Free Circles 03 Practice Test - 9th Grade 

Circle is the set of all the points in a plane which are at a given distance from a fixed point in the plane. That fixed point is known as the centre of the circle. 

A quadrilateral is called cyclic if all the four vertices of it lie on a circle and the sum of its opposite angles is ${180}^{\circ }$.

Chord is a line segment that joins two points on the circumference of a circle. Diameter is the longest chord of a circle which passes through centre joining the two points on the circumference of a circle.

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Given$\angle ABC={50}^{\circ },$ 
$\therefore \angle AOC=2\angle ABC=2\times {50}^{\circ }={100}^{\circ }$

Also, since OA and OC are radii of the circle, OA=OC.
$⟹\angle OAC=\angle OCA$

$\text{In}△AOC,$

$x+x+\angle AOC={180}^{\circ }.$

$⟹2x={180}^{\circ }-\angle AOC={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

$⟹x={40}^{\circ }$

In the figure, $\angle ABD={50}^{\circ }$

Since the angles subtended by an arc in the same segment are equal, we have $\angle ABD=\angle ACD.$

$\Rightarrow X=\angle ACD={50}^{\circ }$

The line drawn through the centre of a circle perpendicularly bisects a chord.
Hence, $\angle x={90}^{\circ }$.

A quadrilateral is cyclic if its opposite angles are supplementary.

In square, rectangle and isosceles trapezium, the opposite angles are supplementary, that is, the sum of the opposite angles is ${180}^{\circ }$.

In a parallelogram, the opposite angles are equal and may not add to ${180}^{\circ }$. Hence, a parallelogram cannot be a cyclic quadrilateral.

It can be observed that $\angle AOC=\angle AOB+\angle BOC={60}^{\circ }+{30}^{\circ }={90}^{\circ }.$

We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

$\text{So},\angle ADC=\frac{1}{2}\angle AOC=\frac{1}{2}\times {90}^{\circ }={45}^{\circ }$

$\therefore \angle ADC={45}^{\circ }$

In $\Delta OAB$, 
AB = OA = OB = radius of the circle.
$\therefore \Delta OAB$ is an equilateral triangle.
Therefore, each interior angle of this triangle will be equal to ${60}^{\circ }$.
$\therefore \angle AOB={60}^{\circ }$.

Since the angle subtended by an arc of the circle at its centre is double the angle subtended by it at any point on the remaining part of the circle, we have 
$\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }.$ 

Now in the cyclic quadrilateral ACBD,
$\angle ACB+\angle ADB={180}^{\circ }.$           [Opposite angles in a cyclic quadrilateral are supplementary]
$\therefore \angle ADB={180}^{\circ }-\angle ACB={180}^{\circ }-{30}^{\circ }={150}^{\circ }$

Therefore, the angles subtended by the chord  AB at a point on the major arc and the minor arc are ${30}^{\circ }$ and ${150}^{\circ }$ respectively.

Given that $\angle BAC={40}^{\circ }\text{and}\angle DBC={80}^{\circ }.$

Since the angles formed by the same segment are equal,
$\angle BDC=\angle BAC={40}^{\circ }.$

$\text{In}\Delta BDC,$

$\angle BDC+\angle DBC+\angle BCD={180}^{\circ }.\text{[Angle sum property]}$

i.e., ${40}^{\circ }+{80}^{\circ }+\angle BCD={180}^{\circ }$

$⟹\angle BCD={180}^{\circ }-{120}^{\circ }={60}^{\circ }$