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The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.
- a)19 cm
- b)20 cm
- c)16 cm
- d)√105
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The radii of two concentric circles are 13 cm and 8 cm. AB is a diamet...
To find the length AD, we can use the Pythagorean theorem.
Since AB is a diameter of the bigger circle, its length is equal to the diameter of the bigger circle, which is twice the radius. So, AB = 2 * 13 cm = 26 cm.
Since BD is tangent to the smaller circle, it is perpendicular to AD. Therefore, triangle ABD is a right triangle.
Using the Pythagorean theorem, we have:
(AD)^2 + (BD)^2 = (AB)^2
(AD)^2 + (8 cm)^2 = (26 cm)^2
(AD)^2 + 64 cm^2 = 676 cm^2
(AD)^2 = 676 cm^2 - 64 cm^2
(AD)^2 = 612 cm^2
AD = √(612 cm^2)
AD ≈ 24.7 cm
Therefore, the length AD is approximately 24.7 cm.
Since AB is a diameter of the bigger circle, its length is equal to the diameter of the bigger circle, which is twice the radius. So, AB = 2 * 13 cm = 26 cm.
Since BD is tangent to the smaller circle, it is perpendicular to AD. Therefore, triangle ABD is a right triangle.
Using the Pythagorean theorem, we have:
(AD)^2 + (BD)^2 = (AB)^2
(AD)^2 + (8 cm)^2 = (26 cm)^2
(AD)^2 + 64 cm^2 = 676 cm^2
(AD)^2 = 676 cm^2 - 64 cm^2
(AD)^2 = 612 cm^2
AD = √(612 cm^2)
AD ≈ 24.7 cm
Therefore, the length AD is approximately 24.7 cm.
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Community Answer
The radii of two concentric circles are 13 cm and 8 cm. AB is a diamet...
Produce BD to meet the bigger circles at E. Join AE. Then
∠AEB = 90° [Angle in a semicircle]
OD ⊥ BE
[∵ BE is tangent to the smaller circle at D and OD is its radius] BD = DE [∵ BE is a chord of the circle and OD ^ BE]
∴ OD || AE [∵ ∠AEB = ∠ODB = 90°]
In ΔAEB, O and D are mid-points of AB and BE. Therefore, by mid-point theorem, we have
OD = 1/2AE
⇒ AE = 2 × 8 = 16 cm
In ΔODB, we have
OB2 = OD2 + BD2 [By Pythagoras Theorem]
⇒ 169 = 82 + BD2
⇒ BD2 = 169 – 64 = 105 ⇒ BD = 105 cm
⇒ DE = √105 cm [∵ BD = DE]
In ΔAED, we have
AD2 = AE2 + ED2 [By Pythagoras Theorem]
⇒ AD2 = 162 + (√105)2 = 256 + 105 = 361
⇒ AD = 19 cm
∠AEB = 90° [Angle in a semicircle]
OD ⊥ BE
[∵ BE is tangent to the smaller circle at D and OD is its radius] BD = DE [∵ BE is a chord of the circle and OD ^ BE]
∴ OD || AE [∵ ∠AEB = ∠ODB = 90°]
In ΔAEB, O and D are mid-points of AB and BE. Therefore, by mid-point theorem, we have
OD = 1/2AE
⇒ AE = 2 × 8 = 16 cm
In ΔODB, we have
OB2 = OD2 + BD2 [By Pythagoras Theorem]
⇒ 169 = 82 + BD2
⇒ BD2 = 169 – 64 = 105 ⇒ BD = 105 cm
⇒ DE = √105 cm [∵ BD = DE]
In ΔAED, we have
AD2 = AE2 + ED2 [By Pythagoras Theorem]
⇒ AD2 = 162 + (√105)2 = 256 + 105 = 361
⇒ AD = 19 cm
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