JIPMAT 2021 (QA) - The radius of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is tangent to the smaller circle touching it at D. The length AD is. <img src="https://balti.afterboards.in/8mo2utsVxt0y7Gf" width=200px> | PYQs + Solutions | AfterBoards
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JIPMAT 2021

Geometry
>
Circles

Conceptual

The radius of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is tangent to the smaller circle touching it at D. The length AD is.

Correct Option: 3
Given:\newline- Radius of outer circle (R)=13(R) = 13 cm\newline- Radius of inner circle (r)=8(r) = 8 cm\newline- AB is diameter of bigger circle\newline- BD is tangent to smaller circle at D
We need to take the triangle OBD into consideration for this, as shown in the diagram above.
Important properties used:\newline1) Since OD⊥BDOD \perp BD (radius ⊥\perp tangent):\newline - D is midpoint of BE\newline - O is midpoint of AB\newline - Therefore, OD=12AEOD = \frac{1}{2}AE
2) Using midpoint theorem:\newline Since OD = 88 cm (inner radius)\newline AE=2×8=16AE = 2 × 8 = 16 cm
3) For right triangle OBD:\newline BD=OB2−OD2BD = \sqrt{OB^2 - OD^2}\newline BD=132−82BD = \sqrt{13^2 - 8^2}\newline BD=169−64BD = \sqrt{169 - 64}\newline BD=105BD = \sqrt{105}\newline Also, DE=BD=105DE = BD = \sqrt{105}
4) For right triangle ADE:\newline AD2=AE2+DE2AD^2 = AE^2 + DE^2\newline AD2=162+(105)2AD^2 = 16^2 + (\sqrt{105})^2\newline AD2=256+105AD^2 = 256 + 105\newline AD2=361AD^2 = 361\newline Therefore, AD=19AD = 19 cm
The answer is 19 cm.

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