jugoslav

You're welcome!

2 * BE = 3 * 4   ==>  BE = 6            

Radius   r = sqrt(8² +1) / 2         r = √65 / 2

Good job, Phill!!!

5/        6² = 4 (4 + x)                     6/      12² = 8 ( 8 + x)

          36 = 16 + 4x                                144 = 64 + 8x

           x = 5                                            x = 10

A triangle has vertices at A(6,8), B(0,0), and C(16,0).  A square is inscribed in triangle ABC so that one side of the square aligns with side AB.  Find the side length of the square.

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BC = 16

AB = sqrt(8² + 6²) = 10

AC = sqrt(8² + 10²) = √164

Use the law of cosine to calculate  ∠ABC and ∠BAC

Use this formula to find "a"               (a / tan∠BAC) + a + (a / tan∠ABC) = AB

corrections::::    ∠Y = tan-1(5 / 12)    

A semicircle with radius 2 and two semicircles with diameter 2 are drawn as shown.  The red circle is tangent to all three semicircles.  Find r.

r = 0.666666667     or     2/3

Clever solution!!!

I was using trig to solve it.

AC * AP = AB * AS

8(8 + PC) = 5 * 30

PC = 10.75

SP = sqrt(AS² - AP²)

SC = sqrt(SP² + PC²)

Area of a circle       A = (SC/2)² * pi

BA is tangent to the circle at A and angle DBA = 90 degrees. If AB = 15 and CD = 10, find the radius of the circle.

r = 1/2 [sqrt(10² + 30²)]             ( 10 and 30 are the lengths of the legs of an inscribed right-angled triangle )