Triangle PQS is similar to triangle ABC
PRS is equilateral with a height of sqrt 3
Dropping a perpendicular to BC from P will be tangent to the circles R and S
Call the point of tangency of this segment T
Triangle QTP is right with QT = 3 TP = sqrt 3
And QP = sqrt (3^2 + (sqrt 3)^2) = sqrt 12
Using the Law of Cosines
QS^2 = QP^2 + PS^2 - 2(QP * PS) cos (QPS)
4^2 = 12 + 4 - 2(QP * PS)cos (QPS)
0 = - 2 (QP * PS) cos (QPS)
cos (QPS) = 0
So angle QPS is right
And sin (PQS) = PS/ QS = 2/4 = 1/2
So angle (PQS) = 30 ° = angle ABC
So angle (PSQ) = 60° = angle ACB
The area of PQS = PS * PQ / 2 = 2(sqrt 12) / 2 = sqrt (12)
Drop a perpendicular from Q to BC and call the intersection U
The triangle BUQ is right and we can find BU as follows
QU / sin (1/2 angle ABC) = BU /sin ( BQU)
1/ sin (15) = BQ/ sin (75)
BQ = sin 75 / sin 15 = cos 15/ sin 15 = cot 15 = 2 + sqrt 3
Similarly......drop a perpendicular from S to BC and call the intersection point V
Using a similar proceedure for finding BU, we can find VC as sqrt 3
So...... BC = BU + QR + RS + VC = 2 + sqrt 3 + 2 + 2 + sqrt 3 = 6 + 2 sqrt 3
The scale factor from ABC to PQS = (BC / QS) = [ 6+ 2 sqrt 3 ] / 4
Then the area of ABC = (scale factor)^2 (area of PQS) =
[ (6 + 2sqrt 3) / 4]^2 * sqrt (12) ≈
19.39 units^2
