An isoceles trianagle with a base of 24 and legs of 15 is inscribed in a circle. Find the radius. (I need more than 1 method to get the solution) Thx in advance
1) Let ABC be an isoseles triangle with vertex at B.
2)Drop (and do not pick it up) the perpendicular from B onto the side AC.
3)Call the intersection point D.
4)Thus, AD=DC=12 and AB=BC=15.
5)By Pythagoren theorem BD=9
6)Therefore, sin A = opposite/hypotenuse = 9/15=3/5
7)Let R be the radius of the circle in which it is inscribed.
8)Law of sines: 2R=BC/sin A = 15/(3/5)=25
9)Thus, 2R=25 thus R=12.5
Hello, arturju!
Another approach . . .
An isoceles triangle with a base of 24 and legs of 15 is inscribed in a circle.
Find the radius.Code:B * * * * | * * | * * |r * | * | * * *O * * / | \ * / | \r * / | \ * A* - - - + - - - *C * D * * * *
Draw sides BA = BC = 15.
As TPHacker pointed out, AD = DC = 12, BD = 9.
. . Then: OD = 9 - r
In right triangle ODC, we have: .OC² .= .OD² + DC²
. . That is: .r² .= .(9 - r)² + 12² . → . 18r = 225
Therefore: .r .= .225/18 .= .25/2 .= .12½
Another one you could try, although it is tedious, is to place your triangle on a cartesian plane.
1. Find the equations of the lines that form the triangle
2. Find the equations of the lines perpendicular to those forming the triangle (they must pass through the midpoints of the original lines)
3. Find the point of intersection of the perpendicular lines (this is the centre-point of the circle)
4. Using the distance formula, find the distance between any vertex of the triangle and the intersection (this is the radius).
My maths teacher said it can be proved that this works for ANY triangle, but the proof is rigorous (involving many similar triangles) and I don't know it.
Hello, arturju!
Divideby0 has a good game plan,
. . and for this problem, it's not too difficult.
An isoceles triangle with a base of 24 and legs of 15
is inscribed in a circle. .Find the radius.Code:| B *(0,9) *|* * | * * | * * | o M * | * * | * * | * - A * * * * + * * * * C - (-12,0) | (12,0)
The midpoint of BC is: .M(6, 9/2)
The slope of BC is: .(0 - 9)/(12 - 0) .= .-3/4
The perpendicular slope is: .m = 4/3
The equation of the perpendicular bisector is:
. . y - 9/2 .= .(4/3)(x - 6) . → . y .= .(4/3)x - 7/2 .[1]
The perpendicular bisector of AC is the y-axis: .x = 0 .[2]
The intersection of [1] and [2] is: .P(0, -3½), the center of the circle.
The radius of the circle is: .PB .= .9 - (-3½) .= .12½