the radii of two concentric circles are 13 cm and 8 cm a b is diameter of bigger circle and BD is tangent to smaller circle touching it at D and intersecting bigger circle at P on producing find AP
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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
☆ Hence AD ( AP ) = 19cm
[ Note: I changed some alphabets note them and convert them according to your Question. plz don't mind ]
HOPE THIS HELPS YOU !
Be Brainly@
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
☆ Hence AD ( AP ) = 19cm
[ Note: I changed some alphabets note them and convert them according to your Question. plz don't mind ]
HOPE THIS HELPS YOU !
Be Brainly@
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