A Very Nice Geometry Problem | 4 Different Methods to Solve

A Very Nice Geometry Problem | 4 Different Methods to Solve
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[@skwest]

Unfortunately, your placement of the R segment in the thumbnail gave a big clue to the solution.
Anyway... tangents from a point, and Pythagoras gives the triangle's sides as (5,12,13). Similar triangles (using the aforementioned R segment) yields the ratio:
5:13::R:(12-R)
So, 13R = 5(12-R)
13R = 60 - 5R
18R = 60
R = 60/18 = 3-1/3
Thanks!

[@quigonkenny]

CA and AD are tangents to the semicircle that intersect at A, so AD = CA = 5. Thus AB = 5+8 = 13.
Triangle ∆BCA:
CA² + BC² = AB²
5² + BC² = 13²
BC² = 169 - 25 = 144
BC = √144 = 12
Let O be the center of the semicircle. Draw OD. Since AB is tangent to semicircle O at D, ∠ODB = 90°. As ∠BCA = ∠ODB and ∠ABC = ∠DBO, ∆ODB and ∆BCA are similar triangles.
OD/DB = CA/BC
r/8 = 5/12
r = 8(5/12) = 10/3 units

[@danielc.martin1574]

Cool!

[@murvetaykac7041]

There is no adding anything .Connect D with Central.BC=12 .İn tringle BDO used Paytagoras formula 8^2+R^2=(12- R )^2 .We can find R=10/3.thank you very much for your methods.

[@devondevon4366]

Answer 10/3 or 3.3333
DA=AC =5 tangent circle theorem
Hence, AB =13 ( 8+5)
Hence, BC =12 Pythaogrean Theorem
Draw the radius (R) from the center of the circle O to D (the point of tangency)
forming a triangle A B O. This line
is perpendicular in which BD and DO are the legs of triangle ABO and , BO is the hypotenuse with
the side 12 - R
Using Pythagorean with the sides R , 8 and 12- R
Hence (12- R)^2 = R^2 + 8^2
144 + R^2 - 24 R = R^2 + 64
144- 64 = 24 R-+R^2 - R^2
80 = 24 R
10 = 3 R (divide both sides by 8)
10/3 = R
or 3.333

[@prossvay8744]

r=10/3 units

[@josealmariferreiraferrreir3374]

Shooow

[@jan-willemreens9010]

... Placing this problem in a Cartesian grid with point B as the origin (0, 0) ... I DA I = I CA I = 5 ... I BC I = SQRT( (8 + 5)^2 - 5^2 ) = 12 (5 - 12 - 13 triplet) ... name intersection point of perpendicular from point D on BC point E ... SIN(Angle B) = 5/13 = I DE I / 8 ... I DE I = 40/13 ... TAN (Angle B) = I DE I / I BE I = 5/12 ... I BE I = (40/13) * 12 / 5 = 96/13 ... so point D is ( 96/13, 40/13 ) ... slope of line through B & A is 5/12 ... create equation of line perpendicular to line BA going through point D ... (1) Slope is - 12/5 & point D (96/13, 40/13) ... so equation of line is Y = - 12/5 * X + 1352/65 ... Center of semicircle M is where Y = 0 ... 12/5 * X = 1352/65 ... X = 26/3 .... finally to find the radius of semicircle we only have to find the distance between M and D applying the Distance Formula ... d(M, D) = 10/3 = 3 1/3 ... thanking you for your clear presentation sir .... this exercise was quite a bit of basic trig. , algebra , and geometry simultaneously, but also fun .... best regards, Jan-W

[@amitavadasgupta6985]

R=6 unit.

[@soli9mana-soli4953]

There would also be a 5th method:
tracing the line OA we get 2 right triangles congruent ADO and AOC, then line AO is the bisector of angle in A, so we can apply the bisector theorem in this way:
AC/AB = OC/OB
5/13 = r/(12 -r)
5*(12 - r) = 13r
r = 10/3

[@leonardonabinger]

That is solved by head.

[@yakupbuyankara5903]

R=10/3

[@alexanderthemeek]

These slow talk videos make me wish UA-cam lets 4x speed

[@comdo777]

answer=56 isit

[@alexniklas8777]

R=10/3