Question

Radii of two concentric circles are 13 and 8. AB is the diameter of circle with a larger radius. BDTouches a circle with smaller radius at D find AD

Answer 1

Correct Question :

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle. Find the length of AP.

AnswEr :

⋆ Reference of Image is in the attachment :

Join OD, and OD will be Radius of Smaller Circle and Perpendicular to the Tangent BD

∴ ∠ ODB = 90° ⠀[ Perpendicular ]

AB is Diameter, and P is point on the Semi Circle. Angle in a Semi Circle is Right Angle.

∴ ∠ APB = 90° ⠀[ Angle in Semi Circle ]

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Now, we will Similar the Triangles.

• In Triangle APB and, Triangle OBD :

⇒ ∠ APB = ∠ ODB ⠀⠀—( Both are 90° )

⇒ ∠ ABP = ∠ OBD ⠀⠀—( Common Angle )

∴ ∆ APB ~ ∆ OBD ⠀⠀—( AA Similarlity )

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• Sides of Similar Triangles are Proportional

⇒ AP /OD = AB /OB

⇒ AP /8 cm = 26 cm /13 cm

• AB is Diameter of Large Circle

⇒ AP /8 cm = 2

⇒ AP = 2 × 8 cm

⇒ AP = 16 cm

∴ Therefore, Value of AP will be 16 cm.

Answer 2

$\bf{\Huge{\underline{\sf{\green{QUESTION\::}}}}}}$

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle. Find the length of AD.

$\bf{\Huge{\underline{\boxed{\sf{\green{ANSWER\::}}}}}}$

Given:

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle.

To find:

The length of AD.

$\bf{\Large{\underline{\rm{\red{Explanation\::}}}}}$

We have,

• Radius of two concentric circles= 13cm & 8cm
• AB is a diameter of bigger circle.
• BD is a tangent to the smaller circle.

We know that tangent of any point of circles is perpendicular to the radius through the point of contact.

∴ OD ⊥ BD

In ΔBOD, we have;

• OB= 13cm
• OD= 8cm

Using Pythagoras Theorem:

⇒ [Hypotenuse]² = [Base]² + [Perpendicular]²

⇒ (13cm)² = BD² + (8cm)²

⇒ 169cm² = BD² + 64cm²

⇒ BD² = (169 - 64)cm²

⇒ BD² = 105cm²

⇒ BD = √105 cm

∴Perpendicular drawn from the centre to bisect the chord.

→ BD = DE = √105 cm

→ BE = 2BD

→ BE = (2× √105)cm

→ BE = 2√105 cm

&

In ΔBOD & ΔBAE

∠BDO = ∠BEA = 90°

∠BEA= 90°          [Angle in semicircle is right angled Δ]

∠OBD = ∠ABE   [common]

by AAS similar criterion

So,

ΔBOD ~ ΔBAE

→ $\frac{BD}{BE} =\frac{OD}{AE}$      [Proportional sides of Δ]

→ $\frac{\cancel{\sqrt{105}} }{2\cancel{\sqrt{105} }} =\frac{8}{AE}$

→ $\frac{1}{2} =\frac{8}{AE}$

→ AE = (2×8)cm

→ AE = 16cm

Now,

In ΔADE,

[Using Pythagoras Theorem]

⇒ AD² = AE² + DE²

⇒ AD² = 16² + (√105)²

⇒ AD² = 256 + 105

⇒ AD² = 361 cm

⇒ AD = √361 cm

⇒ AD = 19cm

Thus,

The length of the AD is 19cm.