A rectangle has sides 6 and eight. Assemble a diagonal of the rectangle, dividing the rectangle into two triangles. Inscribe circle X in a single triangle and circle Y within the different triangle. What’s the distance between the facilities of circles X and Y?
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Assemble a vertical line by the middle of circle X and a horizontal line by the middle of circle Y, as proven under. Let r be the radius of the inscribed circles. Assemble a proper triangle whose legs are the vertical and horizontal distances between X and Y and hypotenuse is the same as XY. The legs are then equal to six – 2r and eight – 2r, so XY = √((6 – 2r)2 + (8 – 2r)2)
All that continues to be is to resolve for r. Half of the rectangle’s space is 6×8/2 = 24. Because the rectangle has sides of 6 and eight, its diagonal is 10 (double a 3-4-5 proper triangle). Assemble three radii from circle X to the tangency factors on the 2 sides and the diagonal. These three radii are the heights of three triangles, as proven under, and the sum of their areas is the same as half the rectangle.
Thus half the rectangle’s space can be equal to:
6r/2 + 8r/2 + 10r/2 r(6 + 8 + 10)/2 12r
This space is the same as 24, so we have now r = 2. Then we substitute into the system for the size of XY to get:
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