The radii of two concentric circles are 3 cm and 8 cm diameter of the bigger circle and PT is a tangent to a smaller circle touching HD and introducing find the length of AP
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The radius of two concentric circles are 13 cm and 8 cm. AB is the diameter of the bigger circle and BD is tangent to the smaller circle touching at D and the bigger circle at E. Point A is joined to D. What is the length of AD?
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Girija Warrier, Passionate about Maths & Music...
Answered Jan 5, 2017 · Author has 2.1k answers and 1.8m answer views

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
Still have a question? Ask your own!
What is your question?
17 ANSWERS

Girija Warrier, Passionate about Maths & Music...
Answered Jan 5, 2017 · Author has 2.1k answers and 1.8m answer views

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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