Ab and cd are two chords of a circle such that ab=6cm ,cd=12cm andab is parallel to cd.the distance between ab and cd is 3cm.find the radius of the circle
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Since CD is parallel to AB then <CBA=<BCD=30 --> <CBE=60 --> <COE=2*60=120 (according to the central angle theorem) --> length of minor arc CE is 120360∗2πr=103∗π120360∗2πr=103∗π;
Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus BCAB=3√2BCAB=32, (BC is opposite to 60 degrees so corresponds to 3√3) --> BC10=3√2BC10=32(AB=diameter=2r=10) --> BC=53√BC=53;
Thus the perimeter of the shaded region is(minor arc CE)+BC+BE: 103∗π+53√+53√=103∗π+103√103∗π+53+53=103∗π+103.
Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus BCAB=3√2BCAB=32, (BC is opposite to 60 degrees so corresponds to 3√3) --> BC10=3√2BC10=32(AB=diameter=2r=10) --> BC=53√BC=53;
Thus the perimeter of the shaded region is(minor arc CE)+BC+BE: 103∗π+53√+53√=103∗π+103√103∗π+53+53=103∗π+103.
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