SOLUTION: AOB and COD are two perpendicular diameters of a circle with radius 4 feet. With center A and radius AB an arc is drawn from B to meet AC extended at P, and with center B and radiu

Algebra ->  Points-lines-and-rays -> SOLUTION: AOB and COD are two perpendicular diameters of a circle with radius 4 feet. With center A and radius AB an arc is drawn from B to meet AC extended at P, and with center B and radiu      Log On


   

Question 732663: AOB and COD are two perpendicular diameters of a circle with radius 4 feet. With center A and radius AB an arc is drawn from B to meet AC extended at P, and with center B and radius BA an arc is drawn from A to meet BC extended at Q. With center C the arc PQ is drawn. DC extended meets this arc at R. Find DR and the perimeter of ADBPRQ.
So far, line segment AO, OC, OB, and OD are all 4 feet. Triangle ACO and OCB are 45-45-90. Line segments AC and CB are 3 root 2. Angles QCR and RCP are each 45 degrees, making arc QP 90 degrees. I really don't know where to go from here. Can someone please help? I have been stuck on this for days!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I did not know how to draw just the arc PQR, so I had to draw the whole green circle.
The sides of the square ADBC are 4sqrt%282%29=about5.657feet in length.
CD=AB=AP=BQ=8 because they are radii of the circles containing the arcs BP and AQ
so CR=CP=CQ=8-4sqrt%282%29 all radii of my green circle containing arc PRQ

So DR=CD%2BCR=8%2B8-4sqrt%282%29=16-4sqrt%282%29=about10.343feet

BPA and QAB are isosceles triangles with a 45%5Eo vertex angle and legs measuring 8 feet.
Based on law of cosines or using the fact that BPC and AQC are right triangles, we can calculate that BP%5E2=AQ%5E2=64%282-sqrt%282%29%29=about37.49feet%5E2
The approximate length would be BP=AQ=sqrt%2864%282-sqrt%282%29%29%29=8sqrt%282-sqrt%282%29%29=about6.123feet
Otherwise we could split those triangles into two congruent right triangles with a 22.5%5Eo angle and 8-foot hypotenuse, and calculate the length of their short legs (in feet) as BP%2F2=AQ%2F2=8sin%2822.5%5Eo%29=about8%2A0.3827=3.0615
Either way the ratio of base to leg length in those isosceles triangles is sqrt%282-sqrt%282%29%29=about0.7654

PRC and RQC are also isosceles triangles with a 45%5Eo vertex angle, so they are similar to BPC and AQC.
We knew that the length of their legs (in feet) were
CR=CP=CQ=8-4sqrt%282%29 and multiplying that times the ratio found above for the similar triangles we can find the length of PR=RQ.
Giving up on accurate value expressions,
CR=CP=CQ=8-4sqrt%282%29=about2.343,
so PR%2BPQ=about1.793feet

Now we can calculate the perimeter of ADBPRQ as the approximate value (in feet) of
5.657%2B5.657%2B6.123%2B6.123%2B1.793%2B1.793=27.146