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Sagot :
since the distance of the the two DE and CE is equal to the radius you can compare the two triangles BDE and CDE. since they have two equal sides and one shared side their third side must be equal as well leaving you with DE = CE
15y - 50 = 17y - 62
62 -50 = 17y - 15y
12 = 2y
6 = y
Knowing that y=6 you can conclude that DE=CE=40mm
Since BD and ED create a right angle (sane as CB and CD) you can use the pythagorean theorem to calculate BE
BE^2= CE^2+CB^2
BE^2 = 30^2 + 40^2
BE^2 = 2500
BE = 50mm
im not sure about the last question sorry
15y - 50 = 17y - 62
62 -50 = 17y - 15y
12 = 2y
6 = y
Knowing that y=6 you can conclude that DE=CE=40mm
Since BD and ED create a right angle (sane as CB and CD) you can use the pythagorean theorem to calculate BE
BE^2= CE^2+CB^2
BE^2 = 30^2 + 40^2
BE^2 = 2500
BE = 50mm
im not sure about the last question sorry
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