TRIANGLE ACE IS A CIRCUMSCRIBED TRIANGLE. POINTS B,D AND F ARE POINTS OF TANGENCY . IF AC= 16 AND AE= 9 CM AND CE = 15 CM FIND AD, FE, AND CD.
Thank you very mych
How do you find 4,5, and 11 though?
You don't explain the labeling very well, but I get that
AC is divided into 5+11
CE is divided into 11+4
AE is divided into 4+5
but whats x ?
Who knows ?
You don't mention x in your description.
You don't mention where B, D, and F are located, but I must have labeled mine the same way as Steve, since I got his answers as well.
What still has to be found is AD
In triangle ACE
9^2 = 16^2 + 15^2 - 2(15)(16)cosC
you should get cosC = 5/6
(don't find the actual angle)
in triangle ACD
AD^2 = 11^2 + 16^2 - 2(11)(16)cosC
= 251/3
AD = √(251/3) or appr 9.15
I assumed AD was a typo, or due to label discrepancy, since the subject is the tangents, which are of length 4,5,11.
Well, well, well, we've got ourselves a circumscribed triangle with some tangents going on. Let's get cracking!
First up, let's find AD, shall we? AD is basically a tangent to the circle, so we know that AD is equal to AE. And what do you know, AE is given as 9 cm! Ta-da!
Moving right along to FE. Well, from what we know, FE is the other tangent to the circle. So, FE is equal to CE. And what does our trusty problem statement say? CE is 15 cm! Boom!
Lastly, let's find CD. To do that, we need to realize that CD is actually 2 times the radius of the circle. But we don't have the radius, you say? Fear not, my friend. Remember that in a circumscribed triangle, the radius is the same length from the center to any of the vertices. So, we can just take our favorite friend AC (which is 16 cm, by the way) and divide it by 2 to get the radius. And voila, CD is twice that! Easy peasy!
So, to review:
AD = AE = 9 cm
FE = CE = 15 cm
CD = 2 * (AC / 2) = 2 * (16 / 2) = 16 cm
There you have it, my mathematical friend. Time to go out and conquer the circles!
To find AD, FE, and CD in a circumscribed triangle, you can use the properties of tangents drawn from the vertices of the triangle to the circumcircle.
First, let's consider triangle ACE. Point B is a point of tangency from vertex A, point D is a point of tangency from vertex C, and point F is a point of tangency from vertex E.
We can start by drawing the radii of the circumcircle from the center O to points A, C, and E. Since triangle ACE is circumscribed, the radii are equal in length. Let's label the points where the radii intersect the triangle as O1, O2, and O3.
We can now observe that triangles ADO1 and CEO2 are similar triangles. This is because they share an angle at O (as both have a right angle), and the other angles are also congruent due to the equal length of the radii. Therefore, we can use the property of similar triangles to find the lengths of AD and CD.
Let's denote AD as x. From the similarity of triangles ADO1 and CEO2, we have the following proportion: AD/16 = x/9, which can be cross-multiplied to get 9x = 16 * 9. Solving for x, we find AD = 16.
Using the same logic, we can find CD. Denote CD as y. From the similarity of triangles ADO1 and CEO2, we have the following proportion: CD/16 = y/15, which can be cross-multiplied to get 16y = 15 * 16. Solving for y, we find CD = 15.
Now, let's find FE. Since point F is a point of tangency from vertex E, it is actually the midpoint of segment CE. Therefore, FE = CE / 2 = 15 / 2 = 7.5.
To summarize:
- AD = 16 cm
- FE = 7.5 cm
- CD = 15 cm