CBSE Questions for Class 9 Maths Circles Quiz 7 - MCQExams.com

Consider given trapezium $$ABCD$$.

Given, $$\angle A={{100}^{\circ}}$$.

Now,

$$ \angle A=\angle B $$ ....[Since, $$DC$$ is the diameter of the circle]

$$\implies$$ $$ \angle B={{100}^{\circ}} $$.

Since, trapezium $$ABCD$$ is a cyclic quadrilateral,

Sum of opposite angle of trapezium $$={{180}^{\circ}}$$....[Opposite angles of cyclic quadrilateral are supplementary].

$$\implies \angle B+\angle D={{180}^{\circ}} $$

$$ {{100}^{\circ}}+\angle D={{180}^{\circ}} $$

$$ \angle D={{80}^{\circ}} $$.

Because angle in a semicircle is always $$90^{\circ},$$ so

So $$\angle RQP = 90^{\circ}$$ and $$RP$$ is the diameter

Thus by pythagorean theorem we have

$$RP^2 = RQ^2 + QP^2$$
          $$  = 12^2 + 5^2 = 169$$

$$RP = 13 = d = 2r$$

$$r = 6.5\ cm$$

Given, $$AD \parallel BC$$.

Since, $$ABCD$$ is a cyclic quadrilateral.

Then, $$\angle OCB + \angle DAB = 180^\circ$$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$$ \implies \angle DAB = 115^\circ$$.

Since, $$AD \parallel BC$$

$$\implies$$ $$\angle DAB = \angle CBE = 115^\circ$$ ...[Corresponding angles].

Hence, option $$C$$ is correct.

In a cyclic quadrilateral, we know, opposite angles are supplementary.

Since $$AECB$$ is a cyclic quadrilateral,

$$\angle E + \angle B = 180^\circ$$

            $$\angle E = 180^\circ - \angle B$$

                   $$ = 180^\circ – 125^\circ$$

                   $$= 55^\circ$$

Hence, option $$A$$ is correct.

In $$\triangle DBC$$

Given that $$CD = CB$$,

$$\therefore \angle CDB = \angle CBD$$.

Also, $$\angle DCB + \angle DBC + \angle CDB = 180^o$$...[Angle sum property]

 $$2\angle CDB = 180^o – 50^o $$

 $$\angle CDB = 65^o$$.

In the first circle, $$AEBD$$ is a cyclic quadrilateral.

Then, opposite angles are supplementary.

 $$\angle EAB + \angle EDB = 180^o$$

 $$\angle EAB + 180^o - \angle CDB = 180^o$$

 $$\angle EAB = \angle CDB = 65^o$$.

Hence, option $$D$$ is correct.

In the cyclic quadrilateral,

angles $$A + C = 180^{\circ}$$, and angles $$B + D = 180^{\circ}$$......(opposite angles of acyclic quadrilaterals are supplimentry).

So, if one of the angle is  $$70^{\circ}$$.

Then, angle opposite to it is $$= 180^{\circ} - 70^{\circ}=110^o$$.

Hence, option $$B$$ is correct.

Opposite angles of a cyclic quadrilateral are supplementary

Then, $$\angle A + \angle C = 5x+4x = 180^{\circ}$$

$$\implies$$ $$9x = 180 ^{\circ}$$

$$\implies$$ $$x= 20^{\circ}$$.

Hence, option $$B$$ is correct.