jugoslav

In triangle PQR, PQ=13, QR=14, and PR=15. Let M be the midpoint of QR. Find PM.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

PM = [sqrt(2 * 13² + 2 * 15² - 14²)] / 2

PM = 12.16552506

PR = a = 9          QR = b = 7           PQ = c = 6          RM = d = ?

d = 1/2[sqrt(2*a² + 2*b² - c²)]

d = √224 / 2 ≈ 7.483

1 * (5 + 8) = 13

1 + (5 * 8) = 41

I made an error! 

4/8 = √80/CQ         CQ = (8√80)/4      Wrong!!!

4/√80 = 7/CQ          CQ = (7√80) / 4      Correct!!!

Let triangle ABC have side lengths AB=13, AC=14, and BC=15. There are two circles located inside angle BAC which are tangent to rays AB, AC, and segment BC. Compute the distance between the centers of these two circles.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AB = 13     AC = 14     BC = 15

s = 1/2(13+14+15) = 21

A = sqrt(21(21-13()21-14)(21-15) = 84

r = A/s = 4

Using the law of cosines, we find that ∠A = 67.380135º

∠PAY = 33.6900675º

AZ = AY = r / tan(A/2) = 6            CW = CY = 8              BW = BZ = 7

CP = sqrt(4² + 8²) = √80

4/8 = √80/CQ         CQ = (8√80)/4

PQ = sqrt(CP² + CQ²)

PQ = 18.02775638

ABC is a right triangle with an angle of C=90 degrees. AC = 10 and BC = 24. Point P is located inside ABC such that the distance from P to AB is twice the distance from P to AC, and the distance from P to AC is equal to the distance from P to BC.  In units, what is the distance from P to AB?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[ABC] = 120

x² + 1/2[x(10-x) + x(24-x)] + 26x = 120

x = 120 / 43

The distance from P to AB = 2x

cosA = 1 / √2

The smallest area of a triangle is when k = 2      A = 3.653 u²

                                                               k = 1      A = 8.956 u² 

In the diagram below, WXYZ is a trapezoid such that WX||ZY and WY ⊥ ZY. If YZ=12, tan Z=1.5, and tan X=2, then what is XY?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

WY = tanZ * 12 = 18

WX = 18 / tanX = 9

XY = sqrt(WX² + WY²)

1/   Semiperimeter       s = 1/2(a + b + c)

2/   Area      A = sqrt[s (s - a)(s - b)(s - c)]

3/   Inradius     r = A / s