x²
x² = 48/3 = 16
x = 4
Triangle ∆FEB:
FE² + EB² = BF²
4² + EB² = 6²
EB² = 36 - 16 = 20
EB = √20 = 2√5
As radii, OA = OB = r. As EB = 2√5, OE = r-2√5. We can now use Intersecting Chords Theorem again.
DE(EG) = AE(EB)
8(8) = (2r-2√5)(2√5)
64 = 4r√5 - 20
4r√5 = 64+20 = 84
r = 84/4√5 = 21/√5 = 21√5/5
Connect C to A and you have a right triangle ABC similar to FBE. Then use proportions to compute AB which is 2r, divide that by 2 and you arrive at the same answer
Twice chrd theorem
x*3x = 8*6
x = 4
Then EB = 2sqrt(5) = a
Then second couple of chords
(2r - a)*a = 2x*2x = 8*8
When EB = 2.sqrt(5) is known we can finish using in the right triangle ADB the formula EA . EB = ED^2,
So: (2R - 2.sqrt(5)). (2.sqrt(5)) = (4 +4)^2, or 4.sqrt(5).R - 20 = 64, or R = (64 + 20) / 4. sqrt(5), or R = 21 / sqrt(5) = (21. sqrt(5)) /5
Imagining a full circle and intersecting chords, 3x^2 = 48.
x^2 = 16. x = 4
EB = sqrt(20)
Another bout of intersecting chords gives 8*8 = (sqrt(20)(2r - sqrt(20)
64 = 2r*sqrt(20) - 20
84 = 2r*sqrt(20)
21 = r*sqrt(5)
r = 21/sqrt(5)
Rationalise to r = (21*sqrt(5))/5
Decimal approx: 9.39
Thank you so much for the tricky problem!
I solved it almost the same as you , except for the last part. Because the triangle ADB is a right one so we can have: sq ED=AExEB or we can use chord theorem again.
Excellent!
You are very welcome!
Thanks ❤️
Could you just apply the chord product rule a second time, once you had found ED, to solve for EB*AE=ED*its symmetrical other side? AE*EB=8*8. Then R = (AE+EB)/2.
Use law of intersecting chords: DF*3DF=6*8=48 yields DF=√16=4
Evaluate leg of right triangle: EB=√(6^2-4^2)=2√5
Use law of intersecting chords: EB*AE=(2DF)(2DF) yields AE=4*4^2/(2√5)=32√5/5
Radius R=(AE+EB)/2=(32√5/5+10√5/5)/2=21√5/5 units
Good night sir
Good night, dear ❤️
Мой ответ : (16/V5)+V5 ~~9.39
*V - sqrt
Posto a =FE,α=CBO.,.le relazioni usate sono a=6sinα,rcosα=7...r^2=(2a)^2+(r-6cosα)^2...facendo le opportune sostituzioni risulta r^2=49*9/5
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😅 a real challenge puzzle. 3a^2=6 8, a=4, EB=2sqrt(5), r^2=64+(r-2sqrt(5))^2=84-4sqrt(5)r+r^2, 84=4sqrt(5)r, r=21/sqrt(5)=21/5 sqrt(5).😊
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R=21√5/5 units.❤❤❤thanks
Excellent!
You are very welcome!
Thanks ❤️
DE = EF + DF = a + a → DK = DE + EK = 2a + 2a; BC = BF + CF = 6 + 8 = 14 = BT + CT = 7 + 7 → sin(BTO) = 1
8(6) = a(3a) → a = 4 → sin(θ) = a/6 = 2/3 → cos(θ) = √(1 - sin^2(θ)) = √5/3 = 7/r → r = 21√5/5
or: ∆ BTO = ∆ TCO → sin(BTO) = 1; BT = CT = 7; TBO = TCO = θ → TOB = COT = δ → sin(θ) = cos(δ) = 2/3 →
cos(θ) = sin(δ) = √5/3 → cos(2δ) = 1 - 2sin^2(δ) = -1/9 → ∆ BCO → 14(14) = 2r^2(1 - cos(2δ)) → r = 21√5/5
or fast lane: sin(θ) = 2/3 → cos(θ) = √5/3 = 14/2r → r = 21√5/5
Thanks ❤️
@ 1:30 , Divergent Thinking is critical too the success of analyzing and evaluating this problem. If only Schrodinger's Cat had the same creative imagination. Meow! 🙂
Excellent!
Thanks ❤️