+0

hard geometry

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35
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In the diagram, four circles of radius 1 with centres P, Q, R, and S are tangent to one another and to the sides of triangle ABC, as shown.  What is the area of triangle ABC?

May 29, 2021

#1
0

ΔABC is a 30º-60º-90º triangle.

Area = [(2 + √3 + 2√3 + 1)(3 + √3)] / 2

May 29, 2021
#2
+119780
+2

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

May 29, 2021
#3
+32612
0

Ugh....and YIKES !    again !

May 29, 2021
#4
+119780
+1

Yeah.....that's  my feeling , too.......LOL!!!!!

CPhill  May 29, 2021