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In a Fine Arts class, students were asked to design triangular tiles in geometric pattern. Neelima made a circular design inside an equilateral triangle $ABC$. The radius of the circle is $4$ cm. Observe the diagram and answer the following questions :
(i) Determine the length $OB$.
(ii) Is $DE \parallel CA$? Give reason for your answer.
(iii) (a) Write all angles of quadrilateral $OEBD$ and show that it is a cyclic quadrilateral.
OR
(iii) (b) Find the perimeter of $\Delta ABC$. (Use $\sqrt{3} = 1.73$)
(i) Determine the length $OB$.
(ii) Is $DE \parallel CA$? Give reason for your answer.
(iii) (a) Write all angles of quadrilateral $OEBD$ and show that it is a cyclic quadrilateral.
OR
(iii) (b) Find the perimeter of $\Delta ABC$. (Use $\sqrt{3} = 1.73$)
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Solution:
(i) In $\Delta ODB, \sin 30^\circ = \frac{4}{OB} \Rightarrow OB = 8$ cm
(ii) Yes, $DE \parallel CA$. $\Delta ABC$ is an equilateral triangle and $AD \perp BC \Rightarrow D$ is the mid point of $BC$. Similarly, $E$ is the mid point of $AB$, so $DE \parallel CA$
(iii) (a) $\angle EBD = 60^\circ \Rightarrow \angle EOD = 120^\circ$. $\angle OEB = \angle ODB = 90^\circ$ (radius is perpendicular to the tangent through the point of contact). $\angle OEB + \angle ODB = 90^\circ + 90^\circ = 180^\circ \therefore$ quad. $OEBD$ is a cyclic quad.
OR
(iii) (b) In $\Delta OBD, \cos 30^\circ = \frac{BD}{8} \Rightarrow BD = 6.92$ cm. $BC = 2 BD = 13.84$ cm. $\therefore$ Perimeter of $\Delta ABC = 41.52$ cm
(i) In $\Delta ODB, \sin 30^\circ = \frac{4}{OB} \Rightarrow OB = 8$ cm
(ii) Yes, $DE \parallel CA$. $\Delta ABC$ is an equilateral triangle and $AD \perp BC \Rightarrow D$ is the mid point of $BC$. Similarly, $E$ is the mid point of $AB$, so $DE \parallel CA$
(iii) (a) $\angle EBD = 60^\circ \Rightarrow \angle EOD = 120^\circ$. $\angle OEB = \angle ODB = 90^\circ$ (radius is perpendicular to the tangent through the point of contact). $\angle OEB + \angle ODB = 90^\circ + 90^\circ = 180^\circ \therefore$ quad. $OEBD$ is a cyclic quad.
OR
(iii) (b) In $\Delta OBD, \cos 30^\circ = \frac{BD}{8} \Rightarrow BD = 6.92$ cm. $BC = 2 BD = 13.84$ cm. $\therefore$ Perimeter of $\Delta ABC = 41.52$ cm