Question

the radii of two concentric circles are 13 and 8cm . AB is a diameter of bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at P producing . find the length of AP.

Answer 1

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

Answer 2

GIVEN: 2 Concentric circles with centre O. Radius OB of bigger circle = 13cm, Radius OD of smaller circle = 8cm.

PB is tangent to the inner circle at point D. So, OD is perpendicular to PB.( as tangent is perpendicular to the radius segment through the point of contact.)

Construction: Join AD & Join AP.

PROOF: In right triangle ODB

DB² = OB² — OD² ( by pythagoras law)

DB² = 13² — 8² = 169 — 64 = 105

=> DB = √105

=> PD = √105 ( as perpendicular OD from the centre of the circle O, to a chord PB , bisects the chord)

Now in triangle APB, angle APB = 90°( as angle on a semi circle is a right angle).

And, OD // AP & OD = 1/2 AP ( as, segment joining the mid points of any 2 sides of a triangle is parallel to the third side & also half of it.

So, AP = 16cm

hope this helps you out!