Find the radius of the semicircle | A Nice Geometry Problem

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[@marioalb9726]

Segment AC= 'a' ; Segment BC= 'b'
Similarity of triangles:
b/10=3/R
b² = 900/R²
Pytagorean theorem:
a² + b² = (2R)²
b² = 4R² - 10²
Equalling :
900/R² = 4R² - 100
4R⁴ - 100R² - 900 = 0
R⁴ - 25R² - 225 = 0
R = 5,659 cm ( Solved √ )

[@user-xz7sv2dn9g]

Very good solved. More easy thsn Our " teacher"

[@marioalb9726]

@@user-xz7sv2dn9g
Thanks for your comment
See below the easiest solution:
cosα = R/c = 10/(2R) --> R² = 5c
Pytagorean theorem:
c² = R² + 3² = 5c + 9
c² - 5c - 9 = 0 --> c= 6,4051cm
R = 6,659 cm ( Solved √ )

[@raghvendrasingh1289]

Nice problem
In triangle AOP
tan A = 3/x
In triangle ACB
sec A = 2x/10 = x/5
eliminating A
x^2/25 - 9/( x^2) = 1
x^4- 25 x^2- 225 =9
rest follows at once.

[@santiagoarosam430]

M es el punto medio de AC ; OM=a ---> 5/r=a/3---> a=15/r ---> 5²=(r-a)(r+a)---> r=√[(25+5√61)/2] =5,66.
Gracias y saludos

[@marioalb9726]

cosα = 10/(2R) ; cos²α = 25/R²
Intersecting chords theorem:
(R+3cosα)(R-3cosα) = (½10)²
R² - 9.cos²α = 5²
R² - 9.(25/R²) = 25
R⁴ - 25R² - 225 = 0
R = 5,659 cm ( Solved √ )

[@marioalb9726]

cosα = R/c = 10/(2R) --> R² = 5c
Pytagorean theorem:
c² = R² + 3² = 5c + 9
c² - 5c - 9 = 0 --> c= 6,4051cm
R = 6,659 cm ( Solved √ )

[@xz1891]

R^2-25=9-[root(9+R^2)-25]^2

[@soli9mana-soli4953]

extending the line AC in the upper part until it intersects the perpendicular to AB from B, at the intersection point D, we obtain the right-angled triangle ABD which has sides twice the sides of AOP. Therefore if OP = 3 it follows that DB = 6
Let DC = x and apply the tangent-secant theorem from the external point D
6² = (10+x)*x
x² + 10x - 36 = 0
x = √ 61 - 5
AD = 10 + √ 61 - 5 = 5 + √ 61
we apply the Pythagorean theorem on ABD
(2r)² = (5 + √ 61)² - 6²
r² = (25 + 5√ 61)/2

[@marioalb9726]

cosα= 10/(2R) ; tanα= 3/R
R = 5/cosα = 3/tanα
5/3 sinα = cos²α = 1 - sin²α
sin²α + 5/3 sinα - 1 = 0
sin α = 0,46837 --> α = 27,929°
R = 5,659 cm ( Solved √ )

[@ДмитрийИвашкевич-я8т]

AP=x.
R^2+9=x^2
∆APO~∆ACB
10/(2R)=R/x
R^2=5x
x^2-9=5x
x=(5+√61)/2
R^2=5x=(25+5√61)/2

[@michaeldoerr5810]

The answer is x = sqrt((25+5*sqrt(61))/2). Also I really have to say that it would be nice if there was a mnemonic device for HL similarity that involve the 90° and the common angle because I have always noticed that the similarity rules are simply the first two letters of the similar triangles are ratioed and then the first and last letters of the similar triangles that get ratioed.

[@SGuerra]

A questão é muito boa. Parabéns.

[@wes9627]

x/3=10/BC and 4x^2=100+BC^2=100+900/x^2.
Thus, 4x^4-100x^2-900=0 or x^4-25x^2-225=0;
x^2=(25+5√61)/2 and x=√[(25+5√61)/2]=5.65912... units.

[@quigonkenny]

Draw BC. By Thales' Theorem, as A and B are ends of a diameter and C is a point on the circumference of semicircle O, then ∠BCA = 90°. As ∠BCA = ∠AOP = 90° and ∠A is common, ∆BCA and ∆AOP are similar triangles.
Triangle ∆AOP:
OA² + OP² = PA²
x² + 3² = PA²
PA² = x² + 9
PA = √(x²+9)
Triangle ∆BCA:
BC² + CA² = AB²
BC² + 10² = (2x)²
BC² + 100 = 4x²
BC² = 4x² - 100
BC = √(4x²-100)
BC/AB = OP/PA
√(4x²-100)/2x = 3/√(x²+9)
√(4x²-100)√(x²+9) = 6x
(4x²-100)(x²+9) = 36x²
4x⁴ + 36x² - 100x² - 900 - 36x² = 0
4x⁴ -100x² - 900 = 0
x⁴ - 25x² - 225 = 0
x² = [-(-25)±√(-25)²-4(1)(-225)]/2(1)
x² = 25/2 ± √(625-900)/2
x² = 25/2 ± √1525/2 = (25±5√61)/2
x² = (25+5√61)/2 | x² = (25-5√61)/2 ❌ x² > 0
x = √((25+5√61)/2)
x = √(50+10√61)/2 ≈ 5.659 units

[@RealQinnMalloryu4]

(10)^2 (3)^2={100+9}=109 180°ABCPO/109=1.71ABCPO 1.71^1 1.71 (ABCPO ➖ 71ABCPO+1).

[@cyruschang1904]

cosθ = 10/2r = 5/r
(sinθ)/(cosθ) = 3/r = (sinθ)r/5
(sinθ) = 15/r^2
(5/r)^2 + (15/r^2)^2 = 1
25/r^2 + 225/r^4 = 1
r^4 - 25r^2 - 225 = 0
let r^2 = x
x^2 - 25x - 225 = 0
x = (25 + 5✓61)/2
r = ✓(50 + 10✓61)/2

[@prossvay8744]

3/R=BC/10
BC=30/R
100+(30/R)^2=(2R)^2
So R=5.66 units.

[@giuseppemalaguti435]

2rcosθ=10..3/r=tgθ...rcosθ=5..r^2=25(secθ)^2=25(1+9/r^2)..r^4=25(r^2+9)..r^2=(25+√1525)/2=(25+5√61)/2

[@jussikunnari9187]

Square root 91

[@mikefoehr235]

Rad= root of 91