if ABCD is a rectangle,AB=8 BC=20&angle BPC=90&r1,r2,r3 be radius of the incircles of triangles APB,BPC,CPD then find r1+r2+r3
Answers
In rectangle ABCD, AB = 8, BC = 20. P is a point on AD so that<BPC = 90 deg. If r1, r2, r3 are the radii of the incircles of APB, BPC and CPD, r1+r2+r3=?
In rectangle ABCD, AB = 8, BC = 20. P is a point on AD.
Let AP = t and PD = 20-t
AB/AP = tan x = 8/t.
CD/PD = tan (90-x) = 8/(20-t) = cot x, or tan x = (20-t)/8
8/t = (20-t)/8, or
64 = t(20-t) = 20t - t^2, or
t^2 -20t +64 = 0, or
(t-4)(t-16) = 0, or
AP = 4 and PD = 16.
In RAT APB, BP^2 = AP^2 + AB^2 = 4^2+8^2 = 16+64 = 80.
Or BP = 80^0.5 = 8.94427191
In RAT CPD, CP^2 = PD^2 + CD^2 = 16^2+8^2 = 256+64 = 320.
Or CP = 320^0.5 = 17.88854382
In RAT APB, AB = 8, AP = 4 and BP = 8.94427191. Therefore, perimeter of APB = 8+4+8.94427191= 20.94427191. Its semi-perimeter, s = 10.47213596.
Its area = 8*4/2 = 16. r1 = Area(APB)/s(APB) = 16/10.47213596 = 1.527864044.
In RAT BPC, BP = 8.94427191, CP = 17.88854382 and BC = 20. Therefore, perimeter of BPC = 8.94427191+17.88854382+20 = 46.83281573. Its semi-perimeter, s = 23.41640787. Its area = 8.94427191*17.88854382/2 = 80. r2 = Area(BPC)/s(BPC) =80/23.41640787 = 3.416407864.
In RAT CPD, CD = 8, DP = 16 and CP = 17.88854382. Therefore, perimeter of CPD = 8+16+17.88854382 = 41.88854382. Its semi-perimeter, s = 20.94427191. Its area = 8*16/2 = 64. r3 = Area(CPD)/s(CPD) =Area/s = 64/20.94427191= 3.05572809.
r1+r2+r3= 1.527864044 + 3.416407864 + 3.05572809 = 7.999999998, or 8.
Hence r1+r2+r3 = 8.