In triangle ABC, it is known that AB = 15, BC = 14, AC = 13, and the median AA 1
In triangle ABC, it is known that AB = 15, BC = 14, AC = 13, and the median AA 1 intersects the bisector BB1 at point P, find the area A1PB1C.
Let us define the area of the triangle ABC by Heron’s theorem.
The semi-perimeter of the triangle ABC is equal to: p = (AB + BC + AC) / 2 = 42/2 = 21 cm.
Then Saс = √21 * (21 – 15) * (21 – 14) * (21 – 13) = √7056 = 84 cm2.
Since AA1 is the median of the ABC triangle, then Saa1b = Saa1s = Saabs / 2 = 42 cm2.
The segment BB1 is the bisector of triangles ABC and AA1B, then AB / BC = AB1 / CB1 = 15/14.
Then Svsv1 / Sawv1 = 15/14.
Svsv1 = 15 * Savv1 / 14.
Svsv1 + Savv1 = Savs = 84.
15 * Savv1 / 14 + Savv1 = 84.
29 * Savv1 / 14 = 84.
Savv1 = 1176/29 = 40.55 cm2.
Similarly:
AB / BA1 = AP / A1P = 15/7.
Then Saur / Sa1vr = 15/7.
Sa1vr = 7 * Saur / 15.
Savr + Savr = Sava1 = 42.
Savr + 7 * Savr / 15 = 42.
22 * Savr / 15 = 42.
Sarv = 630/22 = 28.64 cm2.
Then Sarv1 = Saav1 – Sarv = 40.55 – 28.64 = 11.91 cm2.
Then Sa1rv1s = Saa1s – Saav1 = 42 – 11.91 = 30.09 cm2.
Answer: The area of the quadrangle A1PB1C is 30.09 cm2.