a triangle with side lengths 26, 28, and 30 is constructed so that the longest and shortest sides are tangent to a circle. the third side passes throught the center of the circle. compute the radius of the circle
the area of the triangle is 336 if this helps
I think, since the side that is 28 passes through the center, the radius is 14.
the other two sides are tangent.
draw a pic. mine looks like an ice cream cone, and maybe you will see my thinking.
not a tutor
That sure helps a lot, I was working with 3 different equations, with cosine law equations and it got real messy.
Let the triangle be ABC, where AB=26, AC=30 and BC=28
The circle will have to be on the bisector of angle A, let it fall on BC at D.
Then the radius is the line from D to AB and D to AC.
Area of triangle ABC = area of ABD + area of ACD
= (1/2)(26)r + (1/2)30)r = 28r
but 28r = 336
r = 12
thanks reiny, not even my problem, but I couldn't stop thinking about this problem.
my pic and answer was too easy to be right!!
thank you so much.. this was a huge help.. and if youd like to know how i got the area of the triangle i used the formula... A=sqrt s(s-a)(s-b)(s-c) where s=a+b+c/2...a b and c are the side lengths
Good for you LULU, good old Heron's formula.
And good for your teacher to teach it to you!
To solve this problem, we can use the fact that in a triangle, the length of the side which is tangent to a circle is equal to the radius of the circle.
Let's label the triangle with side lengths 26, 28, and 30 as triangle ABC, where A is the vertex opposite the side with length 26, B is the vertex opposite the side with length 28, and C is the vertex opposite the side with length 30.
Since the shortest side, 26, is tangent to the circle, we can label the center of the circle as O, and the point of tangency on side AB as D.
Now, in order to find the radius of the circle, we need to find the length of the segment OD.
From the problem statement, we know that the third side, CD, passes through the center of the circle. Therefore, CD is the diameter of the circle, and OD is half the length of CD.
To find the length of CD, we can use the fact that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Therefore, we have:
AC + BC > AB
30 + BC > 28
BC > 28 - 30
BC > -2
Since the length of a side cannot be negative, we can conclude that BC > 0.
Similarly, we find:
AB + BC > AC
26 + BC > 30
BC > 30 - 26
BC > 4
Combining these two inequalities, we have:
4 < BC < 30
Since BC is a side of the triangle, it must be smaller than the sum of the other two sides, which are 26 and 28. Therefore, we have:
4 < BC < 26 + 28
4 < BC < 54
Now, we know that OD is half the length of CD. Since CD has a length BC, we can conclude that OD = BC/2.
Therefore, the radius of the circle, which is equal to OD, is BC/2.
Since we found that 4 < BC < 54, we can conclude that the radius of the circle is between BC/2 = 2 and BC/2 = 27.
Hence, the radius of the circle is between 2 and 27 units.