In triangle ABC, point X is on side BC such that AX=13,BX=10,CX=4, and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC. (Someone else asked the same question a bit ago, but nobody answered it, so I'm asking on a new strand)
The circumcircle of a triangle is the circle that passes through all three vertices of the triangle. The radius of the circumcircle of a triangle is the distance from any vertex of the triangle to the center of the circumcircle.
In triangle ABX, the radius of the circumcircle is AB/2=13/2.
In triangle ACX, the radius of the circumcircle is AC/2=14/2=7.
Since the circumcircles of triangles ABX and ACX have the same radius, we have AB/2=AC/2, or AB=AC.
Since AB=AC, we have △ABC is an isosceles triangle.
The area of an isosceles triangle is 21⋅b⋅h, where b is the base of the triangle and h is the height of the triangle.
In triangle ABC, the base is BC=10 and the height is AX=13.
Therefore, the area of triangle ABC is 1/2⋅10⋅13=65.
Let r be the radius of the circumcircles of triangles ABX and ACX. Then the area of triangle ABX is 1/2(AB)(r)=1/2(13)(r), and the area of triangle ACX is 1/2(AC)(r)1/2(14)(r). Since the circumcircles have the same radius, the areas of the triangles are equal. Therefore, 1/2(13)(r)=1/2(14)(r), or r=13.
The area of triangle ABC is1/2(AB)(AC) = 1/2(13)(14) = 91.
