On a semicircle with diameter AD, chord BC is parallel to the diameter. Further; each of the chords AB and CD has length 2, while AD has length 8. What is the length of BC?

Question

Answer 1

Let CE be the perpendicular on AD.
Now AC = 8^2 - 4 = rt60(Angle in a semi cirlce is right angle so we can use pythagoras theorem)
Now area of the triangle ACD = 1/2*2*rt60 = rt60 = 1/2*8*x => x = rt15/2
Now, In triangle EDC,
By pythagoras theorem ED^2 = 4 - 15/4 = .25
=> ED = .5
Thus BC = 8 - 2*.5= 7 cm
Hope that helps.

Answer 2

Let CE be the perpendicular on AD.
Now AC = 8^2 - 4 = rt60(Angle in a semi cirlce is right angle so we can use pythagoras theorem)
Now area of the triangle ACD = 1/2*2*rt60 = rt60 = 1/2*8*x => x = rt15/2
Now, In triangle EDC,
By pythagoras theorem ED^2 = 4 - 15/4 = .25
=> ED = .5
Thus BC = 8 - 2*.5= 7 cm
Hope that helps.

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On a semicircle with diameter AD, chord BC is parallel to the diameter. Further; each of the chords AB and CD has length 2, while AD has length 8. What is the length of BC?