1. A triangle has side lengths of \(10, 24,\) and \(26.\) Let \(a\) be the area of the circumcircle. Let \(b\) be the area of the incircle. Compute \(a-b.\)
2. In triangle \(ABC, AB =BC=17\) and \(AC =16.\) Find the inradius of triangle \(ABC.\)
3. In triangle \(ABC,M\) is the midpoint of \(\overline{AB}\). Let \(D\) be the point on \(\overline{BC}\) such that \(\overline{AD}\) bisects \(\angle{BAC},\) and let the perpendicular bisector of \(\overline{AB}\) intersect \(\overline{AD}\) at \(E.\) If \(AB=44\) and \(ME=12,\) then find the distance from \(E\) to line \(AC.\)
I'll try the first one.
Since the triangle has side lengths of 10, 24, and 26, this triangle is a right triangle (102 + 242 = 262).
The center of the circumcenter of a right triangle is the midpoint of the hypotenuse.
Since the hypotenuse has a length of 26, the distance from this point to any of the vertices is 13.
Therefore, the area of the circumcircle is: pi·132 = 169·pi.
The center of the incircle is the point of intersection of the angle bisectors.
Call the vertex between the 10 and 24 sides = C (it is a right angle).
Call the vertex between the 24 and 26 sides = A.
Call the center of the incircle X.
Draw the perpendicular from X to side AC; call this point Y; this is a radius of the incircle;
call its value 'h'.
Angle C = 90o; therefore, angle(XCY) = 45o.
I will need to find the size of angle(XAY).
I will use the sin(A) in triangle ABC) and find one-half of that value.
sin(A) = 10/26 ---> A = sin-1(10/26) = 22.61986o ---> angle(XAY) = 11.31o.
Now, to find the value of h:
Call the distance from C to Y 'x'; therefore, the distance from A to Y is '24 - x'.
In triangle(XCY), tan(45o) = h/x ---> h = x·tan(45o).
In triangle (XAY), tan(11.31o) = h/(24 - x) ---> h = (24 - x)·tan(11.31o).
Setting these two equation equal to each other: x·tan(45o) = (24 - x)·tan(11.31o)
---> x·tan(45o) = 24·tan(11.31o) - x·tan(11.31o)
---> x·tan(45o) + x·tan(11.31o) = 24·tan(11.31o)
---> x·( tan(45o) + tan(11.31o) ) = 24·tan(11.31o)
---> x = 24·tan(11.31o) / [ tan(45o) + tan(11.31o) ]
---> x = 4
Therefore, the area of the incircle is: pi·42 = 16·pi.
169·pi - 16·pi = 153·pi
#2: Let M be the midpoint of side AC. Draw BM.
Since triangle ABC is an isosceles triangle BM is the angle bisector of angle(B) and also the altitude from B to side AC.
This makes angle(AMB) a right angle.
Since AC = 16, AM = 8.
Using the Pythagorean Theorem on triangle(AMB), we can find that BM = 15.
In right triangle(AMB), cos(A) = 8/17 ---> angle(A) = cos-1(8/17) = 61.9275o.
Draw the bisector of angle(A). Call the point where this angle bisector intersects BM point 'X'.
Angle(MAX) = ½·angle(A) = ½·61.9275o = 30.96o.
To find XM of triangle(MAX) ---> tan(30.96o) = XM/8 ---XM = 8·tan(30.96o) = 4.8.
The radius of the incircle is 4.8.