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From Karnataka Open Educational Resources
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, 16:31, 15 February 2015
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| In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br> | | In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br> |
| ∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC | | ∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC |
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| + | == == Problem-7 [Ex-15.4-A3] == == |
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| + | In the figure AB=10cm,AC=6cm and the radius of the smaller circle is xcm. find x. |