The solution can be divided into 2 parts: The first part is to find radius of the semicircle by Pythagoras theorem (method 1) or tangent-secant theorem (method 2) with construction of line DO as radius of semicircle to point of tangency. The second part is to find x when radius is known by equal tangent theorem plus Pythagoras theorem (your method) or by similar triangle corresponding side proportionality equation as follows: Triangle BDO is similar to triangle BCA (AAA) then CA/DO= BC/BD. Then X = CA = (3 + 2R)(R)/5 Substitute R = 8/3 from part one, X = 40/9.
I think that the second method is a much more simpified version of the first method. I will have to test out my idea first and then upload the picture proof in my follow up comment.
Potencia de B respecto al semicírculo =5²=3*(3+2r)→ r=8/3 → Si "O" es el centro del semicírculo, la razón de semejanza entre los triángulos rectángulos ACB y ODB es: s=(3 +8/3 +8/3)/5=(25/3)/5=5/3→ AC =OD5/3=(8/3)(5/3)=40/9. Gracias y saludos.
Let O be the center of the semicircle. Let r be the radius of semicircle O. First method: Draw OD. As OD is a radius of semicircle O and AB is tangent to semicircle O at D, ∠ODB = 90°. Triangle ∆ODB: OD² + DB² = BO² r² + 5² = (r+3)² r² + 25 = r² + 6r + 9 6r = 25 - 9 = 16 r = 16/6 = 8/3 CA and AD are tangents to semicircle O that intersect at A, therefore CA = AD = x. Triangle ∆BCA: CA² + BC² = AD² x² + (3+16/3)² = (x+5)² x² + (25/3)² = x² + 10x + 25 10x = 625/9 - 25 = 400/9 x = (400/9)/10 = 40/9 units Second method: Same as above, up until r is found to be 8/3. Let ∠ABC α and ∠CAB = β, where β is the complementary angle to α, as α + β = 90°. As ∠DBO = α and ∠ODB is a right angle, ∠BOD = β. Therefore ∆ODB and ∆BCA are similar. Triangle ∆BCA: CA/BC = OD/DB x/(3+2r) = r/5 x = (3+2r)r/5 = (2r²+3r)/5 x = (2(8/3)²+3(8/3))/5 x = (2(64/9)+8)/5 x = (128/9+72/9)/5 x = (200/9)/5 = 40/9 units
40/9? 1. Label the center of the semicircle as O, the length of its radius as R, and construct line segment DO (of length R). 2. BD² + DO² = BO² (Pythagoras), or 5² + R² = (3 + R)², 25 + R² = 9 + 6R + R², 16 = 6R, R = 8/3 2.1 AD = AC (tangents to a circle) 3. BC² + AC² = AB² (Pythagoras), or (2R + 3)² + X² = (X + 5)², substitute R to get: (16/3 + 3)² + X² = X² + 10X + 25, (25/3)² = 10X + 25, 625/9 - 25 = 10X, 400/9 = 10X, 40/9 = X Now to watch the video. - s.west
