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# help

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Point P lies on minor arc AB of the circle circumscribing square ABCD.  If AB = 5 and PA = 4, find the length of PD.

Jan 4, 2020

#1
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I have an answer for this, but it takes a lot of steps -- I welcome a quicker solution.

1) Find the length of the diameter AC:

Since AC is a diameter, I can use triangle ABC to find this length.

Both AB and BC = 5 and triangle ABC is an isosceles right triangle, AC  =  5·sqrt(2).

2) Find the length PC:

Triangle ABC is a right triangle (angle ABC is inscribed in a semicircle), so use the Pythagorean Theorem.

PC2 + PA2  =  AC2     --->     PC2 + 42  =  [5·sqrt(2)]2     --->     PC2  +  16  =  50     --->     PC2  =  34     --->     PC  =  sqrt(34).

3) Find the size of angle(PAC):

Use the Law of Cosines for triangle PAC).

PC2  =  PA2 + AC2  -  2·PA·AC·cos(PAC)     --->     [sqrt(34)]2  =  (4)2 + [5·sqrt(2)]2 - 2·(4)·5(sqrt(2)·cos(PAC)

--->     34  =  16 + 50 - 40·sqrt(2)·cos(PAC)     --->     -32  =  -40·sqrt(2)·cos(PAC)     --->     PAC  =  55.5501o

4) angle(PAD)  =  45o + 55.5501o  =  100.5501o

5) Use the Law of Cosines on Triangle(APD):

--->     DP2  =  (4)2 + (5)2 - 2·(4)·(5)·cos(100.5501)      --->     DP2  =  48.3238      --->      DP  =  6.95

1)  I hope that I haven't made a mistake!

2)  I hope that someone can find an easier way!

Jan 4, 2020
#2
+1

AB = 5

PA = 4

BD = ?                         BD = sqrt [(AB)² + (AD)²]      BD = 7.072      r = 3.536

Let the circle's origin be an "O", and a midpoint of PA an "M"                                                                            ∠POM = q                   sin(q) = (PM)/r     q = 34.44°

∠POA = 2*q                             ∠POA= 68.88°

∠AOB = 90°

∠BOP = ∠AOB - ∠POA            ∠BOP = 21.12°

∠(BOP)/2 = w                      w = 10.56°

PB = ?                             sin(w) =[(PB)/2] / r       (PB)/2 = 0.648      PB = 1.296

Finally...                      PD = sqrt [(BD)² - (PB)²]                PD = 6.952 Jan 4, 2020