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If it is one you can feel with your finger or see with your eyed and it is not the circular ring near the center hole, it is a fault, a blemish, a production error.
First consider triangles ACD and BCD. They share a common base, CD, and their height is the distance between the parallel lines AB and CD. Consequently, area(ACD) = area(BCD) . . . . . . . . eqn 1 But ACD = AOD + OCD and BCD = BOC + OCD So area(ACD) = area(AOD) + area(OCD) and area(BCD) = area(BOC) + area(OCD) Substituting in eqn 1, area(AOD) + area(OCD) = area(BOC) + area(OCD) area(AOD) = area(BOC)
== == 1) Draw a line segment AB of 5 units 2) Draw the perpendicular bisector CD of AB such that Cd meerts AB at C. 3) Mark off CE = 2 units on CD 4) Draw the straight line segments AE & BE. ABE is your triangle. Its base (AB) = 5 and height (CE) = 2, so its area = [base x ht] / 2 = 5 sq units
pi x 12 (diammeter of the CD) = 37.6991118 (the circumference of the CD)
It is: CD = 500-100 = 400