In the given figure, the radii of two concentric circles are 12 cm12 cm and 7 cm7 cm. ABAB is the diameter of the bigger circle and BDBD is a tangent to the smaller circle touching it at DD. Find the length ADAD.
Answer:
17.06 cm17.06 cm
- We know that angle in a semicircle is of 90∘.90∘. So, ∠AEB=90∘.∠AEB=90∘.
- We also know that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
So, OD⊥BEOD⊥BE and ODOD bisects BEBE. - Using Pythagoras' theorem in right △OBD△OBD, we have OB2=OD2+BD2OB2=OD2+BD2 It is given that OB=12 cmOB=12 cm and OD=7 cm.OD=7 cm. ⟹BD=√OB2−OD2=√(12)2−(7)2 cm=√95 cm⟹BD=√OB2−OD2=√(12)2−(7)2 cm=√95 cm Now, BE=2BD=2√95 cm. [ D is the midpoint of BE ] BE=2BD=2√95 cm. [ D is the midpoint of BE ]
- Using Pythagoras' theorem in right △AEB△AEB, we have AB2=AE2+BE2 As, AB is the diameter of the circle, AB = 2×OB=2×12 cm=24 cm ⟹AE=√AB2−BE2=√(24)2−(2√95)2 cm=√196 cm
- Using Pythagoras' theorem in right △AED, we have
AD2=AE2+DE2
We know that DE=BD=√95 cm [As, OD bisects BE]
⟹AD=√AE2+DE2=√(√196)2+(√95)2 cm=17.06 cm