I thank Divyesh for the suggestion!
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?
As regular, watch the video for an answer.
Distance Between Two Inscribed Circles In A Rectangle
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“All can be properly in the event you use your thoughts to your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport concept and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts due to group assist! Assist out and get early entry to posts with a pledge on Patreon.
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Reply To Distance Between Two Inscribed Circles In A Rectangle
(Just about all posts are transcribed shortly after I make the movies for them–please let me know if there are any typos/errors and I’ll appropriate them, thanks).
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:
√((6 – 2r)2 + (8 – 2r)2)
= √(22 + 42)
= √20
≈ 4.472
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