+29 Points A , B, and C are drawn, with the following:
AB = 6
AC = 16
BC = 10
Point B is on AC
Semicircles are constructed on AB, AC, and BC, and another circle is constructed such that it is tangent to all three semicircles. Find the radius of the circle.
+1418 Let's label the center of the circle O. We can see that triangle ABC is a right triangle with angle B being the right angle.
First, let's find the area of triangle ABC using Heron's formula:
s = (AB + BC + CA) / 2
s = (6 + 10 + 16) / 2
s = 16
Area of triangle ABC = sqrt (s(s-AB)(s-AC)(s-BC))
Area of triangle ABC = sqrt (16(16-6)(16-16)(16-10))
Area of triangle ABC = sqrt (16 * 10 * 6 * 6)
Area of triangle ABC = sqrt (5760)
Area of triangle ABC = 24
Since the radius of the circle is the same as the height of triangle ABC drawn from point O, and the area of a triangle is equal to 0.5 * base * height, we can find the height by dividing the area by the base:
Height = 2 * Area / AB
Height = 2 * 24 / 6
Height = 8
Now, we have a right triangle OBC with OB as the radius of the circle, BC as the base, and height as 8. Using the Pythagorean theorem:
OB^2 = BC^2 + Height^2
OB^2 = 10^2 + 8^2
OB^2 = 100 + 64
OB^2 = 164
Therefore, the radius of the semicircle is sqrt(164) = 2*sqrt(41).
+29 Sorry, but I think you missed the fact that B is on AC. Perhaps you could try again?
+677 Since B is on segment AC, we can divide triangle ABC into two right triangles: right triangle ABI and right triangle BCI.
Find the lengths of AI and CI:
We know AB = 6 and BC = 10. Since the semi-circle is constructed on AB, diameter AB is twice the radius of the semi-circle, so the radius of the semi-circle on AB is 6 / 2 = 3. This radius (3) is also the height of right triangle ABI with hypotenuse AB (6) and base AI. Using the Pythagorean Theorem on right triangle ABI:
AI^2 + 3^2 = 6^2 AI^2 = 27 AI = √27 = 3√3
Similarly, the radius of the semi-circle on BC is 10 / 2 = 5. This is the height of right triangle BCI with hypotenuse BC (10) and base CI. Using the Pythagorean Theorem:
CI^2 + 5^2 = 10^2 CI^2 = 75 CI = √75 = 5√3
Radius of the Tangent Circle:
Since the circle is tangent to all three semi-circles, the center of this circle must be at the intersection point of the angle bisectors of triangle ABC. In a right triangle, the angle bisector of the right angle coincides with the median drawn to the hypotenuse, which divides the hypotenuse into two segments with a 1:1 ratio. Therefore, point I, the intersection of AI and CI, is the center of the tangent circle.
We now have a right triangle with legs AI = 3√3 and CI = 5√3, and the hypotenuse is the diameter of the tangent circle (which is twice the radius). Using the Pythagorean Theorem:
(diameter of tangent circle)^2 = (3√3)^2 + (5√3)^2 (diameter of tangent circle)^2 = 48 diameter of tangent circle = √48 = 4√3
Since the diameter is twice the radius, the radius of the tangent circle is:
radius = diameter / 2 = (4√3) / 2 = 2√3
Therefore, the radius of the circle tangent to all three semi-circles is 2√3.
+129771 Let the center of the circle = O
Let the midpoint of the large semi-circle = F
Let the midpoint of the the right semi-circle = D
Let the center of the semi-circle on the left = E
Form triangles EOF and DOF
DO = (5 + r)
FO = (8 - r)
EO = (3 + r)\
DF = 3
EF = 5
By the Law of Cosines we have two equations
EO^2 = EF^2 + FO^2 - 2(EF * FO) cos (EFO)
DO^2 = DF^2 + FO^2 - 2(DF * FO) cos ( DFO)
Because angles EFO and DFO are supplementary cos (DFO) = - cos (EFO)
So re-writing....we have
EO^2 = EF^2 + FO^2 - 2(EF * FO) cos (EFO)
DO^2 = DF^2 + FO^2 - 2(DF * FO) -cos ( EFO)
EO^2 = EF^2 + FO^2 - 2(EF * FO) cos (EFO)
DO^2 = DF^2 + FO^2 + 2(DF * FO) cos ( EFO)
(3+r)^2 = 5^2 +(8-r)^2 - 2 (5 * (8-r)) cos (EFO)
(5+ r)^2 = 3^2 + (8-r)^2 + 2( 3 (8-r)) cos EFO
cos EFO = [ (3 + r)^2 - 25 - (8-r)^2] /[ -2 * (40 - 5r) ]
cos EFO = [ (5 + r)^2 - 9 - (8-r)^2] / [ 2 *(24 - 3r) ]
Then
[ (3 + r)^2 - 25 - (8-r)^2] /[ -2 * (40 - 5r) ] = [ (5 + r)^2 - 9 - (8-r)^2] / [2 *(24 - 3r) ]
This is a little sticky to simplify, but WolframAlpha helps
We get that
272 / (r-8) + 49 = 0
272 / (r -8) = -49
272 = -49 ( r -8)
272 = -49r + 392
-120 =-49r
r = 120 / 49 ≈ 2.449
