In the given figure, ADEC is a cyclic quadrilateral. CE and AD are ext

1.	In the given figure, ADEC is a cyclic quadrilateral. CE and AD are extended to meet at B. ∠CAD = 60° and ∠CBA = 30°. BD = 6 cm and CE = 5 \(\sqrt{3}\) cm. What is the ratio AC : AD?(a) \(\frac{3}{4}\) (b) \(\frac{4}{5}\) (c) \(\frac{2\sqrt{3}}{5}\)(d) cannot be determined
Answer» Answer : (a) \(\frac{3}{4}\) ∠CED = 120° (\(\because\) CEDA is a cyclic quad.) ⇒ ∠BED = 60° \(\therefore\) In ∆ EDB, ∠EDB = 90° \(\therefore\) \(\frac{BD}{BE}\) = cos 30° ⇒ \(\frac{6}{BE}\) = \(\frac{\sqrt{3}}{2}\) ⇒ BE = 4\(\sqrt{3}\) cm. BC = BE + CE = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) = 9\(\sqrt{3}\) cm \(\because\) AB and CB are secants of the given circle, BD × BA = BE × EC ⇒ 6 × BA = 4\(\sqrt{3}\) × 9\(\sqrt{3}\) ⇒ BA = 18 cm. \(\because\) ∠ACB = 90°, ∆ ABC is a rt. ∠d ∆ ⇒ AC2 = \(\sqrt{AB^2 -BC^2}\) = \(\sqrt{{18}^2 -(9\sqrt{3})^2}\) = \(\sqrt{324 -243}\) = \(\sqrt{81}\) = 9 cm. \(\therefore\) AD = AB – BD = 12 cm. \(\therefore\) AC : AD = 9 : 12 = 3 : 4 = \(\frac{3}{4}\)

In the given figure, ADEC is a cyclic quadrilateral. CE and AD are extended to meet at B. ∠CAD = 60° and ∠CBA = 30°. BD = 6 cm and CE = 5 \(\sqrt{3}\) cm. What is the ratio AC : AD?(a) \(\frac{3}{4}\) (b) \(\frac{4}{5}\) (c) \(\frac{2\sqrt{3}}{5}\)(d) cannot be determined

Answer»

Answer : (a) \(\frac{3}{4}\)

∠CED = 120° (\(\because\) CEDA is a cyclic quad.)

⇒ ∠BED = 60°

\(\therefore\) In ∆ EDB, ∠EDB = 90°

\(\therefore\) \(\frac{BD}{BE}\) = cos 30° ⇒ \(\frac{6}{BE}\) = \(\frac{\sqrt{3}}{2}\) ⇒ BE = 4\(\sqrt{3}\) cm.

BC = BE + CE = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) = 9\(\sqrt{3}\) cm

\(\because\) AB and CB are secants of the given circle,

BD × BA = BE × EC

⇒ 6 × BA = 4\(\sqrt{3}\) × 9\(\sqrt{3}\)

⇒ BA = 18 cm.

\(\because\) ∠ACB = 90°, ∆ ABC is a rt. ∠d ∆

⇒ AC2 = \(\sqrt{AB^2 -BC^2}\) = \(\sqrt{{18}^2 -(9\sqrt{3})^2}\)

= \(\sqrt{324 -243}\) = \(\sqrt{81}\) = 9 cm.

\(\therefore\) AD = AB – BD = 12 cm.

\(\therefore\) AC : AD = 9 : 12 = 3 : 4

= \(\frac{3}{4}\)