line BC is tangent to circle A at B and to circle D at C (not drawn to scale) AB=11, BC=23,and DC=2. find AD to the nearest tenth.
Image: i.imgur(dot)com/f4ThxTu.jpg
To find AD, we need to use the properties of tangents and secants of circles.
We can start by finding the length of AB. The length of AB is given as 11 units.
Next, we need to find the length of BD. Since line BC is tangent to circle D at C, the line segment BC is perpendicular to the radius of circle D drawn to the point of tangency (C). Therefore, triangle BCD is a right triangle, and BC is its hypotenuse. The length of BC is given as 23 units.
Using the Pythagorean theorem, we can calculate the length of BD:
BD^2 = BC^2 - CD^2
BD^2 = 23^2 - 2^2
BD^2 = 529 - 4
BD^2 = 525
Taking the square root of both sides, we find:
BD ≈ √525
BD ≈ 22.9 (rounded to the nearest tenth)
Now, we can find the length of AD. Since line BD is tangent to circle A at B, we can form a right triangle ABD. AD is the hypotenuse of this right triangle, and AB and BD are the legs.
Using the Pythagorean theorem:
AD^2 = AB^2 + BD^2
AD^2 = 11^2 + 22.9^2
AD^2 = 121 + 525.21
AD^2 = 646.21
Taking the square root of both sides, we find:
AD ≈ √646.21
AD ≈ 25.4 (rounded to the nearest tenth)
Therefore, AD is approximately 25.4 units to the nearest tenth.
In the given diagram, we have two circles - Circle A and Circle D, and line BC is tangent to both circles at points B and C respectively.
We are given the lengths AB = 11, BC = 23, and DC = 2.
To find AD, we can use the property that the tangents drawn from an external point to a circle are equal in length. Therefore, we can consider triangle ABC and triangle DBC.
In triangle ABC,
AB = 11
BC = 23
In triangle DBC,
BC = 23
CD = 2
Since BC is the common side, triangle ABC and triangle DBC are right-angled triangles.
Using Pythagoras' theorem in triangle ABC:
AC^2 = AB^2 + BC^2
AC^2 = 11^2 + 23^2
AC^2 = 121 + 529
AC^2 = 650
Taking the square root of both sides:
AC ≈ √650
AC ≈ 25.5 (to the nearest tenth)
Similarly, using Pythagoras' theorem in triangle DBC:
DC^2 = BC^2 + BD^2
2^2 = 23^2 + BD^2
4 = 529 + BD^2
529 + BD^2 = 4
Subtracting 529 from both sides:
BD^2 = -525
Since BD^2 cannot be negative, there is no real solution for BD.
Now, to find AD, we can add AC and DC:
AD = AC + DC
AD ≈ 25.5 + 2
AD ≈ 27.5 (to the nearest tenth)
Therefore, AD is approximately 27.5 (to the nearest tenth).
I looked at your diagram and noticed that you sketched a line from D to AB
That was good, now let the point of contact be E
so clearly triangle ADE is right-angled
with AE = 9 and DE = 23
by Pythagoras:
AD^2 = 9^2 + 23^2 = 610
AD = √610 = appr 24.7