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The radii of two concentric circles are 13 cm and & 8 cmAB is a diameter of the bigger circle. BD is a tangento the smaller circle touching it at D. Find thelength ADA. 19 cmB. 15 cmC. 17 cmD 13 cm |
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Answer» GIVEN: 2 Concentric circles with centre O. Radius OB of bigger circle = 13cm, Radius OD of smaller circle = 8cm. EB is tangent to the inner circle at point D. So, OD is perpendicular to EB.( as tangent is perpendicular to the radius segment through the point of contact.) Construction: Join AD & Join AE. PROOF: In right triangle ODB DB² = OB² — OD² ( by pythagoras law) DB² = 13² — 8² = 169 — 64 = 105 => DB = √105 => ED = √105 ( as perpendicular OD from the centre of the circle O, to a chord EB , bisects the chord) Now in triangle AEB, angle AEB = 90°( as angle on a semi circle is a right angle). And, OD // AE & OD = 1/2 AE ( as, segment joining rhe mid points of any 2 sides of a triangle is parallel to the third side & also half of it. So, AE = 16cm Now, in right triangle AED, AD² = AE² + ED² => AD² = 16² + (√105)² => AD² = 256 +105 => AD² = 361 => AD = √361 = 19cm |
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