the reddi of two concentric circle are 13cm and 8cm ab is a diameter of the bigger circle and bd is tangent to the smaller circle rouching it at d and intersecting the large circke at p on producing .fine length of ab
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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
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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