the radii of two concentric circles are 16cm and 10cm AB is a diameter of the bigger circle BD is - Maths - Circles

Question

the radii of two concentric circles are 16cm and 10cm AB is a
diameter of the bigger circle BD is a tangent to the smaller circle
touching it at D find length AD

Priyanka Kedia · Accepted Answer

Given that : The radii of two concentric circles are 16 cm and 10 cm AB is the diameter of bigger circle and BD is tangent to smaller circle Thus, radius of bigger circle = OA = OB = 13 cm To Find: length of AD Construction : join AE and OD Let BD intersect bigger circle at E In bigger circle AB is the diameter, So, $∠ A E B = 90^{0} (A n g l e i n s e m i c i r c l e)$ In smaller circle, O is the center and BD is the tangent at point D So, $O D ⊥ B D \Rightarrow ∠ O D B = 90^{0}$ $∠ A E B = ∠ O D B = 90^{0}$ And, $A E | | O D \Rightarrow Δ A E B \sim Δ O D B$ So, $\frac{A B}{O B} = \frac{A E}{O D} \Rightarrow \frac{2 O B}{O B} = \frac{A E}{8} \Rightarrow A E = 16 c m$ And, $O D = \frac{1}{2} A E = \frac{1}{2} \times 16 = 8 \Rightarrow O D = 8 c m$ $D E = D B = \sqrt{{O B}^{2} - {O D}^{2}} = \sqrt{{(13)}^{2} - {(8)}^{2}} = \sqrt{169 - 64} = \sqrt{105} \Rightarrow D E = \sqrt{105} c m$ Now, $I n Δ A E D, ∠ A E D = 90^{0} S o, {A D}^{2} = {A E}^{2} + {D E}^{2} \Rightarrow {A D}^{2} = {(16)}^{2} + {(\sqrt{105})}^{2} \Rightarrow {A D}^{2} = 256 + 105 = 361 \Rightarrow A D = 19 c m$ Hence, the length of AD is 19 cm

the radii of two concentric circles are 16cm and 10cm.AB is a diameter of the bigger circle.BD is a tangent to the smaller circle touching it at D. find length AD.