The radii of two concentric circles are 12 cm and 13 cm. AB is a diameter of the bigger circle. BD is a tangent to a smaller circle touching it at D. Find the length (in cm) of AD? (correct to one decimal place)
Answer & explanation
Correct answer: option 1
By pythagoras theorem, in triangle ODB
\( {OB }^{2 } \) = \( {OD }^{2 } \) + \( {BD }^{2 } \)
= 169 = 144 + \( {BD }^{2 } \)
= BD = 5 cm
BD = DE = \(\frac{1}{2}\) x BE [Perpendicular drawn from the center on a chord bisects it in two equal parts]
= DE = BD = 5 cm
= BE = 10 cm
In triangle ABE,
\(\angle\)AEB = 90 [Angle made in semicircle]
= \( {AB }^{2 } \) = \( {EB }^{2 } \) + \( {AE }^{2 } \)
= 676 = 100 + \( {AE }^{2 } \)
= AE = \(\sqrt {576 }\)
= AE = 24 cm
In triangle ADE
= \( {AD }^{2 } \) = \( {ED }^{2 } \) + \( {AE }^{2 } \)
= \( {AD }^{2 } \) = 25 + 576
= \( {AB }^{2 } \) = 601
= AD = \(\sqrt {601 }\)
= AD = 24.51 cm
Therefore, length of AD is 24.5 cm.